§ 4. Partial operators

1. Partial operators

In this number, K is a commutative field.

We recall (E, II, §3, p. 9–10) that a graph[^1] is a set all of whose elements are pairs. If A and B are sets, a correspondence between A and B is a triplet $(\Gamma ,A,B)$ , where Γ is a graph contained in $A\times B$ ; its set of definition (also called its domain) is pr $_1$ (Γ), and the set of its values is pr $_2(\Gamma )$ . A correspondence is a function (E, II, p. 13, def. 9) if its graph is functional and if its initial set coincides with its set of definition. Every subset of a functional graph is a functional graph.

Definition 1 032J

Let E and F be vector spaces over K. A partial operator $u$ from E into F is a correspondence $(\Gamma ,E,F)$ between E and F satisfying the following conditions:

(i) The graph Γ is a vector subspace of $E\times F$ ;

(ii) The graph Γ is functional.

If E = F, one says that $u$ is a partial operator on E.

Let $u$ be a partial operator from E into F. The graph Γ of the correspondence $u$ is called the graph of the partial operator $u$ , and is also denoted by $\Gamma_u$ . We denote by $\mathscr{P}(E; F)$ the set of partial operators from E into F; we write simply $\mathscr{P}(E) =\mathscr{P}(E; E)$ .

To give a partial operator from E into F amounts to giving a vector subspace D of E and a linear mapping $u$ from D into F, the associated partial operator being the correspondence $(\Gamma ,E,F)$ where $\Gamma \subset D\times F$ is the graph of $u$ .

The domain of definition of a partial operator $u$ is called simply the domain of $u$ , and denoted by dom( $u$ ).

Every linear mapping $u$ from E into F is a partial operator from E into F.

If $D\subset E$ is a vector subspace, we shall denote by $1_D$ the partial operator with domain D which is the identity mapping on D, that is, the correspondence $(\Delta_D,E,E)$ where $\Delta_D$ is the diagonal of $D\times D$ (E, II, p. 13, def. 8). We shall denote by $0_D$ the partial operator with domain D which is zero on D, that is, the correspondence $(D\times \{0\},E,F)$ .

Two partial operators $u= (\Gamma ,E,F)$ and $u'= (\Gamma ',E,F)$ from E into F are equal if and only if dom( $u$ ) $=$ dom( $u'$ ) and if the linear mappings $u$ and $u'$ from dom( $u$ ) into F coincide.

Following E, II, §3, the following notions are defined:

(i) Let $u$ be a partial operator from E into F; let D be its domain and $u: D\rightarrow F$ the associated linear mapping. The image of a subset A of E by $u$ is the subset $u(A\cap D)$ of F; it is denoted simply by $u(A)$ . The inverse image by $u$ of a subset B of F by $u$ is the subset $\overset{-1}{u}(B)$ of D.

If A (resp. B) is a vector subspace of E (resp. of F), then its image by $u$ (resp. its inverse image) is a vector subspace of F (resp. of E).

The image of $u$ is the vector subspace $u(D)$ of F, also denoted by Im( $u$ ). One says that $u$ is a surjective partial operator if Im( $u$ ) $= F$ . The kernel of $u$ is the vector subspace $\overset{-1}{u}(\{0\})$ of E, also denoted by Ker( $u$ ). The kernel of $u$ is reduced to 0 if and only if the linear mapping $u$ from dom( $u$ ) into F is injective. One then says that $u$ is injective. If $u$ is injective and surjective, one says that it is bijective.

(ii) If E, F and G are K-vector spaces and $u= (\Gamma ,E,F)$ , $v= (\Gamma ',F,G)$ are partial operators from E into F and from F into G, respectively, the composite correspondence $v\circ u= (\Gamma '\circ \Gamma ,E,G)$ is a partial operator from E into G. Its domain is $\overset{-1}{u}$ (dom( $v$ )). If H is a K-vector space and $w= (\Gamma '',G,H)$ a partial operator from G into H, one has $w\circ (v\circ u) = (w\circ v)\circ u$ . One will sometimes write $vu$ instead of $v\circ u$ .

(iii) In particular, for every partial operator $u$ from E into F and every $a\in K$ , the partial operators $au= (a1_F)\circ u$ and $ua=u\circ (a1_E)$ are defined. They are equal if $a\not = 0$ , or if the domain of $u$ is equal to E; one has $u0 = 0_E$ and $0u= 0_{dom(u)}$ .

Let E be a vector space. From what precedes, the set $\mathscr{P}$ (E), endowed with the law of composition defined by $(u, v)\mapsto u\circ v$ , is a unital associative magma (A, I, p. 4, Def. 5 and A, I, p. 12, Def. 2) with identity element $1_E$ . For every $n\in \mathbf{N}$ , one denotes by $u^n$ the composite $\overset{n}{\circ}u$ (A, I, p. 13).

Moreover, one defines the following notions:

(i) If $u= (\Gamma ,E,F)$ is a partial operator from E into F, and if G is a subspace of E, the reduction of $u$ to G is the partial operator $(\Gamma \cap (G\times F),E,F)$ from E into F. Its domain is dom( $u$ ) $\cap G$ ; one will sometimes denote it by $u|G$ , when no confusion with the restriction of $u$ to the subspace G is to be feared.

(ii) Let $v$ be an injective partial operator from F into E. Then the inverse correspondence $v^{-1}= (\Gamma^{-1},E,F)$ of $v$ is a partial operator such that dom( $v^{-1}$ ) $=$ Im( $v$ ). One says that $v^{-1}$ is the inverse partial operator of $v$ . One has the equalities $v\circ v^{-1}= 1_{dom(v^{-1})}$ in $\mathscr{P}(E)$ and $v^{-1}\circ v= 1_{dom(v)}$ in $\mathscr{P}(F)$ . The partial operator $v^{-1}$ is injective and one has $(v^{-1})^{-1}=v$ .

(iii) Let E, F and G be vector spaces. Let $u$ (resp. $v$ ) be an injective partial operator from E into F (resp. from F into G). Then the partial operator $v\circ u$ is injective and $(v\circ u)^{-1}=u^{-1}\circ v^{-1}$ .

(iv) If $u$ and $v$ are partial operators from E into F, one says that $v$ is an extension of $u$ , and one writes $u\subset v$ , if the graph of $u$ is contained in the graph of $v$ . This implies that dom( $u$ ) $\subset$ dom( $v$ ) and that $u$ is the restriction of $v$ to dom( $u$ ). The relation “ $u\subset v$ ” is an order relation in $\mathscr{P}(E; F)$ . For example, one has $au\subset ua$ for every $a\in K$ and every $u\in \mathscr{P}(E; F)$ .

(v) Let E be a vector space over K and $(F_i)_{i\in I}$ a family of vector spaces over K. For $i\in I$ , let $u_i$ be a partial operator from E into $F_i$ . The product partial operator of the $u_i$ is the partial operator from E into the product vector space of the spaces $F_i$ whose domain is the intersection D of the spaces dom( $u_i$ ) and which associates to $x\in D$ the family $(u_i(x))_{i\in I}$ . It is denoted by $(u_i)_{i\in I}$ .

(vi) Let $A : F\times F\rightarrow F$ be the linear mapping $(x, y)\mapsto x+y$ . Let $u$ and $v$ be partial operators from E into F. The sum $u+v$ is the partial operator $A\circ (u, v)$ from E into F. Its domain is dom( $u$ ) $\cap$ dom( $v$ ). For $u,v,w$ in $\mathscr{P}(E; F)$ , one has $(u+v) +w=u+ (v+w)$ .

Let G be a vector space over K. For every $u$ and $v$ in $\mathscr{P}(E; F)$ and every $w\in \mathscr{P}(F; G)$ , one has $w\circ u+w\circ v\subset w\circ (u+v)$ . In general, there is no equality in this formula (exercise 1 of IV, p. 344), but this is the case when the domain of $w$ is equal to F. For $w\in \mathscr{P}(G; E)$ , one has $u\circ w+v\circ w= (u+v)\circ w$ .

(vii) Let L be an extension of the field K. Let E and F be vector spaces over K and $E_{(L)}= L\otimes_KE, F_{(L)}= L\otimes_KF$ the L-vector spaces obtained by extension of scalars from K to L (A, II, p. 82). For every partial operator $u$ from E into F, let $u_{(L)}$ denote the partial operator from $E_{(L)}$ into $F_{(L)}$ whose graph is the vector subspace $L\otimes_K\Gamma_u$ of $E_{(L)}\times F_{(L)}$ ; its domain is $L\otimes_K$ dom( $u$ ), and it coincides on this with the unique linear mapping sending $1\otimes x$ to $1\otimes u(x)$ for every $x\in$ dom( $u$ ).

Let $v$ be a partial operator from E into F. One has $u\subset v$ if and only if $u_{(L)}\subset v_{(L)}$ .

(viii) Let $E_1, F_1, E_2, F_2$ be K-vector spaces. Let $u$ (resp. $v$ ) be a partial operator from $E_1$ into $F_1$ (resp. from $E_2$ into $F_2$ ). Let $u\otimes v$ denote the partial operator from $E_1\otimes F_1$ into $E_2\otimes F_2$ with domain dom( $u$ ) $\otimes$ dom( $v$ ) such that $(u\otimes v)(x\otimes y) =u(x)\otimes v(y)$ for every $(x, y)\in E_1\times E_2$ .

2. Closed operators, closable operators, and operators with dense domain

In this No., K denotes a commutative topological field (TG, III, p. 54).

Definition 2 032K

Let E and F be topological vector spaces over K (EVT, I, p. 1, def. 1). Let $u\in \mathscr{P}(E; F)$ be a partial operator from E into F.

One says that $u$ is an operator with dense domain if the domain of $u$ is dense in E.

One says that $u$ is closed if the graph of $u$ is closed in the topological vector space $E\times F$ . One says that $u$ is closable if it possesses a closed extension.

Let E, F and G be topological vector spaces over K. Every extension of an operator $u\in \mathscr{P}(E; F)$ with dense domain has dense domain. Moreover, if $v\in \mathscr{L}(E; F)$ , then $u+v$ is an operator with dense domain. If $v: F\rightarrow G$ (resp. $w: G\rightarrow E$ ) is an isomorphism of topological vector spaces, then $v\circ u$ (resp $u\circ w$ ) is an operator with dense domain.

Example 032L

Let E and F be topological vector spaces over K.

1) Suppose the space F separated. Let $u$ be a linear mapping from E into F. If $u$ is continuous, then the partial operator $u$ is closed (TG, I, p. 53, Corollary 2). Suppose moreover that K is a non-discrete valued field and that E and F are metrisable complete topological vector spaces over K. By the closed graph theorem (EVT, I, p. 19, Corollary 5), the partial operator defined by $u$ is then closed if and only if $u$ is continuous.

2) Suppose the space E separated. Let $v$ be an injective continuous linear mapping from F into E. The partial operator $v^{-1}\in \mathscr{P}(E; F) ($ cf. IV, p. 226) is then closed, since its graph is the image of the graph of $v$ , which is closed, under the isomorphism of topological vector spaces from $F\times E$ onto $E\times F$ defined by $(y, x)\mapsto (x, y)$ .

Proposition 1 032M

Let E and F be topological vector spaces over K. A partial operator $u$ of E into F is closable if and only if the closure of the graph $\Gamma_u$ of $u$ in $E\times F$ is a functional graph. There then exists a unique partial operator $v$ of E into F whose graph is $\overline{\Gamma}_u$ , and it is the smallest closed extension of $u$ .

If the closure of the graph of $u$ in $E\times F$ is a functional graph, it is the graph of a partial operator, and this is a closed extension of $u$ , so that $u$ is closable. Conversely, suppose that $u\subset w$ with $w$ closed. The closure $\overline{\Gamma}_u$ of the graph of $u$ in $E\times F$ is contained in $\Gamma_w$ , hence $\Gamma_u$ is a functional graph.

The last assertion follows from the fact that if $w$ is a closed extension of $u$ , then the graph of $w$ contains $\overline{\Gamma}_u$ .

Definition 3 032N

Let E and F be topological vector spaces over K. Let $u$ be a closable operator of E into F. The closed operator whose graph is $\Gamma_u$ is called the closure of $u$ . It is denoted by $\overline{u}$ .

Remark 032O

Let E and F be topological vector spaces over K. Let $u$ be a closable operator of E into F. The domain of the closure of $u$ is contained in the closure of the domain of $u$ in E. In general, it is distinct from it (exercise 1 of IV, p. 344, b)).

If $u\in \mathscr{P}(E; F)$ is closable and dom( $u$ ) $= E$ , then $u=\overline{u}$ is closed, since then dom( $\overline{u}$ ) $=$ dom( $u$ ).

Proposition 2 032P

Let $K =\mathbf{R}$ and let E and F be topological vector spaces over $\mathbf{R}$ . Let $u$ be a partial operator of E into F. Then $u$ has dense domain (resp. is closed, is closable) if and only if $u_{(\mathbf{C})}$ has dense domain (resp. is closed, is closable).

Suppose that the domain of $u$ is dense in E. Every neighbourhood of 0 in $E_{(\mathbf{C})}$ contains a neighbourhood of the form $V +iV$ (TVS, II, p. 65), where V is a neighbourhood of 0 in E, and therefore contains an element of the domain of $u_{(\mathbf{C})}$ ; this partial operator therefore has dense domain. The converse is also true since the mapping of $E_{(\mathbf{C})}$ into E which associates $x$ to $x+iy$ for every $(x, y)\in E\times E$ is continuous and surjective.

The graph of $u_{(\mathbf{C})}$ identifies with the complexified topological vector space of the graph of $u$ . It is therefore closed in $E_{(\mathbf{C})}\times E_{(\mathbf{C})}$ if the graph of $u$ is closed in $E\times E$ . Conversely, one has $\Gamma_u= \Gamma_{u_{(\mathbf{C})}}\cap (E\times E)$ in $E_{(\mathbf{C})}\times E_{(\mathbf{C})}$ ; since $E\times E$ is closed in $E_{(\mathbf{C})}\times E_{(\mathbf{C})}$ , the partial operator $u$ is closed when $u_{(\mathbf{C})}$ is.

A partial operator $v$ from E into F is an extension of $u$ if and only if $v_{(\mathbf{C})}$ is an extension of $u_{(\mathbf{C})}$ , hence the partial operator $u_{(\mathbf{C})}$ is closable if $u$ is closable. Conversely, if $u_{(\mathbf{C})}$ is closable, the partial operator $u$ is also, since $\Gamma_u= \Gamma_{u_{(\mathbf{C})}}\cap (E\times E)$ , which is then a functional graph (prop. 1).

Lemma 1 032Q

Let E, F and G be topological vector spaces over K. Let $u$ be a closed operator from E into F.

a) For every $v\in \mathscr{L}(E; F)$ , the partial operator $u+v$ is closed.

b) For every $v\in \mathscr{L}(G; E)$ , the partial operator $u\circ v$ is closed.

Let us prove a). Let $\gamma$ be the mapping $(x, y)\mapsto (x, y-v(x))$ of $E\times F$ into itself; it is continuous. For every $(x, y)\in E\times F$ , one has $\gamma (x, y)\in \Gamma_u$ if and only if $x\in$ dom( $u$ ) and $y=u(x) +v(x)$ , that is to say that $\overset{-1}{\gamma}(\Gamma_u) = \Gamma_{u+v}$ . The assertion follows.

Let us prove b). The mapping $\eta = (v,1_F)$ of $G\times F$ into $E\times F$ is continuous; for every $(z, y)\in G\times F$ , one has $\eta (z, y) = (v(z), y)$ , hence $\overset{-1}{\eta}(\Gamma_u) = \Gamma_{u\circ v}$ . The assertion follows.

Let E and F be topological vector spaces over K, F being separated. Let $a\in K$ . If $u\in \mathscr{P}(E; F)$ is closable, the same is true of $au$ . If $a\not = 0$ , one has $\overline{au}=au$ , and $u$ is closed if and only if $au$ is. If $a= 0$ , the closure of $au$ is $0_{\overline{dom(u)}}$ , and $au$ is equal to $0_{dom(\overline{u})}$ ; it may therefore happen that $u$ is closed but that $au$ is not.

Proposition 3 032R

Let E and F be topological vector spaces over K, F being separated. Let $u$ be a closed partial operator from E into F. The kernel of $u$ is a closed subspace of E.

In fact, the kernel of $u$ is the inverse image of the closed subspace $\Gamma_u\cap (E\times \{0\})$ of $E\times F$ by the continuous linear mapping $x\mapsto (x,0)$ of E into $E\times F$ .

In the rest of this No., it is assumed that K is a non-discrete valued field.

Definition 4 032S

Let E and F be normed spaces over K and let $u$ be a partial operator from E into F. For $x$ in dom( $u$ ), one writes

\|x\|_u= (\|x\|^2_E+\|u(x)\|^2_F)^{1/2}

The mapping $x\mapsto \|x\|_u$ resulting is a norm on dom( $u$ ). The normed space thus obtained is denoted by $E_u$ .

Remark 032T

Let E and F be normed spaces over K and let $u$ be a partial operator from E into F.

1) The canonical injection of $E_u$ into E is continuous since one has $\|x\|\leqslant \|x\|_u$ for every $x\in E$ . In particular, every subspace of dom( $u$ ) which is closed in E is closed in $E_u$ .

2) If E and F are Hilbert spaces, then the space $E_u$ is a prehilbertian space, since the norm on $E_u$ arises from the positive Hermitian form

(x, y)\mapsto (x|y)_u=\langle x|y\rangle +\langle u(x)|u(y)\rangle

on dom( $u$ ).

Proposition 4 032U

Let E and F be Banach spaces (resp. Hilbert spaces), and let $u$ be a partial operator from E into F. Then $u$ is closed if and only if the normed space $E_u$ is a Banach space (resp. a Hilbert space).

It is enough to treat the case of Banach spaces. The norm of $E_u$ is obtained, by transport of structure by means of the bijective linear mapping $(x, y)\mapsto x$ of $\Gamma_u$ onto dom( $u$ ), from the norm obtained by restriction to the subspace $\Gamma_u$ of the norm $(x, y)\mapsto (\|x\|^2_E+\|y\|^2_F)^{1/2}$ on the Banach space $E\oplus F$ . Thus the space $E_u$ is a Banach space if and only if the subspace $\Gamma_u$ of $E\oplus F$ is closed.

If $u$ and $v$ are partial operators from E into F and from F into G, respectively, and if $u$ is closed, then the partial operator $v\circ u$ is not closed in general, even if $v$ is continuous (exercise 1 of IV, p. 344, c)). One nevertheless has the following sufficient condition:

Lemma 2 032V

Let E, F and G be normed spaces over K, F being a Banach space. Let $u$ be a closed partial operator from E into F and let $v\in \mathscr{L}(F; G)$ . If there exists $C\in \mathbf{R}_+$ such that

\|u(x)\|\leqslant C(\|x\|+\|(v\circ u)(x)\|)

for every $x\in$ dom( $v\circ u$ ) $=$ dom( $u$ ), that is to say, if the linear mapping $x\mapsto u(x)$ from $E_{v\circ u}$ into F is continuous, then $v\circ u$ is closed.

Let us write $w=v\circ u$ . Let $(x_n, w(x_n))_{n\in\mathbf{N}}$ be a sequence in the graph of $w$ which converges in $E\times G$ . Let $x$ be the limit of the sequence $(x_n)$ . The hypothesis implies that the sequence $(u(x_n))_{n\in\mathbf{N}}$ is then a Cauchy sequence in F; it converges to an element $y$ of F. Since $u$ is closed, one has therefore $x\in$ dom( $u$ ) and $y=u(x)$ . Then $w(x_n) =v(u(x_n))$ tends to $v(y) =$ $v(u(x))$ since $v$ is continuous, hence the graph of $w$ is closed.

Let $u$ be a closed partial operator on a Banach space E and F a subspace of dom( $u$ ). If F is dense in the Banach space $E_u$ , then the restriction of $u$ to F is closable, and its closure is equal to $u$ . One then says that F is a core for $u$ .

3. Examples of partial operators

In this No., $K =\mathbf{R}$ or $\mathbf{C}$ .

Example 1 032W

Let X be a locally compact topological space and let $\mu$ be a positive measure on X. Let $p_1$ and $p_2$ be fixed elements of $[1,+\infty [$ .

Let $g$ be a $\mu$ -measurable function on X with values in K. Let D denote the subspace of $\mathscr{L}_K^{p_1}(X, \mu)$ formed by the functions $f$ in $\mathscr{L}_K^{p_1}(X, \mu)$ such that $gf\in \mathscr{L}_K^{p_2}(X, \mu)$ . The linear mapping from D into $\mathscr{L}_K^{p_2}(X, \mu)$ defined by $f\mapsto gf$ determines a partial operator from $\mathscr{L}_K^{p_1}(X, \mu)$ into $\mathscr{L}_K^{p_2}(X, \mu)$ , which is denoted by $m_g$ .

The vector subspace of $\mu$ -negligible functions in $\mathscr{L}_K^{p_1}(X, \mu)$ is contained in D, and the image under $m_g$ of a $\mu$ -negligible function is again $\mu$ -negligible. We shall denote by $\widetilde{m}_g$ the partial operator from $L^{p_1}_K(X, \mu)$ into $L^{p_2}_K(X, \mu)$ deduced from $m_g$ by passing to quotients. It is said to be the multiplication operator by $g$ from $L^{p_1}_K(X, \mu)$ into $L^{p_2}_K(X, \mu)$ . Functions $g_1$ and $g_2$ which are locally equal $\mu$ -almost everywhere define the same multiplication operator.

Proposition 5 032X

The multiplication operator $\widetilde{m}_g$ from $L^{p_1}_K(X, \mu)$ into $L^{p_2}_K(X, \mu)$ is a closed operator with dense domain.

Let us first prove that the partial operator $\widetilde{m}_g$ is closed. Let $(f_n, h_n)_{n\in\mathbf{N}}$ be a sequence in $\mathscr{L}_K^{p_1}(X, \mu)\times \mathscr{L}_K^{p_2}(X, \mu)$ such that the sequence $(f_{\widetilde{n}},\widetilde{h}_n)$ of the classes of $f_n$ and $h_n$ belongs to the graph of $\widetilde{m}_g$ and converges in $L^{p_1}_K(X, \mu)\times L^{p_2}_K(X, \mu)$ when $n$ tends to infinity. Let $(f, h)$ be a pair in $\mathscr{L}_K^{p_1}(X, \mu)\times \mathscr{L}_K^{p_2}(X, \mu)$ such that the pair $(\widetilde{f} ,\widetilde{h})$ of their classes is the limit of $(\widetilde{f}_n,\widetilde{h}_n)$ .

There exists a sequence $(f_{n_k})_{k\in\mathbf{N}}$ extracted from the sequence $(f_n)_n$ such that $f_{n_k}(x)$ converges to $f(x)$ for $\mu$ -almost every $x$ (INT, IV, p. 131, § 3, n $^o4$ , th. 3). This implies that $h_{n_k}(x) =g(x)f_{n_k}(x)$ converges $\mu$ -almost everywhere to $g(x)f(x)$ . Moreover the sequence $(h_{n_k})$ converges to $h$ in the space $\mathscr{L}_K^{p_2}(X, \mu)$ . The functions $h$ and $gf$ are therefore equal $\mu$ -almost everywhere (loc. cit.). Thus $\widetilde{f}$ belong to the domain of $\widetilde{m}_g$ and $\widetilde{h}=\widetilde{m}_g(\widetilde{f})$ . This proves that $\widetilde{m}_g$ is closed.

Let us prove that the domain of $\widetilde{m}_g$ is dense in $L^{p_1}_K(X, \mu)$ . It is enough to verify that the classes of functions $f\in \mathscr{K}(X; K)$ belong to the closure of the domain of $\widetilde{m}_g$ in $L^{p_1}_K(X, \mu)$ . Let $f\in \mathscr{K}(X; K)$ and let $\widetilde{f}$ be its class in $L^{p_1}_K(X, \mu)$ . For every integer $n\in \mathbf{N}$ , let $\varphi_n$ denote the characteristic function of the set of elements $x\in X$ such that $|g(x)|\leqslant n$ , and put $f_n=f \varphi_n$ . Then $|gf_n|\leqslant n|f|$ , which belong to $\mathscr{L}_K^{p_2}(X, \mu)$ , hence $f_n$ belong to the domain of $\widetilde{m}_g$ . For every element $x$ of X, the sequence $(f_n(x))_{n\in\mathbf{N}}$ converges to $f(x)$ when $n\rightarrow +\infty$ ; moreover, one has $|f_n|\leqslant |f|$ , which belong to $\mathscr{L}_K^{p_1}(X, \mu)$ . By Lebesgue's theorem (INT, IV, p. 137, § 3, n $^o7$ , th. 6), the sequence of classes of $f_n$ converges to $\widetilde{f}$ in $L^{p_1}_K(X, \mu)$ . Thus, the class of $f$ belong to the closure of the domain of $\widetilde{m}_g$ .

In the following proposition, it is supposed that $p_1=p_2= 2$ .

Proposition 6 032Y

a) Let $g'$ be a $\mu$ -measurable function on X such that $|g|\leqslant |g'|$ . One has dom( $\widetilde{m}_{g'}$ ) $\subset$ dom( $\widetilde{m}_g$ ) and dom( $\widetilde{m}_{g'}$ ) is a core of the partial operator $\widetilde{m}_g$ ;

b) Let F be a subspace of $\mathscr{L}_K^2(X, \mu)$ whose intersection with $\mathscr{K}(X; K)$ is dense in $\mathscr{K}(X; K)$ and whose image G in $L^2_K(X, \mu)$ is contained in dom( $\widetilde{m}_g$ ). If $|g|^2$ is locally $\mu$ -integrable, then $\mathscr{K}(X; K)$ is contained in the domain of $\widetilde{m}_g$ and G is a core of $m_g$ .

Let us prove a). If $f\in \mathscr{L}_K^2(X, \mu)$ belong to the domain of $m_{g'}$ , so that $f g'\in \mathscr{L}_K^2(X, \mu)$ , the assumption implies that $f g\in \mathscr{L}_K^2(X, \mu)$ , whence the result.

Let us prove that dom( $\widetilde{m}_{g'}$ ) is a core of $\widetilde{m}_g$ , that is to say that the domain of $\widetilde{m}_{g'}$ is dense in the Hilbert space $E_{\widetilde{m}_g}$ . Let $h\in \mathscr{L}_K^2(X, \mu)$

whose class $\widetilde{h}$ belong to $E_{\widetilde{m}_g}$ and is orthogonal to dom( $\widetilde{m}_{g'}$ ). This means that

(\widetilde{h}|\widetilde{h}')_{\widetilde{m}_g}=\int_Xh h'(1 +|g|^2)d\mu= 0

for every function $h'\in \mathscr{L}_K^2(X, \mu)$ whose class $\widetilde{h}'$ belongs to dom( $\widetilde{m}_{g'}$ ).

Let C be a compact subset of X and let $\varphi$ be its characteristic function. Let $n\in \mathbf{N}$ . Denote by $\varphi_n$ the characteristic function of the $\mu$ -integrable set $C_n$ of the $x\in C$ such that $|h(x)|\leqslant n$ and put $h'_n=\varphi_nh$ . The class of $h'_n$ belongs to the domain of $\widetilde{m}_{g'}$ since $|g'h'_n|\leqslant n\varphi$ ; hence

0 =\int_X\overline{h}h'_n(1 +|g|^2)d\mu=\int_X|h|^2\varphi_n(1 +|g|^2)d\mu

This implies that $h$ is zero for $\mu$ -almost every $x\in C_n$ and therefore, since $n$ is arbitrary, that $h$ is zero for $\mu$ -almost every $x\in C$ . It follows finally that $h$ is zero $\mu$ -almost everywhere, since C is arbitrary and $h$ is moderate (INT, V, p. 9, § 1, n $^o3$ , cor.).

Let us now consider assertion b). Since $|g|^2$ is locally $\mu$ -integrable, the function $f g$ belongs to $\mathscr{L}_K^2(X, \mu)$ if $f\in \mathscr{K}(X; K)$ , hence $\mathscr{K}(X; K)$ is contained in the domain of $\widetilde{m}_g$ .

Let $h\in \mathscr{L}_K^2(X, \mu)$ whose class $\widetilde{h}$ belongs to $E_{m_g}$ and is orthogonal to G. One then has

0 = (\widetilde{h}|\widetilde{h}')_{\widetilde{m}_g}=\int_Xh\overline{h}'(1 +|g|^2)d\mu

for every $h'\in F$ of class $\widetilde{h}'$ . In view of the assumption on F, this means that the measure $h(1 +|g|^2)\cdot \mu$ is zero, hence that $h$ is zero $\mu$ -almost everywhere since $h$ is moderate.

Let $p$ be a real number $\geqslant 1$ . Let $h$ be an element of $\mathscr{L}_K^{\infty}(X, \mu)$ . The multiplication operator $\widetilde{m}_h$ by $h$ is an endomorphism of $L^p_K(X, \mu)$ (IV, p. 186, No. $^o5$ ). Suppose that the set Y of the $x\in X$ such that $h(x) = 0$ is locally $\mu$ -negligible. Then the endomorphism $\widetilde{m}_h$ is injective (Lemma 7 of IV, p. 186). Let us denote by $h^{-1}$ the function on X equal to 0 on Y and to $x\mapsto 1/h(x)$ on X- Y. The reciprocal partial operator $\widetilde{m}^{-1}_h$ is the multiplication operator by $h^{-1}$ from $L^p_K(X, \mu)$ into $L^p_K(X, \mu)$ , that is to say, $\widetilde{m}^{-1}_h=\widetilde{m}_{h^{-1}}$ . In fact, the image of $\widetilde{m}_h$ is the space of classes of functions $g\in \mathscr{L}_K^p(X, \mu)$ of the form $g=hf$ for $f\in \mathscr{L}_K^p(X, \mu)$ . This condition is equivalent to $g(x)/h(x) =f(x)$ for every $x\in X$ - Y and $g(x) = 0$ if $x\in Y$ . This implies that the domain of $\widetilde{m}^{-1}_h$ in $L^p_K(X, \mu)$ is the domain of $\widetilde{m}_{h^{-1}}$ , and that the equality $\widetilde{m}^{-1}_h=\widetilde{m}_{h^{-1}}$ is valid.

In what follows, we shall sometimes denote simply by $m_h$ the partial multiplication operator by $h$ from $L^{p^1}_K(X, \mu)$ into $L^{p_2}_K(X, \mu)$ .

Example 2 032Z

Let E be a Hilbert space over K and $B = (e_i)_{i\in I}$ an orthonormal basis of E. Let $(\lambda_i)_{i\in I}$ be a family of elements of K. Let D be the vector subspace of E consisting of the elements $x\in E$ such that the family $(\lambda_i\langle e_i|x\rangle )_{i\in I}$ is square-summable in K. The space D is dense in E since it contains the vector $e_i$ for every $i\in I$ . The partial operator $u$ with domain D given by

x\mapsto \sum_{i\in I}\lambda_i\langle e_i|x\rangle e_i

is called a diagonal partial operator in the basis B, and $(\lambda_i)_{i\in I}$ is called the family of eigenvalues of $u$ .

The operator $u$ is closed. In fact, let $(x_n, u(x_n))_{n\in\mathbf{N}}$ be a sequence of elements of the graph of $u$ which converges in $E\times E$ , and let $(x, y)$ be its limit. Then $\langle e_i|x_n\rangle \rightarrow \langle e_i|x\rangle$ for every $i\in I$ and

\langle e_i|u(x_n)\rangle =\lambda_i\langle e_i|x_n\rangle \rightarrow \langle e_i|y\rangle

for every $i\in I$ . Consequently, $\lambda_i\langle e_i|x\rangle =\langle e_i|y\rangle$ for every $i\in I$ , which proves that $x\in D$ and $u(x) =y$ , that is to say that $u$ is closed.

This example is in fact a particular case of the preceding one, applied to the topological space X = I endowed with the discrete topology and the counting measure $\mu$ , since E identifies with the space $\ell^2(I) = L^2(I, \mu)$ by the mapping $x\mapsto (\langle e_i|x\rangle )_{i\in I}$ (EVT, V, p. 23, cor. 2) and $u$ then identifies with the multiplication operator $m_{\lambda}$ , where $\lambda$ is the function $i\mapsto \lambda_i$ .

Example 3 0330

The set $\mathbf{N}_{\mathbf{R}}=\mathbf{N}\cup \{\infty , \omega \}$ is endowed with the total ordering described in VAR, R2, p. 10, such that $n <\infty < \omega$ for every $n\in \mathbf{N}$ . Let $r\in \mathbf{N}_{\mathbf{R}}$ . Let $n\in \mathbf{N}$ and let U be an open subset of $\mathbf{R}^n$ . Let $k\in \mathbf{N}$ be such that $k\leqslant r$ . Let $(n_{\alpha})_{|\alpha|\leqslant k}$ be a family of elements of $\mathscr{C}^r$ (U), where the multi-indices considered belong to $\mathbf{N}^n$ . The family $(n_{\alpha})$ defines a scalar differential operator D of order $\leqslant k$ on U (cf. VAR, R2, 14.1.6, 14.1.4). For every integer $m$ such that $k\leqslant m\leqslant r$ , the differential operator D defines a linear mapping of $C^m(U)$ into $C^{m-k}(U)$ which sends $f\in C^m(U)$ to

D(f) =\sum_{|\alpha|\leqslant k}n_{\alpha}\partial^{\alpha}(f)

The same formula defines a continuous linear mapping of $\mathscr{D}(U)$ into $\mathscr{D}'(U)$ (def. 2 of IV, p. 208).

Definition 5 0331

Let E be a vector subspace of $\mathscr{D}'(U)$ containing $\mathscr{D}(U)$ . A differential operator associated with D on E is any partial operator on E which is an extension of the partial operator with domain $\mathscr{D}(U)$ defined by $\varphi \mapsto D(\varphi )$ .

Suppose for example that the coefficients $n_{\alpha}$ are bounded functions on U. Let $\mu$ be the Lebesgue measure on U. If $p$ is an element of $[1,+\infty [$ , one can then define on $L^p(U)$ a differential operator associated with D whose domain is the Sobolev space $W^{k,p}(U)$ (No. 14 of IV, p. 221), since in this case one has $n_{\alpha}\partial^{\alpha}(f)\in L^p(U)$ for every $f\in W^{k,p}(U)$ and every $|\alpha |\leqslant k$ .

4. Adjoint

In this No., K is one of the fields $\mathbf{R}$ or $\mathbf{C}$ , and E and F denote Hilbert spaces over K.

Let $u$ be an operator with dense domain from E into F. Let D denote the domain of $u$ . For $y\in F$ , let $\lambda_y$ be the linear form on D such that $\lambda_y(x) =\langle y|u(x)\rangle$ for every $x\in D$ . We denote by $D^*$ the set of vectors $y\in F$ such that $\lambda_y$ is continuous on D. It is a vector subspace of F. Let $y\in D^*$ ; since D is dense in E, the linear form $\lambda_y$ extends in a unique way to a continuous linear form on E, which we again denote by $\lambda_y$ . By EVT, V, p. 15, Theorem 3, there exists a unique element $u^*(y)$ in E such that $\lambda_y(x) =\langle u^*(y)|x\rangle$ for every $x\in E$ . The mapping $y\mapsto u^*(y)$ is linear from $D^*$ into E.

Definition 6 0332

The partial operator from F into E with domain $D^*$ defined by $y\mapsto u^*(y)$ is called the adjoint of $u$ . It is denoted by $u^*$ .

Remark 0333

Thus $y\in D^*$ if and only if there exists $c\in \mathbf{R}_+$ such that $|\langle y|u(x)\rangle |\leqslant c\|x\|$ for every $x\in D$ . The element $u^*(y)$ is then characterised by the relation

\langle y|u(x)\rangle =\langle u^*(y)|x\rangle \tag{1}

In the case where $u$ is a continuous linear mapping from E into F, its adjoint in the sense of the preceding definition coincides with the adjoint defined in EVT, V, p. 38, Definition 1, since $D^*$ is equal to F in this case.

Let $v\in \mathscr{P}(E; F)$ be such that $u\subset v$ . Then $v^*\subset u^*$ .

Let $v\in \mathscr{P}(E; F)$ be such that $u+v$ has dense domain. Then the partial operator $v$ has dense domain and we have $u^*+v^*\subset (u+v)^*$ . In general, there is no equality (exercise 9 of IV, p. 347). If $v\in \mathscr{L}(E; F)$ , then $u+v$ has dense domain and $(u+v)^*=u^*+v^*$ . This is so, for example, if F = E and if $v=\lambda 1_E$ where $\lambda \in K$ .

Let G be a Hilbert space over K and let $v\in \mathscr{P}(F; G)$ be an operator with dense domain. If $v\circ u$ has dense domain, then $u^*\circ v^*\subset (v\circ u)^*$ . In general, there is no equality (loc. cit.). If $u$ (resp. $v$ ) is an isomorphism, then $v\circ u$ has dense domain and we have $(v\circ u)^*=u^*\circ v^*$ . This is so, for example, if E = F (resp. F = G) and $u=\lambda 1_E$ (resp. $v=\lambda 1_F$ ) where $\lambda \in K^*$ .

Let $s$ denote the isometric isomorphism of Hilbert spaces from $E\oplus F$ into $F\oplus E$ defined by $s(x, y) = (-y, x)$ for every $(x, y)\in E\oplus F$ .

Proposition 7 0334

Let $u$ be a partial operator with dense domain from E into F.

a) The graph of $u^*$ is equal to $s(\Gamma^{\circ}_u) =s(\Gamma_u)^{\circ}$ ;

b) The partial operator $u^*$ is closed;

c) The kernel of $u^*$ is the orthogonal of the image of $u$ .

Let us put $W =s(\Gamma_u)^{\circ}$ . Since the linear mapping $s$ is unitary, we have $W =s(\Gamma^{\circ}_u)$ .

We have $(y, x)\in W$ if and only if

\langle (y, x)|(-u(x'), x')\rangle = 0

for all $x'\in$ dom( $u$ ), that is to say, if

\langle y|u(x')\rangle =\langle x|x'\rangle

for all $x'\in$ dom( $u$ ). When $y\in$ dom( $u^*$ ) and $x=u^*(y)$ , this property holds (cf. formula (1), p. 236). Conversely, if this condition is satisfied, it follows that $|\langle y|u(x')\rangle |\leqslant \|x\| \|x'\|$ for all $x'\in$ dom( $u$ ), whence it follows that $y$ belong to dom( $u^*$ ); one then has $u^*(y) =x$ . Hence $W = \Gamma_{u^*}$ .

The operator $u^*$ is closed, because the space $s(\Gamma_u)^{\circ}$ is closed in $F\oplus E$ .

Let us prove assertion c). If $y$ is orthogonal to the image of $u$ , the linear form $\lambda_y:x\mapsto \langle y|u(x)\rangle$ on D is zero, hence $y\in$ dom( $u^*$ ) and $u^*(y) = 0$ . Conversely, let $y\in$ dom( $u^*$ ). Then $u^*(y) = 0$ if and only if $y$ is orthogonal to $u(x)$ for all $x\in D$ (formula (1), p. 236).

Proposition 8 0335

Let $u$ be an operator with dense domain from E into F. Then $u^*$ has dense domain if and only if $u$ is closable. When this is so, the closure $\overline{u}$ of $u$ is equal to $u^{**}$ , and the adjoint of $\overline{u}$ is equal to $u^*$ .

By Prop. 7, the partial operator $u^*$ is closed. Suppose that the domain $D^*$ of $u^*$ is dense in F. Let $u^{**}$ be the adjoint of $u^*$ ; it is a closed partial operator from E into F. Let us prove that $u\subset u^{**}$ , which will imply that $u$ is closable. Let $x\in$ dom( $u$ ). By definition of $u^*$ , the linear forms on $D^*$ given by $y\mapsto \langle x|u^*(y)\rangle$ and $y\mapsto \langle u(x)|y\rangle$ are equal; hence $x\in$ dom( $u^{**}$ ) and $u^{**}(x) =u(x)$ , whence the assertion.

Conversely, suppose that $u$ is closable; one has $\Gamma_{\overline{u}}=\overline{\Gamma}_u$ (Prop. 1 of IV, p. 228). Let $y\in F$ be a vector orthogonal to dom( $u^*$ ). The element $(y,0)$ of $F\oplus E$ then belongs to the orthogonal complement of the graph of $u^*$ . But, by Prop. 7, a), one has

\Gamma^{\circ}_{u^*}= (s(\Gamma_u)^{\circ})^{\circ}=s(\Gamma_u) =s(\Gamma_u)

It follows therefore that $(0, y)\in \Gamma_{\overline{u}}$ , whence $y=\overline{u}(0) = 0$ . Since the orthogonal complement of dom( $u^*$ ) is reduced to 0, the space dom( $u^*$ ) is dense in F.

Finally, Prop. 7, applied to $u^*$ , implies that

\Gamma_{u^{**}}=s^{-1}(\Gamma^{\circ}_{u^*}) =s^{-1}(s(\Gamma^{\circ \circ}_u)) =\overline{\Gamma}_u

hence $u^{**}=\overline{u}$ , then $\overline{u}^*= (u^*)^{**}=\overline{u^*}=u^*$ since $u^*$ is closed.

Corollary 0336

If $u$ is a closed partial operator with dense domain from E into F, then $u^*$ has dense domain and one has $u^{**}=u$ .

Definition 7 0337

Let $u$ be a partial operator on E. One says that $u$ is symmetric if $u$ has dense domain and if $u^*$ is an extension of $u$ . One says that $u$ is self-adjoint if $u$ has dense domain and $u^*=u$ . One says that $u$ is essentially self-adjoint if it is closable and if the closure $\overline{u}$ of $u$ is self-adjoint.

One says that $u$ is a partial operator bounded below if $u$ is symmetric and if there exists a real number $c$ such that $\langle x|u(x)\rangle \geqslant c\|x\|^2$ for every $x$ belonging to the domain of $u$ . One then says that $c$ is a lower bound of $u$ . If $c= 0$ , one also says that $u$ is a positive partial operator.

We denote by $\mathscr{A}(E)$ the set of self-adjoint partial operators on E.

Remark 1 0338

For an operator $u$ with dense domain on E to be symmetric, it is necessary and sufficient that one have

\langle x|u(y)\rangle =\langle u(x)|y\rangle \tag{2}

for every $(x, y)\in$ dom( $u$ ) $^2$ . This formula in fact shows that the domain of $u$ is contained in that of $u^*$ , and then that $u^*$ and $u$ coincide on the domain of $u$ . In particular, it follows that $\langle x|u(x)\rangle \in \mathbf{R}$ for every $x\in$ dom( $u$ ).

As will be seen in different examples, formula (2) can often be verified by a straightforward calculus. On the other hand, the exact determination of the domain of the adjoint, which alone makes it possible to know whether or not a symmetric operator is self-adjoint, may be very delicate.

Remark 2 0339

A self-adjoint partial operator $u$ is essentially self-adjoint (cf. prop. 8).

Remark 3 033A

Let $u$ be a symmetric partial operator on E. The operator $u$ is closable (prop. 7, b)). It satisfies dom( $u$ ) $\subset$ dom( $u^*$ ), and $u$ is self-adjoint if and only if dom( $u$ ) $=$ dom( $u^*$ ). Moreover, the closure $\overline{u}$ of $u$ is symmetric since $\overline{u}\subset u^*=\overline{u}^*$ (prop. 8).

Remark 4 033B

Suppose $K =\mathbf{C}$ . Let $u\in \mathscr{P}(E; E)$ be a partial operator with dense domain. The condition $\langle x|u(x)\rangle \in \mathbf{R}$ for every $x\in$ dom( $u$ ) implies that $u$ is symmetric (EVT, V, p. 2, remark); in particular, if $\langle x|u(x)\rangle \in \mathbf{R}_+$ for every $x\in$ dom( $u$ ), then $u$ is positive.

Remark 5 033C

Let $u$ and $v$ be symmetric partial operators on E. If $u$ is self-adjoint and if $u\subset v$ , then $v\subset v^*\subset u^*=u$ , hence $u=v$ .

Remark 6 033D

An essentially self-adjoint partial operator $u$ is symmetric, since $u\subset \overline{u}$ implies $\overline{u}=\overline{u}^*\subset u^*$ , hence $u\subset u^*$ .

Remark 7 033E

Let $u$ and $v$ be symmetric partial operators on E. If $u+v$ has dense domain, for example if $u$ or $v$ belong to $\mathscr{L}$ (E), then $u+v$ is symmetric. In general, the partial operator $u+v$ is not self-adjoint, even if $u$ and $v$ are (exercise 9 of IV, p. 347).

Remark 8 033F

Let $u$ be a symmetric partial operator on E. A real number $c$ is a lower bound of $u$ if and only if the operator $u-c\cdot 1_E$ is positive.

Lemma 3 033G

Suppose that $K =\mathbf{R}$ . Let $u$ be a partial operator with dense domain from E into F.

a) The adjoint of $u_{(\mathbf{C})}$ is $(u^*)_{(\mathbf{C})}$ ;

b) Suppose that E = F; the partial operator $u$ is symmetric (resp. self-adjoint) if and only if the partial operator $u_{(\mathbf{C})}$ is symmetric (resp. self-adjoint).

Let us prove a). Let $y\in F_{(\mathbf{C})}$ and write $y=y_1+iy_2$ with $y_1,y_2\in F$ . For all $(x_1, x_2)\in E\times E$ , we have

\langle u_{(\mathbf{C})}(x_1+ix_2)|y\rangle =\langle u(x_1)|y_1\rangle +i\langle u(x_1)|y_2\rangle

-i\langle u(x_2)|y_1\rangle +\langle u(x_2)|y_2\rangle

If $y\in$ dom( $u^*$ ) $_{(\mathbf{C})}$ , it follows that $y\in$ dom(( $u_{(\mathbf{C})}$ ) $^*$ ) and that $u^*_{(\mathbf{C})}(y) = (u_{(\mathbf{C})})^*(y)$ , hence $u^*_{(\mathbf{C})}\subset (u_{(\mathbf{C})})^*$ .

Conversely, suppose that $y$ belongs to dom(( $u_{(\mathbf{C})}$ ) $^*$ ). Taking $x_2= 0$ (resp. $x_1= 0$ ) in the above formula, we verify that $y_1\in$ dom( $u^*$ ) (resp. that $y_2\in$ dom( $u^*$ )), whence $y\in$ dom( $u^*$ ) $_{(\mathbf{C})}$ .

Assertion a) implies that $u_{(\mathbf{C})}$ is symmetric (resp. self-adjoint) if $u$ is.

Conversely, suppose that $u_{(\mathbf{C})}$ is symmetric. The relation $\langle u(x)|y\rangle =\langle x|u(y)\rangle$ for all $(x, y)\in$ dom( $u_{(\mathbf{C})}$ ) $\times$ dom( $u_{(\mathbf{C})}$ ) implies that $u$ is symmetric on taking $x$ and $y$ in the subspace dom( $u$ ) of dom( $u_{(\mathbf{C})}$ ). If $u_{(\mathbf{C})}$ is self-adjoint, what precedes proves that $u$ is symmetric; since dom( $u^*$ ) $=$ dom( $u^*_{(\mathbf{C})}$ ) $\cap F$ , assertion a) implies that dom( $u^*$ ) $=$ dom( $u_{(\mathbf{C})}$ ) $\cap F =$ dom( $u$ ), therefore $u$ is self-adjoint.

5. Elementary Criteria for Self-Adjoint Operators

Proposition 9 033H

Let $v\in \mathscr{L}(F; E)$ be an injective continuous linear mapping of F into E whose image is dense in E. The adjoint of $v$ is an injective continuous linear mapping of E into F and one has $(v^*)^{-1}= (v^{-1})^*$ . In particular, if E = F, the endomorphism $v$ is hermitian if and only if the partial operator $v^{-1}$ is self-adjoint.

The partial operator $v^{-1}$ is a closed operator with dense domain from E into F (Example 2 of IV, p. 228), and the adjoint $v^*$ of $v$ is a continuous linear mapping from E into F; it is injective, since the image of $v$ is dense in E (EVT, V, p. 41, prop. 4). Let $s$ (resp. $s'$ ) be the isometric isomorphism $(x, y)\mapsto (-y, x)$ of $E\oplus F$ onto $F\oplus E$ (resp. the isometric isomorphism $(y, x)\mapsto (-x, y)$ of $F\oplus E$ onto $E\oplus F$ ), and let $\iota$ (resp. $\iota '$ ) be the isometric isomorphism $(y, x)\mapsto (x, y)$ of $F\oplus E$ onto $E\oplus F$ (resp. the isometric isomorphism $(x, y)\mapsto (y, x)$ of $E\oplus F$ onto $F\oplus E$ ). Then $s\circ \iota =-\iota '\circ s'$ , whence

\Gamma_{(v^{-1})^*}=s(\Gamma_{v^{-1}})^{\circ}=s(\iota (\Gamma_v))^{\circ}=-\iota '(s'(\Gamma_v))^{\circ}=-\iota '(\Gamma_{v^*}) = \Gamma_{(v^*)^{-1}}

by prop. 7 of IV, p. 236. The proposition follows.

Proposition 10 (Hellinger–Toeplitz) 033I

Let $u$ be a symmetric partial operator on the Hilbert space E. If the domain of $u$ is equal to E, then $u\in \mathscr{L}(E)$ and $u$ is hermitian.

In fact, the partial operator $u$ is closable (prop. 8 of IV, p. 237), and therefore closed since its domain is E (IV, p. 228, Remark). We then conclude by invoking EVT, I, p. 19, cor. 5.

Corollary 033J

Let $u$ be a symmetric partial operator on E. If the partial operator $u$ induces a bijective linear mapping of dom( $u$ ) onto E, then $u$ is self-adjoint.

In fact, the inverse partial operator $u^{-1}$ of $u$ is symmetric with domain E (prop. 9), hence $u^{-1}$ is a self-adjoint element of $\mathscr{L}(E)$ (prop. 10), and $u$ is therefore hermitian (prop. 9).

Proposition 11 033K

Let $u$ be a symmetric partial operator on E, and let $\lambda \in \mathbf{C}$ . If $u+\lambda 1_E$ and $u+\lambda 1_E$ are surjective, then $u$ is self-adjoint.

It is enough to prove that dom( $u^*$ ) $\subset$ dom( $u$ ). Let $x\in$ dom( $u^*$ ). By hypothesis there exists $y\in$ dom( $u$ ) such that $u(y) +\lambda y=u^*(x) +\lambda x$ . Let us prove that $y=x$ . For every $z\in$ dom( $u$ ), we have

\langle (u+\lambda 1_E)(z)|x\rangle =\langle z|(u^*+\overline{\lambda}1_E)(x)\rangle

=\langle z|(u+\overline{\lambda}1_E)(y)\rangle =\langle (u+\lambda 1_E)(z)|y\rangle

since $u$ is symmetric. Since the operator $u+\lambda 1_E$ is surjective, we indeed have $y=x$ , hence $x\in$ dom( $u$ ).

We shall see later (cf. prop. 17 of IV, p. 248) that if $\lambda \in \mathbf{C}-\mathbf{R}$ , then the converse is true.

Proposition 12 033L

Let $u$ be a closed operator with dense domain from E into F. The partial operator $u^*\circ u$ on E is self-adjoint and positive. Its domain is a core for $u$ .

Let us denote by $v$ the partial operator $1_E+u^*\circ u$ . Its domain is dom( $u^*\circ u$ ), which is contained in dom( $u$ ). For all $x\in$ dom( $u$ ) and $y\in$ dom( $v$ ), one has $u(y)\in$ dom( $u^*$ ) and

\langle x|v(y)\rangle =\langle x|y\rangle +\langle x|(u^*\circ u)(y)\rangle =\langle x|y\rangle +\langle u(x)|u(y)\rangle \tag{3}

\langle y|v(y)\rangle =\|y\|^2+\|u(y)\|^2 \tag{4}

Formula (4) implies that the partial operator $v$ is injective. Moreover, by prop. 7 of IV, p. 236, a), one has $F\oplus E = \Gamma_{u^*}\oplus s(\Gamma_u)$ . Let $x\in E$ . There exist $y'\in$ dom( $u^*$ ) and $x'\in$ dom( $u$ ) such that

(0, x) = (y', u^*(y')) + (-u(x'), x') = (y'-u(x'), x'+u^*(y'))

Hence one has $y'=u(x')$ , whence $x'\in$ dom( $u^*\circ u$ ) $=$ dom( $v$ ) and

x=x'+u^*(y') =x'+ (u^*\circ u)(x') =v(x')

The partial operator $v$ on E is therefore surjective, and induces a bijective linear mapping of dom( $v$ ) onto E.

Let $x\in E$ be orthogonal to dom( $v$ ). Write $x=v(x')$ where $x'\in$ dom( $v$ ). By formula (4), one obtains

0 =\langle x'|x\rangle =\langle x'|v(x')\rangle =\|x'\|^2+\|u(x')\|^2

whence $x'= 0$ , and then $x= 0$ . The domain of $v$ is therefore dense in E.

The partial operator $v$ has dense domain; it is bijective and formula (3) shows that it is symmetric. One concludes that $v$ is self-adjoint by applying the corollary to proposition 10. Consequently, $u^*\circ u=v-1_E$ is self-adjoint. Moreover, the formula

\langle x|(u^*\circ u)(x)\rangle =\|u(x)\|^2

for all $x\in$ dom( $u^*\circ u$ ) implies that $u^*\circ u$ is positive.

Finally, let $y\in E_u$ be orthogonal to dom( $u^*\circ u$ ). There exists an element $x$ in the domain of $u^*\circ u$ such that $y=v(x) =x+ (u^*\circ u)(x)$ . One then has $0 = (x|y)_u=\langle y|y\rangle$ , whence $y= 0$ .

6. Differential operators

Let $n\in \mathbf{N}$ and U be an open set in $\mathbf{R}^n$ . We endow $\mathbf{R}^n$ and U with the Lebesgue measure denoted by $\mu$ .

Let $k\in \mathbf{N}$ and $h\in \mathbf{N}$ be such that $h\geqslant k$ . Let D be a scalar differential operator on U of order $\leqslant k$ , with coefficients $(n_{\alpha})_{|\alpha|\leqslant k}$ of class $C^h$ on U. Suppose that for every $\alpha$ such that $|\alpha |\leqslant k$ and every $\beta$ such that $0\leqslant \beta \leqslant \alpha$ , the function $\partial^{\beta}n_{\alpha}$ is bounded on U.

Let $^tD$ be the scalar differential operator on U transpose of D (VAR, R2, 14.3.2); it is of order $\leqslant k$ and of class $C^{h-k}$ ; for $\varphi \in \mathscr{D}$ (U), one has

^tD(\varphi ) =\sum_{|\alpha|\leqslant k}(-1)^{|\alpha|}\partial^{\alpha}(\overline{n}_{\alpha}\varphi )

(loc. cit.); in particular, the coefficients of $^tD$ are bounded on U.

We denote by $D_-$ the partial operator on $L^2(U)$ with domain $\mathscr{D}(U)$ defined by

\varphi \mapsto D(\varphi ) =\sum_{|\alpha|\leqslant k}n_{\alpha}\partial^{\alpha}\varphi \tag{5}

Let $H_D$ be the space of $f\in L^2(U)$ such that the distribution

D(f) =\sum_{|\alpha|\leqslant k}n_{\alpha}\partial^{\alpha}f

belongs to $L^2(U)$ ; we denote by $D_+$ the partial operator with domain $H_D$ defined by $f\mapsto D(f)$ .

Since one has $\partial^{\alpha}f\in L^2(U)$ if $f\in H^k(U)$ and $|\alpha |\leqslant k$ , the Sobolev space $H^k(U)$ is contained in $H_D$ ; in general, these spaces are distinct.

One has $D_-\subset D_+$ , and these are differential operators associated with D on $L^2(U)$ (def. 5 of IV, p. 235).

Proposition 13 033M

Let $u$ be a partial operator on $L^2(U)$ . If $D_-\subset u\subset D_+$ , then $u$ is closable and $(^tD)_-\subset u^*\subset (^tD)_+$ .

Let $\varphi$ and $\psi$ belong to $\mathscr{D}(U)$ . Then one has

\langle \varphi |D(\psi )\rangle =\sum_{|\alpha|\leqslant}\int_{kU}n_{\alpha}\varphi \partial^{\alpha}\psi d\mu

=\sum_{|\alpha|\leqslant k}(-1)^{|\alpha|}\langle \partial^{\alpha}(\overline{n}_{\alpha}\varphi )|\psi \rangle =\langle^tD(\varphi )|\psi \rangle

(cf. VAR, R2, 14.3.8). Since $\mathscr{D}(U)$ is dense in $L^2(U)$ (prop. 4 of IV, p. 202), this implies that $\varphi \in$ dom( $u^*$ ) and $u^*(\varphi ) =^tD(\varphi )$ . Hence one has $(^tD)_-\subset u^*$ ; in particular, $u^*$ has dense domain and $u$ is closable (prop. 8 of IV, p. 237).

Let $f\in$ dom( $u^*$ ) and $\varphi \in \mathscr{D}(U)$ . Since $\mathscr{D}(U)\subset$ dom( $u$ ), the distribution associated with $u^*(f)$ satisfies

\langle u^*(f), \varphi \rangle =\langle \overline{\varphi}|u^*(f)\rangle =\langle u(\overline{\varphi})|f\rangle

Since $D_-\subset u$ , $u(\overline{\varphi})$ is calculated by formula (5), whence

\langle u^*(f), \varphi \rangle =\sum_{\alpha}\langle n_{\alpha}\partial^{\alpha}\overline{\varphi}|f\rangle =\sum_{\alpha}\langle \partial^{\alpha}\overline{\varphi}|\overline{n}_{\alpha}f\rangle

=\sum_{\alpha}\langle \overline{n}_{\alpha}f, \partial^{\alpha}\varphi \rangle =\sum_{\alpha}(-1)^{|\alpha|}\langle \partial^{\alpha}(\overline{n}_{\alpha}f), \varphi \rangle =\langle^tD(f), \varphi \rangle

The distributions $u^*(f)$ and $^tD(f)$ are therefore equal; the distribution $f$ therefore belongs to $H_{_tD}$ and $u^*(f) =^tD(f)$ , whence $u^*\subset (^tD)_+$ .

Remark 033N

The proposition means that the adjoint of a partial operator $u$ such that $D_-\subset u\subset D_+$ can always be calculated in the sense of distributions: the elements $f$ of the domain of $u^*$ are elements of $L^2(U)$ such that the distribution $^tD(f)$ belongs to $L^2$ (U), and one has $u^*(f) =^tD(f)$ .

One says that D is formally symmetric if $^tD = D$ as a scalar differential operator. If this is the case, the partial operator $D_-$ is symmetric.

Consider the particular case of the scalar differential operator of order 2 defined by

\Delta =-\sum_{i=1}^n\partial_i^2

A Laplacian on U is any partial operator $u$ , self-adjoint on $L^2$ (U), such that $\Delta_-\subset u($ cf. VAR, R2, 14.4.3, p. 83). We shall see later (IV, p. 261, example) that there always exists at least one Laplacian on $L^2(U)$ ; there may exist more than one (exercise 17 of IV, p. 358).

7. Spectrum and Resolvent

Lemma 4 033O

Let E be a complex Banach space and let $u$ be a closed injective partial operator on E such that $u^{-1}\in \mathscr{L}(E)$ . Let $v\in \mathscr{L}(E)$ be such that $\|v\|<\|u^{-1}\|^{-1}$ . Then the partial operator $u+v$ is injective, one has $(u+v)^{-1}\in \mathscr{L}(E)$ , and

(u+v)^{-1}=u^{-1}\circ \sum_{k=0}^{+\infty}(-vu^{-1})^k \tag{6}

where the series is absolutely convergent in $\mathscr{L}(E)$ . Moreover, one has

\|(u+v)^{-1}\|\leqslant \|u^{-1}\|_{-1}

1- \|v\| \|u\|

Since $\|v\| \|u^{-1}\|<1$ , the series with general term $(-vu^{-1})^k$ is absolutely convergent in $\mathscr{L}$ (E), and its sum is the inverse of the endomorphism $1_E+vu^{-1}$ (prop. 2 of I, p. 22). Consequently, the partial operator $(1_E+vu^{-1})\circ u=u+v($ IV, p. 226, rem. vi) is injective and the reciprocal partial operator $(u+v)^{-1}=u^{-1}\circ (1_E+vu^{-1})^{-1}($ IV, p. 226, rem. iii) belongs to $\mathscr{L}(E)$ . Since

$\|(1_E+vu^{-1})^{-1}\|\leqslant \sum^{+\infty}(\|v\| \|u^{-1}\|)^k=$ 1 $_{-1}$ ,

1- \|v\| \|u\|

k=0

the lemma is proved.

Definition 8 033P

Let $u$ be a closed operator with dense domain on a complex Banach space E. The resolvent set of $u$ is the set of complex numbers $\lambda$ such that the partial operator $\lambda 1_E-u$ is injective and its inverse $(\lambda 1_E-u)^{-1}$ belongs to $\mathscr{L}(E)$ .

The spectrum of $u$ , denoted by Sp( $u$ ), is the complement of the resolvent set in $\mathbf{C}$ .

If $\lambda \in \mathbf{C}-$ Sp( $u$ ), we denote by $R(u, \lambda )\in \mathscr{L}(E)$ the inverse of $\lambda 1_E-u$ . The mapping of $\mathbf{C}-$ Sp( $u$ ) into $\mathscr{L}(E)$ which associates $R(u, \lambda )$ to $\lambda$ is called the resolvent of $u$ .

Remark 033Q

Let E be a complex Banach space and let $u$ be a closed operator with dense domain on E.

1) If $u\in \mathscr{L}$ (E), its spectrum coincides with the spectrum of the element $u$ of the algebra $\mathscr{L}(E)$ (I, p. 2, Definition 1).

2) Let D be the domain of $u$ . For every $\lambda \in \mathbf{C}-$ Sp( $u$ ), one has

1_E= (\lambda 1_E-u)\circ R(u, \lambda ),1_D= R(u, \lambda )\circ (\lambda 1_E-u) \tag{7}

Moreover, for $\lambda_1$ and $\lambda_2$ in the resolvent set of $u$ , one has

R(u, \lambda_1)-R(u, \lambda_2) = (\lambda_2-\lambda_1)R(u, \lambda_2)\circ R(u, \lambda_1) \tag{8}

R(u, \lambda_1)\circ R(u, \lambda_2) = R(u, \lambda_2)\circ R(u, \lambda_1) \tag{9}

In fact, one has

R(u, \lambda_1)-R(u, \lambda_2) = R(u, \lambda_1)\circ 1_E-1_D\circ R(u, \lambda_2)

Since $1_E= (\lambda_21_E-u)\circ R(u, \lambda_2)$ and $1_D= R(u, \lambda_1)\circ (\lambda_11_E-u)$ , one obtains

R(u, \lambda_1)-R(u, \lambda_2) =\lambda_2R(u, \lambda_1)\circ R(u, \lambda_2)

-R(u, \lambda_1)\circ u\circ R(u, \lambda_2)-\lambda_1R(u, \lambda_1)\circ R(u, \lambda_2)

+ R(u, \lambda_1)\circ u\circ R(u, \lambda_2)

whence the first formula. By interchanging the roles of $\lambda_1$ and $\lambda_2$ , one deduces the second formula.

3) Let $\lambda \in \mathbf{C}$ . By the closed graph theorem (EVT, I, p. 19, Corollary 5), if the linear mapping of dom( $u$ ) into E defined by $x\mapsto (\lambda 1_E-u)(x)$ is bijective, then its inverse, whose graph is closed, is continuous from E into E. Therefore $\lambda$ belongs to the resolvent set of $u$ if and only if the mapping $x\mapsto (\lambda 1_E-u)(x)$ of dom( $u$ ) into E is bijective.

4) If $\lambda$ belongs to the spectrum of $u$ , one of the following properties holds:

(i) The kernel of $\lambda 1_E-u$ is not reduced to 0; one then says that $\lambda$ is an eigenvalue of $u$ , and that the dimension of Ker( $\lambda 1_E-u$ ) is its multiplicity;

(ii) The partial operator $\lambda 1_E-u$ is injective and its image is not dense in E; one then says that $\lambda$ belongs to the residual spectrum of $u$ ;

(iii) The partial operator $\lambda 1_E-u$ is injective, its image is dense in E, but $\lambda 1_E-u$ is not surjective; one then says that $\lambda$ belongs to the continuous spectrum of $u$ .

5) Let $\lambda$ be a complex number belonging to the resolvent set of $u$ . The resolvent $R(u, \lambda )$ is an injective linear mapping of E into E; its image is the domain of $u$ and $u=\lambda 1_E-R(u, \lambda )^{-1}($ cf. IV, p. 226). Conversely, this property characterizes the resolvent set and the resolvent. Precisely, let $\lambda \in \mathbf{C}$ ; if there exists a continuous injective linear mapping $w$ of E into E such that $u=\lambda 1_E-w^{-1}$ , then $\lambda$ belongs to the resolvent set of $u$ and $w= R(u, \lambda )$ .

In particular, if $v$ is a closed operator with dense domain on E, and if $\lambda \in \mathbf{C}$ is a complex number not belonging to Sp( $u$ ) $\cup$ Sp( $v$ ), then the equality $R(u, \lambda ) = R(v, \lambda )$ implies that $u=v$ .

6) The spectrum of a closable partial operator $u$ is defined as the spectrum of its closure.

There exist closed operators whose spectrum is empty, or whose spectrum is equal to $\mathbf{C}$ (exercise 12 of IV, p. 347).

Let E be a complex Banach space and $u$ a closed operator with dense domain on E. If F is a complex Banach space and if $v: E\rightarrow F$ is an isomorphism, then one has Sp( $v\circ u\circ v^{-1}$ ) $=$ Sp( $u$ ) and $R(v\circ u\circ v^{-1}, \lambda ) =v\circ R(u, \lambda )\circ v^{-1}$ for every $\lambda \notin$ Sp( $u$ ).

Proposition 14 033R

Let E be a complex Banach space. Let $u$ be a closed operator with dense domain on E and U its resolvent set.

a) For every $\lambda \in U$ , the open disc in $\mathbf{C}$ with center $\lambda$ and radius $\|R(u, \lambda )\|^{-1}$ is contained in U;

b) The set U is open in $\mathbf{C}$ ;

c) Suppose that Sp( $u$ ) is not empty. Let $\lambda \in U$ and denote by $\delta$ the distance in $\mathbf{C}$ from $\lambda$ to the spectrum of $u$ . One has $\delta >0$ and $\|R(u, \lambda )\|\geqslant 1/\delta$ ;

d) The mapping $\lambda \mapsto R(u, \lambda )$ is a holomorphic mapping of U into $\mathscr{L}(E)$ . For every integer $k\in \mathbf{N}$ and every $\lambda \in U$ , one has

$\frac{\partial^k}{\partial \lambda^k}R(u, \lambda ) = (-1)^kk$ !R( $u, \lambda$ ) $^{k+1}$ .

Let $\lambda \in U$ . For every $\mu\in \mathbf{C}$ such that $\|(\mu-\lambda )1_E\|<\|R(u, \lambda )\|^{-1}$ , lemma 4 applied to the injective partial operator $\lambda 1_E-u$ and to $v= (\mu-\lambda )1_E$ implies that the partial operator $\mu1_E-u=\lambda 1_E-u+v$ is injective and has a continuous inverse. This implies a). According to loc. cit., one also has the formula

R(u, \mu) = R(u, \lambda )\circ \sum_{k\in\mathbf{N}}(\lambda -\mu)^kR(u, \lambda )^k

which implies that the resolvent of $u$ is holomorphic in U.

Assertion b) results at once from a). If Sp( $u$ ) is nonempty, the distance from $\lambda$ to Sp( $u$ ) is strictly positive (TG, IX, p. 13, Prop. 2), whence c).

The last part of assertion d) is proved by induction on $k$ , the case $k= 1$ being a consequence of formula (8), p. 245.

Proposition 15 033S

Let $u$ be a closed operator with dense domain on a complex Banach space E and $\lambda$ a complex number belonging to the resolvent set of $u$ .

a) The subset Sp(R( $u, \lambda$ )) $-\{0\}$ of $\mathbf{C}$ is the image of the spectrum of $u$ under the mapping $\mu\mapsto (\lambda -\mu)^{-1}$ of $\mathbf{C}-\{\lambda \}$ into $\mathbf{C}^*$ ;

b) For every $\mu\not =\lambda$ in $\mathbf{C}$ , one has

Ker( $\mu1_E-u$ ) $=$ Ker(( $\lambda -\mu$ ) $^{-1}1_E-R(u, \lambda )$ ).

Let us prove assertion a). For every $\mu\not =\lambda$ , we compute

\mu1_E-u= (\lambda -\mu) ((\lambda -\mu)^{-1}1_E-R(u, \lambda )) (\lambda 1_E-u)

Since $\lambda \notin$ Sp( $u$ ) and $\mu\not =\lambda$ , the linear mapping $(\lambda -\mu)(\lambda 1_E-u)$ is a bijection of dom( $u$ ) onto E. Consequently, this formula implies that $\mu1_E-u$ is a bijection of dom( $u$ ) onto E if and only if $(\lambda -\mu)^{-1}1_E-R(u, \lambda )$ is a bijection of E onto E, which implies the assertion.

Let us prove b). If $\mu\not =\lambda$ and $x\in$ Ker(( $\lambda -\mu$ ) $^{-1}1_E-R(u, \lambda )$ ), then $x\in$ dom( $u$ ) and the formula $1_E= (\lambda 1_E-u)\circ R(u, \lambda )$ implies that $x\in$ Ker( $\mu1_E-u$ ). Conversely, if $x\in$ Ker( $\mu1_E-u$ ) and $\mu\not =\lambda$ , the formula $1_{dom(u)}= R(u, \lambda )\circ (\lambda 1_E-u)$ implies $R(u, \lambda )(x) = (\lambda -\mu)^{-1}x$ .

Proposition 16 033T

Let $u$ be a closed operator with dense domain on a complex Hilbert space E. The spectrum of $u^*$ is the image of the spectrum of $u$ under complex conjugation and, for every element $\lambda$ of the resolvent set of $u$ , one has $R(u, \lambda )^*= R(u^*, \lambda )$ . In particular, if $u$ is self-adjoint, the endomorphism $R(u, \lambda )$ is normal for every $\lambda \notin$ Sp( $u$ ).

Let $\lambda \in \mathbf{C}-$ Sp( $u$ ) be an element of the resolvent set of $u$ . One has $u=\lambda 1_E-R(u, \lambda )^{-1}$ , hence

u^*=\overline{\lambda}1_E-(R(u, \lambda )^{-1})^*=\overline{\lambda}1_E-(R(u, \lambda )^*)^{-1}

(IV, p. 236 and prop. 9 of IV, p. 239). Applying remark 5, we deduce that $\overline{\lambda}\in \mathbf{C}-$ Sp( $u^*$ ) and that $R(u, \lambda )^*= R(u^*, \lambda )$ . Consequently, the spectrum of $u^*$ is contained in the image of Sp( $u$ ) under complex conjugation. We obtain equality by applying this property to $u^*$ , since $u^{**}=u$ (cor. to prop. 8 of IV, p. 237). The last assertion then results from formula (9), p. 245.

Corollary 033U

Let $u$ be a self-adjoint partial operator on a complex Hilbert space E. If E is not zero, then the spectrum of $u$ is not empty.

Suppose that Sp( $u$ ) is empty. Then $u$ is injective and $u^{-1}=-R(u,0)$ is an injective endomorphism of E such that Sp( $u^{-1}$ ) $\subset \{0\}$ (prop. 15, a)), hence Sp( $u^{-1}$ ) $=\{0\}($ I, p. 26, cor. 1). Since $u^{-1}$ is normal (prop. 16), this implies that $u^{-1}$ is zero (I, p. 110, example 1), which is a contradiction.

Lemma 5 033V

Let E be a complex Hilbert space and let $u$ be a closed partial operator with dense domain on E. Let $\lambda \in \mathbf{C}$ . Suppose that there exists a real number $c >0$ such that

(10) $\|u(x)-\lambda x\|\geqslant c\|x\|$ for every $x\in$ dom( $u$ ),

(11) $\|u^*(x)-\lambda x\|\geqslant c\|x\|$ for every $x\in$ dom( $u^*$ ).

Then $\lambda$ belongs to the resolvent set of $u$ and $\|R(u, \lambda )\|\leqslant c^{-1}$ .

The assumption implies that $u-\lambda 1_E$ and $u^*-\overline{\lambda}1_E$ are injective. Let F be the image of $u-\lambda 1_E$ . The space F is dense in E, since its orthogonal is equal to Ker( $u^*-\overline{\lambda}1_E$ ) (prop. 7, c) of IV, p. 236), which is zero.

Let us prove that the space F is closed. Let $(x_n)_{n\in\mathbf{N}}$ be a sequence in dom( $u$ ) such that the sequence $(u(x_n)-\lambda x_n)_{n\in\mathbf{N}}$ converges to $y\in F$ . Inequality (10) implies that the sequence $(x_n)_{n\in\mathbf{N}}$ is a Cauchy sequence in E. Let $x\in E$ be its limit. The sequence $(x_n, u(x_n))$ converges to $(x, y+\lambda x)$ in $E\times E$ ; since the graph of $u$ is closed, we therefore have $x\in$ dom( $u$ ) and $u(x) =y+\lambda x$ , which proves that $y\in F$ .

We conclude that F = E. Thus, the partial operator $u-\lambda 1_E$ is bijective, whence $\lambda \notin$ Sp( $u$ ) (remark 3). Inequality (10) then implies that $\|R(u, \lambda )\|\leqslant c^{-1}$ .

Proposition 17 033W

Let E be a complex Hilbert space and let $u$ be a self-adjoint partial operator on E.

a) The spectrum of $u$ is contained in $\mathbf{R}$ ;

b) If $u$ is positive, then the spectrum of $u$ is contained in $\mathbf{R}_+$ ;

c) Suppose that E is nonzero. Let $\lambda \notin$ Sp( $u$ ) and let $\delta >0$ be the distance from $\lambda$ to the spectrum of $u$ . One has $\|R(u, \lambda )\|=\delta^{-1}$ .

Let $(a, b)\in \mathbf{R}\times \mathbf{R}$ and $\lambda =a+ib$ . Let $x\in$ dom( $u$ ). Since $u$ is self-adjoint, one has $\langle x|u(x)\rangle \in \mathbf{R}$ , whence

\|u(x)-\lambda x\|^2=\|u(x)\|^2-2a\langle x|u(x)\rangle + (a^2+b^2)\|x\|^2

=\|u(x)-\lambda x\|^2

Suppose that $b\not = 0$ . One then obtains

\|u(x)-\lambda x\|^2=\|u(x)-\lambda x\|^2\geqslant (\|u(x)\| -a\|x\|)^2+b^2\|x\|^2

\geqslant b^2\|x\|^2

By Lemma 5, one therefore has $\lambda \notin$ Sp( $u$ ), whence assertion a).

Suppose that $u$ is also positive. If $b= 0$ and $a <0$ , one analogously obtains for $x\in$ dom( $u$ ) the inequality

\|u(x)-\lambda x\|^2=\|u(x)-\lambda x\|^2\geqslant (\|u(x)\| -a\|x\|)^2\geqslant a^2\|x\|^2

therefore $\lambda \notin$ Sp( $u$ ) $($ loc. cit. $)$ , which proves b).

Finally, let us prove c). By Prop. 16, the resolvent $R(u, \lambda )$ is a normal endomorphism of E. Its norm is therefore equal to its spectral radius (Cor. 1 of I, p. 108), whence

$\|R(u, \lambda )\|=$ sup $_{\mu\in Sp(R(u,\lambda))}|\mu|$

(Th. 1 of I, p. 24). The spectrum of $R(u, \lambda )$ cannot be reduced to $\{0\}$ , for in that case one would have $\|R(u, \lambda )\|= 0$ , hence the image dom( $u$ ) of $R(u, \lambda )$ would be zero, and E likewise. Prop. 15 therefore implies that

$\|R(u, \lambda )\|=$ sup $_{\mu\in Sp(u)}\frac{1}{|\lambda-\mu|}=\frac{1}{\delta}$ .

Corollary 033X

Let $u$ be a self-adjoint operator on E.

a) The residual spectrum of $u$ is empty;

b) For all $\lambda \not =\mu$ in $\mathbf{C}$ , the eigenspaces of $u$ corresponding to $\lambda$ and $\mu$ are orthogonal.

Let us prove a). Let $\lambda$ belong to the spectrum of $u$ ; it is a real number (Prop. 17). One has Ker( $\lambda 1_E-u$ ) $=$ Im( $\lambda 1_E-u$ ) $^{\circ}$ (Prop. 7 of IV, p. 236), therefore $\lambda$ is not an eigenvalue of $u$ if the partial operator $\lambda 1_E-u$ has dense image. By definition this implies that the residual spectrum of $u$ is empty.

Let us prove b). By the proposition, we may suppose that $\lambda$ and $\mu$ are real. Let $x\in$ dom( $u$ ) be such that $u(x) =\lambda x$ and $y\in$ dom( $u$ ) be such that $u(y) =\mu y$ . We then have

\lambda \langle x|y\rangle =\langle u(x)|y\rangle =\langle x|u(y)\rangle =\mu\langle x|y\rangle

whence $\langle x|y\rangle = 0$ .

Remark 033Y

If $u$ is a closed symmetric non-self-adjoint operator, its spectrum is not contained in $\mathbf{R}($ cf. cor. 10 of IV, p. 257 below), and it is possible that the eigenspaces of $u$ relative to $\lambda$ and $\overline{\lambda}$ are not orthogonal (exercise 11 of IV, p. 347).

8. Pseudospectrum

Definition 9 033Z

Let E be a complex Banach space and let $u$ be a closed partial operator on E. Let $\varepsilon$ be a strictly positive real number. The $\varepsilon$ -pseudospectrum of $u$ is by definition the union of the spectrum of $u$ and of the set of complex numbers $\lambda$ belonging to the resolvent set of $u$ such that $\|R(u, \lambda )\|> \varepsilon^{-1}$ . This set is denoted by PSp $_{\varepsilon}(u)$ .

Some authors define the $\varepsilon$ -pseudospectrum of $u$ as the set $T_{\varepsilon}(u)$ equal to the union of Sp( $u$ ) and of the set of $\lambda \in \mathbf{C}-$ Sp( $u$ ) such that $\|R(u, \lambda )\|\geqslant \varepsilon^{-1}$ . The closure of PSp $_{\varepsilon}(u)$ is contained in $T_{\varepsilon}(u)$ , but this inclusion may be strict, even if E is a Hilbert space (cf. exercises 18 of IV, p. 348 and 19 of IV, p. 349).

Proposition 18 0340

Let E be a complex Banach space and $u$ a closed partial operator on E. Let $\varepsilon \in \mathbf{R}^*_+$ . The set PSp $_{\varepsilon}(u)$ is an open subset of $\mathbf{C}$ . It is nonempty if E is nonzero.

If E is zero, then PSp $_{\varepsilon}(u)$ is empty. Suppose that E is nonzero. By prop. 14 of IV, p. 246, the set PSp $_{\varepsilon}(u)$ is an open subset of $\mathbf{C}$ .

The set PSp $_{\varepsilon}(u)$ is nonempty if the spectrum of $u$ is nonempty. If Sp( $u$ ) is empty, then Liouville's theorem (VAR, R1, p. 29, 3.3.6) implies that the holomorphic function on $\mathbf{C}$ defined by $\lambda \mapsto R(u, \lambda )$ is not bounded, hence there exists $\lambda$ in $\mathbf{C}$ such that $\|R(u, \lambda )\|> \varepsilon^{-1}$ .

Proposition 19 0341

Let E be a complex Banach space and $u$ a closed partial operator on E. One has PSp $_{\varepsilon}(u)\subset$ PSp $_{\delta}(u)$ if $0< \varepsilon < \delta$ and

$\bigcap_{\varepsilon\in\mathbf{R}^*_+}$ PSp $_{\varepsilon}(u) =$ Sp( $u$ ).

The first assertion results from the definition. For the second, the spectrum of $u$ is contained in PSp $_{\varepsilon}(u)$ for every $\varepsilon >0$ by definition, and if $\lambda \notin$ Sp( $u$ ), then $\lambda \notin$ PSp $_{\varepsilon}(u)$ when $\varepsilon <\|R(u, \lambda )\|^{-1}$ .

Proposition 20 0342

Let E be a complex Banach space and $u$ a closed partial operator on E. Let $\varepsilon \in \mathbf{R}^*_+$ . For every $\lambda \in \mathbf{C}$ , the following conditions are equivalent:

(i) One has $\lambda \in$ PSp $_{\varepsilon}(u)$ ;

(ii) Either $\lambda \in$ Sp( $u$ ), or there exists $x\in$ dom( $u$ ) such that $\|x\|= 1$ and $\|(\lambda 1_E-u)(x)\|< \varepsilon$ ;

(iii) There exists $v\in \mathscr{L}(E)$ such that $\|v\|< \varepsilon$ and $\lambda \in$ Sp( $u+v$ ).

One may suppose that E is nonzero. Condition (ii) is a reformulation of the definition, and hence of condition (i).

Suppose that condition (i) is satisfied and let us prove (iii). If $\lambda$ belongs to the spectrum of $u$ , one can take $v= 0$ in (iii).

Suppose therefore that $\lambda \notin$ Sp( $u$ ). By definition of PSp $_{\varepsilon}(u)$ , there exists $y\in E$ such that $\|y\|= 1$ and $\|R(u, \lambda )y\|> \varepsilon^{-1}$ . Put $x= R(u, \lambda )y$ . One has $x\not = 0$ . By the Hahn–Banach theorem (EVT, II, p. 67, cor. 2), the linear form $\ell$ on $\mathbf{C}x$ such that $\ell (x) = 1$ admits a continuous extension $\ell_1\in E'$ such that $\|\ell_1\|=\|\ell \|$ ; hence $\|\ell_1\|=\|x\|^{-1}< \varepsilon$ . For every $e\in E$ , put $v(e) =\ell_1(e)y$ . One has $v\in \mathscr{L}(E)$ and $v(x) =y$ . It follows that $(u+v)x=u(x) +y=\lambda x$ , therefore $\lambda \in$ Sp( $u+v$ ). Since moreover $\|v\|=\|\ell_1\|< \varepsilon$ , condition (iii) is satisfied.

Conversely, let $v\in \mathscr{L}(E)$ such that $\|v\|< \varepsilon$ and $\lambda \in$ Sp( $u+v$ ). The partial operator $\lambda 1_E-(u+v)$ is therefore not injective with a continuous inverse; by Lemma 4 of IV, p. 243, applied to the injective partial operator $\lambda 1_E-u$ and to $-v$ , one therefore has $\|v\|\geqslant \|R(u, \lambda )\|^{-1}$ . It follows that condition (iii) implies (i).

Corollary 0343

Let E be a complex Banach space and $u$ a closed partial operator on E. Let $\varepsilon >0$ .

a) For every $v\in \mathscr{L}(E)$ , one has PSp $_{\varepsilon}(u)\subset$ PSp $_{\varepsilon+\|v\|}(u+v)$ ;

b) Let $\delta >0$ and let $D_{\delta}$ be the open disk with center 0 and radius $\delta$ in $\mathbf{C}$ . One has PSp $_{\varepsilon}(u) + D_{\delta}\subset$ PSp $_{\varepsilon+\delta}(u)$ .

Let $\lambda \in$ PSp $_{\varepsilon}(u)$ . There exists an endomorphism $w\in \mathscr{L}(E)$ such that $\|w\|< \varepsilon$ and $\lambda \in$ Sp( $u+w$ ) (Prop. 20). Since $u+w= (u+v)+(w-v)$ and $\|w-v\|< \varepsilon +\|v\|$ , one has $\lambda \in$ PSp $_{\varepsilon+\|v\|}(u+v) ($ loc. cit.).

Let $\mu\in D_{\delta}$ ; one has $\lambda +\mu\in$ Sp( $u+ (w+\mu1_E)$ ) and $\|w+\mu1_E\|< \varepsilon +\delta$ , hence $\lambda +\mu\in$ PSp $_{\varepsilon+\delta}(u) ($ loc. cit.).

Proposition 21 0344

Let E be a complex Banach space and $u$ a closed partial operator on E. Let $\varepsilon \in \mathbf{R}^*_+$ . Every bounded connected component of PSp $_{\varepsilon}(u)$ meets the spectrum of $u$ .

Let U be a connected component of PSp $_{\varepsilon}(u)$ which does not meet Sp( $u$ ). The set U is open and closed in PSp $_{\varepsilon}(u)$ , and its closure U in $\mathbf{C}$ therefore satisfies $U\cap$ PSp $_{\varepsilon}(u) = U$ .

Since Sp( $u$ ) is contained in PSp $_{\varepsilon}(u)$ and U does not meet Sp( $u$ ), this equality proves that the set U is disjoint from Sp( $u$ ), hence contained in the resolvent set of $u$ . Moreover, it implies that the set $\overline{U}-$ U does not meet PSp $_{\varepsilon}(u)$ .

One therefore has $\|R(u, \lambda )\|\leqslant \varepsilon^{-1}$ for every $\lambda$ in $\overline{U}-$ U, whereas $\|R(u, \lambda )\|> \varepsilon^{-1}$ for $\lambda \in U$ . If the set U is bounded, its closure $\overline{U}$ is compact and there exists $\lambda_0\in \overline{U}$ such that $\|R(u, \lambda )\|\leqslant \|R(u, \lambda_0)\|$ for $\lambda \in U$ . The foregoing implies that $\lambda_0\in U$ , which contradicts the maximum principle (VAR, R1, p. 29, 3.3.7) since the resolvent of $u$ is holomorphic in the resolvent set of $u$ (Prop. 14 of IV, p. 246).

9. Multiplication operators

Let X be a locally compact space and $\mu$ a positive measure on X. We consider the multiplication operators on $L^2(X, \mu)$ ; these are closed operators with dense domain (Prop. 5 of IV, p. 232). For every $\mu$ -measurable function $g$ on X, we shall denote by $m_g$ the partial operator of multiplication by $g$ on $L^2(X, \mu)$ .

Proposition 22 0345

Let $g$ be a $\mu$ -measurable function on X.

a) The spectrum of $m_g$ is the essential $\mu$ -image S of $g$ ;

b) Let $\lambda \in \mathbf{C}-$ Sp( $m_g$ ). The resolvent $R(m_g, \lambda )$ is the multiplication operator $m_h$ , where $h$ is the function on X defined by $h(x) = 0$ if $g(x) =\lambda$ and $h(x) = (\lambda -g(x))^{-1}$ otherwise.

Let us prove that $\mathbf{C}-$ S is contained in the resolvent set of $m_g$ . Let $\lambda \in \mathbf{C}-$ S. There exists an open neighbourhood U of $\lambda$ such that the subset $Y =\overset{-1}{g}(U)$ of X is locally $\mu$ -negligible. The function $k$ defined on X by $k(x) = (\lambda -g(x))^{-1}$ if $x \notin Y$ and $k(x) = 0$ if $x\in Y$ then belong to $\mathscr{L}^{\infty}(X, \mu)$ (Lemma 5 of IV, p. 184); the multiplication operator by $k$ is therefore an endomorphism of $L^2(X, \mu)$ .

Since $|gk|\leqslant 1 +|\lambda k|$ , we have

|gkf|\leqslant |f|+|\lambda kf|

for $f\in \mathscr{L}^2(X, \mu)$ , which implies that the image of $m_k$ is contained in the domain of $m_g$ . Conversely, let $f\in \mathscr{L}^2(X, \mu)$ whose class $\widetilde{f}$ belongs to the domain of $m_g$ . Outside the locally $\mu$ -negligible set Y, we have $f(x) =k(x)(\lambda -g(x))f(x)$ , hence $\widetilde{f}$ is in the image of $m_k$ . The same formula proves that $\lambda$ belongs to the resolvent set of $m_g$ and that $R(m_g, \lambda ) =m_k$ . Since the set Y is locally $\mu$ -negligible, the multiplication operator $m_k$ coincides with the operator $m_h$ described in assertion b).

Let us prove conversely that $\mathbf{C}-$ Sp( $m_g$ ) is contained in $\mathbf{C}-$ S. Let $\lambda \in \mathbf{C}-$ Sp( $m_g$ ). Let $M>\|R(m_g, \lambda )\|$ be a real number. Let us denote by Y the set of $x\in X$ such that $|\lambda -g(x)|<M^{-1}$ . Let us prove that Y is locally $\mu$ -negligible, which will imply that $\lambda$ does not belong to S, and will complete the proof.

Let K be a compact subset of X. Let $\varphi$ be the characteristic function of $Y\cap K$ ; it is an element of $\mathscr{L}^2(X, \mu)$ , and we denote by $\widetilde{\varphi}$ its class in $L^2(X, \mu)$ . Let $\psi$ be a function in $\mathscr{L}^2(X, \mu)$ whose class in $L^2(X, \mu)$ is $R(m_g, \lambda )(\widetilde{\varphi})$ . We have $R(m_g, \lambda )(\widetilde{\varphi})\in$ dom( $m_g$ ) and $(\lambda -m_g)(R(m_g, \lambda )(\widetilde{\varphi})) =\widetilde{\varphi}$ , hence $(\lambda -g(x))\psi (x) = 1$ for $\mu$ -almost all $x\in Y\cap K$ . This implies

\|R(m_g, \lambda )\|^2\|\widetilde{\varphi}\|^2\geqslant \|R(m_g, \lambda )(\widetilde{\varphi})\|^2\geqslant M^2\mu(Y\cap K) = M^2\|\widetilde{\varphi}\|^2

In view of the choice of M, this means that $\varphi$ is zero $\mu$ -almost everywhere. Thus, the set $Y\cap K$ is $\mu$ -negligible. Therefore the set Y is locally $\mu$ -negligible (INT, IV, p. 172, § 5, n $^o2$ , Prop. 5).

Proposition 23 0346

Let $g$ be a $\mu$ -measurable function on X. The adjoint of the multiplication operator $m_g$ is $m_{\overline{g}}$ .

For every integer $n\geqslant$ 1, let $\varphi_n$ be the characteristic function of the set of elements $x\in$ X such that $|g(x)|\leqslant n$ , and let $\widetilde{\varphi}_n$ be its class in $L^2(X, \mu)$ . Let $f\in \mathscr{L}^2(X, \mu)$ be such that its class $\widetilde{f}$ belongs to dom( $m^*_g$ ), and let $\psi$ be a function whose class is $m^*_g(\widetilde{f})$ .

For every $h\in \mathscr{L}^2(X, \mu)$ whose class $\widetilde{h}$ belongs to dom( $m_g$ ), we have also $\widetilde{h}\widetilde{\varphi}_n\in$ dom( $m_g$ ), and therefore $\langle \widetilde{f}|m_g(\widetilde{h}\widetilde{\varphi}_n)\rangle =\langle m^*_g(\widetilde{f})|\widetilde{h}\widetilde{\varphi}_n\rangle$ . This yields the equality

\int_X\overline{(f g-\psi)}\varphi_nh d\mu= 0

Since the domain of $m_g$ is dense in $L^2(X, \mu)$ , it follows that $(f g-\psi )\varphi_n$ is zero $\mu$ -almost everywhere. Since $n$ is arbitrary, this means that $m^*_g(\widetilde{f})$ is the class in $L^2(X, \mu)$ of $f g$ . In particular, since $m^*_g(\widetilde{f})\in L^2(X, \mu)$ , one concludes that $f$ belongs to the domain of $m_{\overline{g}}$ and that $m^*_g(\widetilde{f}) =m_{\overline{g}}(\widetilde{f})$ .

The adjoint of $m_g$ is therefore an extension of $m_{\overline{g}}$ . Moreover, one has

\langle f|m_g(h)\rangle =\int_X\overline{f}\cdot (gh)d\mu=\int_Xf\overline{g}h d\mu

for every $f\in L^2(X, \mu)$ and $h\in$ dom( $m_g$ ), which proves that the linear form $h\mapsto \langle f|m_g(h)\rangle$ is continuous whenever $m_{\overline{g}}(f)g$ belongs to $L^2(X, \mu)$ . Consequently the domain of $m_{\overline{g}}$ is contained in that of $m^*_g$ , which concludes the proof.

Corollary 0347

Let $g$ be a $\mu$ -measurable function on X. The multiplication operator $m_g$ on $L^2(X, \mu)$ is self-adjoint (resp. positive ) if and only if the function $g$ is locally $\mu$ -almost everywhere real-valued (resp. locally $\mu$ -almost everywhere positive).

The first assertion results from the proposition. If $g$ is locally $\mu$ -almost everywhere positive, one has $\langle f|m_g(f)\rangle =\int_Xg|f|^2d\mu\geqslant 0$ for every $f\in L^2(X, \mu)$ , hence the partial operator $m_g$ is positive.

Conversely, if $m_g$ is positive, then its spectrum is contained in $\mathbf{R}_+$ (prop. 17 of IV, p. 248); since this is the $\mu$ -essential image of $g$ (prop. 22, a)), this means that $g$ is locally $\mu$ -almost everywhere positive.

Lemma 6 0348

Let $g_1$ and $g_2$ be $\mu$ -measurable functions on X.

a) The partial operator $m_{g_1}+m_{g_2}$ is closable and its closure is the multiplication operator $m_{g_1+g_2}$ ;

b) One has $m_{g_1}\circ m_{g_2}\subset m_{g_1g_2}$ ;

c) Suppose that $g_1$ is bounded. Then one has $m_{g_2}\circ m_{g_1}=m_{g_1g_2}$ . Moreover, the domain of $m_{g_2}$ is contained in the domain of $m_{g_1g_2}$ , and $m_{g_1}\circ m_{g_2}$ is the restriction of $m_{g_1g_2}$ to dom( $m_{g_2}$ ).

It is elementary that $m_{g_1+g_2}$ is an extension of $m_{g_1}+m_{g_2}$ ; the latter operator is therefore closable, and $\overline{m_{g_1} + m_{g_2}}\subset m_{g_1+g_2}$ .

Let $f\in \mathscr{L}^2(X, \mu)$ be such that the function $h= (g_1+g_2)f$ belongs to $\mathscr{L}^2(X, \mu)$ . For every integer $n\geqslant 1$ , let $X_n$ denote the set of $x\in X$ such that $|g_1(x)|+|g_2(x)|\leqslant n$ , and let $\varphi_n$ be the characteristic function of $X_n$ . We have $\varphi_nf\in$ dom( $m_{g_1}+m_{g_2}$ ). Since $(\varphi_nf)(x)$ tends to $f(x)$ for every $x\in X$ , and since for every $n\in \mathbf{N}$ one has $|\varphi_nf|\leqslant |f|$ and $|(g_1+g_2)\varphi_nf|\leqslant$ $|(g_1+g_2)f|=|h|$ , with $h\in \mathscr{L}^2(X, \mu)$ by hypothesis, Lebesgue's theorem (INT, IV, p. 137, § 3, No. $^o7$ , Theorem 6) implies that the sequence of pairs of classes in $L^2(X, \mu)\times L^2(X, \mu)$ of $(\varphi_nf,(g_1+g_2)\varphi_nf)$ , which belong to the graph of $m_{g_1}+m_{g_2}$ , tends to the pair of classes of $(f, h)$ in $L^2(X, \mu)$ . The closure of $m_{g_1}+m_{g_2}$ is therefore indeed equal to $m_{g_1+g_2}$ .

It is elementary that $m_{g_1}\circ m_{g_2}\subset m_{g_1g_2}$ . Suppose that $g_1$ is bounded, so that $m_{g_1}$ is a continuous linear mapping on $L^2(X, \mu)$ . Then the domain of $m_{g_2}\circ m_{g_1}$ is the set of classes of functions $f\in \mathscr{L}^2(X, \mu)$ such that $g_2(g_1f)\in \mathscr{L}^2(X, \mu)$ , that is to say the domain of $m_{g_1g_2}$ . Hence $m_{g_2}\circ m_{g_1}=m_{g_1g_2}$ .

It is likewise elementary that dom( $m_{g_1}\circ m_{g_2}$ ) $=$ dom( $m_{g_2}$ ) is contained in dom( $m_{g_1g_2}$ ), and that the restriction of $m_{g_1g_2}$ to this space is equal to $m_{g_1}\circ m_{g_2}$ .

One must beware of believing that $m_{g_1}\circ m_{g_2}=m_{g_1g_2}$ in general (exercise 10 of IV, p. 347). Nevertheless, one has the following partial result:

Proposition 24 0349

Let $g$ be a $\mu$ -measurable function on X.

a) One has $m_{\overline{g}}\circ m_g=m_{|g|^2}$ ;

b) For all integers $k, \ell \in \mathbf{N}$ , one has $m_{g^k\overline{g}^{\ell}}=m^k_gm^{\ell}_{\overline{g}}$ .

One has $m_{\overline{g}}\circ m_g\subset m_{|g|^2}$ by Lemma 6. Conversely, one deduces from the inequality $|g|\leqslant 1 +|g|^2$ that dom( $m_{|g|^2}$ ) $=$ dom( $m_{\overline{g}}\circ m_g$ ), whence the first assertion.

Let $k, \ell \in \mathbf{N}$ . We have $m^k_gm^{\ell}_{\overline{g}}\subset m_{g^k\overline{g}^{\ell}}($ loc. cit.). The domain of $m^k_gm^{\ell}_{\overline{g}}$ is the set of classes in $L^2(X, \mu)$ of functions $h\in \mathscr{L}^2(X, \mu)$ such that $|g|^jh$ belong to $\mathscr{L}^2(X, \mu)$ for every integer $j$ such that $0\leqslant j\leqslant k+\ell$ . The inequalities

|g^jh|\leqslant |h|+|g^{k+\ell}h|

valid for $0\leqslant j\leqslant k+\ell$ , show that dom( $m^k_gm^{\ell}_{\overline{g}}$ ) is equal to dom( $m_{g^k\overline{g}^{\ell}}$ ), whence assertion b).

10. Self-adjoint extensions of a symmetric operator

In this No., we shall classify the self-adjoint extensions of symmetric operators on a complex Hilbert space, and in particular determine conditions ensuring the existence of a self-adjoint extension of such an operator.

Let E be a complex Hilbert space. For every partial operator $u$ on E, in this No. we shall write $u+i$ and $u-i$ instead of $u+i1_E$ and $u-i1_E$ .

Let $u$ be a closed operator with dense domain on E. We recall (cf. definition 4 of IV, p. 230 and proposition 4 of IV, p. 230) that $E_u$ denotes the Hilbert space dom( $u$ ) endowed with the Hermitian form

(x|y)_u=\langle x|y\rangle +\langle u(x)|u(y)\rangle

We denote by $\|x\|_u$ the norm of an element $x$ of the Hilbert space $E_u$ . Every subspace of dom( $u$ ) which is closed in E is closed in $E_u($ IV, p. 230, remark 1).

Definition 10 034A

Let $u$ be a closed symmetric operator on a complex Hilbert space E. Let $E_+=$ Ker( $u^*-i$ ) and $E_-=$ Ker( $u^*+i$ ). The pair $(E_+,E_-)$ of subspaces of dom( $u^*$ ) is called the deficiency pair of $u$ . The subspaces $E_+$ and $E_-$ are closed subspaces of E (prop. 3 of IV, p. 229). The pair (dim(E $_+$ ), dim(E $_-$ )) of the Hilbert dimensions of these subspaces is called the deficiency index of $u$ .

Proposition 25 034B

Let $u$ be a closed symmetric operator on a complex Hilbert space E. We have

Ker( $u^*-i$ ) $^{\circ}=$ Im( $u+i$ ),

Ker( $u^*+i$ ) $^{\circ}=$ Im( $u-i$ ).

By prop. 7, c) of IV, p. 236, it is enough to prove that the image of $u+i$ (resp. of $u-i$ ) is closed in E. Since $u$ is symmetric, we have $\langle (u+i)(x)|(u+i)(y)\rangle = (x|y)_u$ for all $x$ and $y$ in dom( $u$ ). The mapping $x\mapsto u(x) +ix$ of $E_u$ into E is therefore isometric. Its image is closed in E (lemma 8 of I, p. 107). Since this is also the image of $u+i$ , the latter is closed in E. Analogously, the image of $u-i$ is closed in E.

Lemma 7 034C

Let $u$ be a closed symmetric operator on a complex Hilbert space E and $v$ a closed symmetric extension of $u$ . The domain of $u$ is a closed subspace of the Hilbert space $E_v$ .

Since $v$ is an extension of $u$ , we have $(x|y)_v= (x|y)_u$ for $x$ and $y$ in dom( $u$ ). The canonical injection of $E_u$ into $E_v$ is therefore isometric, and the conclusion follows (lemma 8 of I, p. 107).

Proposition 26 034D

Let $u$ be a closed symmetric operator on a complex Hilbert space E. Let $(E_+,E_-)$ be the deficiency pair of $u$ . The spaces $E_+,E_-$ and dom( $u$ ) are closed mutually orthogonal subspaces of $E_{u^*}$ whose Hilbert sum is equal to $E_{u^*}$ .

Since $u$ is symmetric, we have $u\subset u^*$ , hence the space dom( $u$ ) is closed in $E_{u^*}$ (lemma 7). The subspaces $E_+$ and $E_-$ are closed in E and contained in dom( $u^*$ ), hence closed in $E_{u^*}$ .

Let $x\in E_+$ . We have $u^*(x) =ix$ , whence $\langle x|u(y)\rangle =-i\langle x|y\rangle$ for every $y\in$ dom( $u$ ). Consequently, for every $y\in$ dom( $u$ ), we have

(x|y)_{u^*}=\langle x|y\rangle +\langle u^*(x)|u^*(y)\rangle =i(\langle x|u(y)\rangle - \langle x|u^*(y)\rangle )= 0

since $u$ is symmetric. The spaces $E_+$ and dom( $u$ ) are therefore orthogonal in $E_{u^*}$ . Analogously, $E_-$ and dom( $u$ ) are orthogonal in $E_{u^*}$ .

Let $x\in E_+$ and $y\in E_-$ . Then

(x|y)_{u^*}=\langle x|y\rangle +\langle u^*(x)|u^*(y)\rangle =\langle x|y\rangle - \langle x|y\rangle = 0

therefore $E_+$ and $E_-$ are orthogonal in $E_{u^*}$ .

Let $x\in E_{u^*}$ be orthogonal to the closed subspace dom( $u$ ) $\oplus E_+\oplus E_-$ . For every $y\in$ dom( $u$ ), we have $u^*(y) =u(y)$ since $u\subset u^*$ , whence

0 = (x|y)_{u^*}=\langle x|y\rangle +\langle u^*(x)|u^*(y)\rangle =\langle x|y\rangle +\langle u^*(x)|u(y)\rangle

This implies that $z=u^*(x)$ belongs to the domain of $u^*$ and satisfies $u^*(z) =-x$ . Let $x_-=z-ix$ . This is an element of dom( $u^*$ ) satisfying $u^*(x_-) =-x-iz=-ix_-$ . Hence $x_-$ belongs to $E_-$ . But, for every $w\in E_-$ , we have

-i\langle x_-|w\rangle =-i\langle z|w\rangle +\langle x|w\rangle

=\langle u^*(x)|u^*(w)\rangle +\langle x|w\rangle = (x|w)_{u^*}= 0

Taking $w=x_-$ , we deduce that $x_-= 0$ , that is to say $z=ix$ ; since $z=u^*(x)$ , we therefore have $x\in E_+$ , whence $x$ = 0 since $x$ is orthogonal to $E_+$ in $E_{u^*}$ . This completes the proof that the direct sum dom( $u$ ) $\oplus E_+\oplus E_-$ is equal to $E_{u^*}$ .

Corollary 034E

Let $u$ be a closed symmetric operator on E. Then $u$ is self-adjoint if and only if the deficiency index of $u$ is equal to $(0,0)$ .

In fact, the symmetric operator $u$ is self-adjoint if and only if dom( $u^*$ ) $=$ dom( $u$ ); now the proposition proves that dom( $u^*$ ) is the Hilbert sum of dom( $u$ ) and of the deficiency subspaces of $u$ .

Let $u$ be a closed symmetric partial operator. The spectrum of $u$ is contained in $\mathbf{R}$ if and only if $u$ is self-adjoint. In fact, it is known that Sp( $u$ ) $\subset \mathbf{R}$ if $u$ is self-adjoint (Prop. 17 of IV, p. 248); if $u$ is not self-adjoint, one of the subspaces Ker( $u^*+i$ ) or Ker( $u^*-i$ ) is nonzero, hence the image of $u+i$ or of $u-i$ is a proper subspace of E (Prop. 25 of IV, p. 256), so that $i$ or $-i$ belongs to Sp( $u$ ).

Definition 11 034F

Let $u$ be a closed symmetric operator on a complex Hilbert space E. A boundary condition for $u$ is a partially isometric linear mapping (TVS, V, p. 41, Def. 3) of Ker( $u^*-i$ ) into Ker( $u^*+i$ ).

Let $u$ be a closed symmetric operator on a complex Hilbert space E and $b$ a boundary condition for $u$ . Let I = Ker( $b$ ) $^{\circ}$ be the initial subspace of $b$ . We denote by $u_b$ the reduction of $u^*$ to the subspace of dom( $u^*$ ) which is the direct sum of dom( $u$ ) and of the graph in $E_+\oplus E_-$ of the restriction of $b$ to I. Since $E_{u^*}=$ dom( $u$ ) $\oplus E_+\oplus E_-$ (Prop. 26), the partial operator $u_b$ is well defined.

In other words, the domain of $u_b$ is the space of elements $x\in E$ of the form $x=x_0+y+b(y)$ , where $x_0\in$ dom( $u$ ) and $y$ belongs to the initial subspace of $b$ . One then has $u_b(x_0+y+b(y)) =u(x_0) +iy-ib(y)$ .

Theorem 1 (von Neumann) 034G

Let $u$ be a closed symmetric operator on a complex Hilbert space E and $(E_+,E_-)$ its deficiency pair.

a) For every boundary condition $b$ for $u$ , the partial operator $u_b$ is a closed symmetric extension of $u$ ;

b) The mapping $b\mapsto u_b$ is a bijection of the set of boundary conditions for $u$ onto the set of closed symmetric extensions of $u$ ;

c) For every boundary condition $b$ for $u$ , one has

Im( $u_b+i$ ) $=$ Im( $u+i$ ) $\oplus$ Im( $b$ )

Im( $u_b-i$ ) $=$ Im( $u-i$ ) $\oplus$ Ker( $b$ ) $^{\circ}$ ;

d) For every boundary condition $b$ for $u$ , the deficiency pair of $u_b$ is (Ker( $b$ ), Ker( $b^*$ )).

Let $b$ be a boundary condition for $u$ and $I\subset E_+$ its initial subspace. The restriction of $u^*$ to $\Gamma_{b|I}\subset E_+\oplus E_-$ is the continuous linear mapping defined by $x+b(x)\mapsto ix-ib(x)$ for $x\in I$ . The graph of $u_b$ is the direct sum of the graph of $u$ and the graph of this linear mapping; it is therefore closed. The closed operator $u_b$ is an extension of $u$ such that $u_b\subset u^*$ .

Let $\gamma_1$ and $\eta_1$ be elements of I. Consider the elements

\gamma =\gamma_1+b(\gamma_1),\eta =\eta_1+b(\eta_1)

of the graph $\Gamma_{b|I}$ . We calculate

\langle \gamma |u^*(\eta )\rangle =i(\langle b(\gamma_1)|\eta_1\rangle - \langle \gamma_1|b(\eta_1)\rangle )+i(\langle \gamma_1|\eta_1\rangle - \langle b(\gamma_1)|b(\eta_1)\rangle )

=i(\langle b(\gamma_1)|\eta_1\rangle - \langle \gamma_1|b(\eta_1)\rangle )

since $b$ is isometric on I. It follows that one has

\langle u^*(\gamma )|\eta \rangle =\langle \gamma |u^*(\eta )\rangle \tag{12}

for all $\gamma$ and $\eta$ in $\Gamma_{b|I}$ .

Let $x$ and $y$ be elements of dom( $u$ ) $,\gamma$ and $\eta$ elements of $\Gamma_{b|I}$ . Then

\langle x+\gamma |u_b(y+\eta )\rangle =\langle x|u(y)\rangle +\langle x|u^*(\eta )\rangle +\langle \gamma |u(y)\rangle +\langle \gamma |u^*(\eta )\rangle

=\langle u(x)|y\rangle +\langle u(x)|\eta \rangle +\langle u^*(\gamma )|y\rangle +\langle u^*(\gamma )|\eta \rangle

=\langle u_b(x+\gamma )|y+\eta \rangle

where we have used the fact that $u$ is symmetric and that $\gamma$ and $\eta$ belong to the domain of $u^*$ , as well as formula (12). The operator $u_b$ is therefore a closed symmetric extension of $u$ . Assertion a) is proved.

The mapping $b\mapsto u_b$ is injective. In fact, a partially isometric mapping of $E_+$ into $E_-$ is uniquely determined by its initial subspace and by its restriction to the latter. Now the domain of $u_b$ determines the initial subspace I of $b$ and the graph of the restriction of $b$ to I.

Let us prove that the mapping $b\mapsto u_b$ is surjective. Let $w$ be a closed symmetric extension of $u$ . One has $w\subset w^*\subset u^*$ . The domain of $w$ is therefore a subspace of dom( $u^*$ ) containing dom( $u$ ), and $w$ is the restriction of $u^*$ to this subspace. Let G be the intersection of the domain of $w$ and of $E_+\oplus E_-$ . It is a closed subspace of $E_{u^*}$ (lemma 7) and one has dom( $w$ ) $=$ dom( $u$ ) $\oplus G$ by proposition 26.

Let $x\in E_+$ and $y\in E_-$ be such that $x+y\in G$ . Since $w$ is symmetric, and is the restriction of $u^*$ to dom( $w$ ), we have

\langle x+y|u^*(x+y)\rangle =\langle u^*(x+y)|x+y\rangle

This equality is equivalent to

\langle x|ix\rangle +\langle x| -iy\rangle +\langle y|ix\rangle +\langle y| -iy\rangle

=\langle ix|x\rangle +\langle ix|y\rangle +\langle -iy|x\rangle +\langle -iy|y\rangle

that is, to $\|x\|^2=\|y\|^2$ . In particular, the canonical projection $p_+: G\rightarrow E_+$ is injective and, if I denotes its image, the space G is the graph of an isometric mapping $b_0$ of I into $E_-$ . For $x$ in I and $y$ in $E_-$ such that $x+y\in G$ , we have

\|x+y\|^2_{u^*}=\|x\|^2_{u^*}+\|y\|^2_{u^*}= 2(\|x\|^2+\|y\|^2)

= 4\|x\|^2= 4\|p_+(x+y)\|^2

Thus, considering G as a closed subspace of the Hilbert space $E_{u^*}$ , the mapping $x\mapsto \frac{1}{2}p_+(x)$ of G into I is an isometry. In particular, I is closed in $E_+$ (lemma 8 of I, p. 107). There then exists a unique boundary condition $b$ for $u$ whose initial space is I and which coincides with $b_0$ on I. We have dom( $w$ ) $=$ dom( $u_b$ ), whence $w=u_b$ . This establishes assertion b).

Let us prove assertions c) and d). Let $b$ be a boundary condition for $u$ and I its initial subspace. For $x_0\in$ dom( $u$ ) and $y\in I$ , we have

u_b(x_0+y+b(y))-i(x_0+y+b(y)) = (u-i)(x_0)-2ib(y)

We have Im( $u-i$ ) $= E^{\circ}_-$ (prop. 25). Since the image of $b$ is contained in $E_-$ , this formula shows that Im( $u_b-i$ ) $=$ Im( $u-i$ ) $\oplus$ Im( $b$ ). By loc. cit., we thus have Ker( $u^*_b+i$ ) $= E_-\cap$ Im( $b$ ) $^{\circ}=$ Ker( $b^*$ ).

Analogously, one verifies that Im( $u_b+i$ ) $=$ Im( $u+i$ ) $\oplus I$ and hence (loc. cit.) that Ker( $u^*_b-i$ ) $= I^{\circ}\cap E_+=$ Ker( $b$ ).

Corollary 1 034H

Let $u$ be a closed symmetric operator on a complex Hilbert space E. The mapping $b\mapsto u_b$ induces a bijection of the set of isometries of Ker( $u^*-i$ ) onto Ker( $u^*+i$ ) onto the set of self-adjoint extensions of $u$ . In particular, there exists a self-adjoint extension of $u$ if and only if the two components of the deficiency index of $u$ are equal.

This follows from the corollary of proposition 26 and the theorem, together with EVT, V, p. 25, cor. 2.

Corollary 2 034I

Let E be a real Hilbert space. Let $u$ be a closed symmetric operator on E. The partial operator $u_{(\mathbf{C})}$ on $E_{(\mathbf{C})}$ is symmetric and closed, and admits at least one self-adjoint extension.

The partial operator $u_{(\mathbf{C})}$ is closed and symmetric by prop. 2 of IV, p. 228 and lemma 3 of IV, p. 239. The $\mathbf{R}$ -linear isomorphism $j$ of $E_{(\mathbf{C})}$ onto $E_{(\mathbf{C})}$ such that $j(z\otimes x) =\overline{z}\otimes x$ for every $z\in \mathbf{C}$ and every $x\in E$ induces an isomorphism of Hilbert spaces of Ker( $u^*_{(\mathbf{C})}-i$ ) onto Ker( $u^*_{(\mathbf{C})}+i$ ), and one can apply corollary 1.

Corollary 3 034J

Let $u$ be a symmetric operator on E. The following conditions are equivalent:

(i) The operator $u$ is essentially self-adjoint;

(ii) The spaces Ker( $u^*+i$ ) and Ker( $u^*-i$ ) are zero;

(iii) The spaces Im( $u+i$ ) and Im( $u-i$ ) are dense in E;

(iv) The partial symmetric operator $u$ has a unique self-adjoint extension.

Conditions (ii) and (iii) are equivalent by prop. 7, c) of IV, p. 236.

The partial operator $u$ is essentially self-adjoint if and only if $\overline{u}$ is self-adjoint, that is to say if Ker( $\overline{u}^*-i$ ) $=$ Ker( $\overline{u}^*+i$ ) $=\{0\}$ (cor. of prop. 26). Since $\overline{u}^*=u^*$ (prop. 8 of IV, p. 237), condition (i) is therefore equivalent to condition (ii). Moreover, the preceding corollary shows that (ii) implies that $u$ has a unique self-adjoint extension, which is condition (iv).

Finally, suppose that $u$ has a unique self-adjoint extension $v$ . Then the same is true of $\overline{u}$ , since every self-adjoint extension of $\overline{u}$ is one of $u$ , hence is equal to $v$ . By the preceding corollary, the spaces Ker( $\overline{u}^*+i$ ) and Ker( $\overline{u}^*-i$ ) must be zero, whence condition (ii), using again $\overline{u}^*=u^*$ .

Example 034K

Let U be an open set of $\mathbf{R}^n$ endowed with the Lebesgue measure $\mu$ . Let Δ be the scalar differential operator

\Delta =-\sum_{i=1}^n\partial_i^2

on U. Let us denote by $u$ the partial operator on the real Hilbert space $L^2_{\mathbf{R}}(U, \mu)$ with domain $\mathscr{D}_{\mathbf{R}}(U) ($ IV, p. 201, remark) defined by $\varphi \mapsto -\sum^n_{i=1}\partial_i^2\varphi$ . One has $u_{(\mathbf{C})}= \Delta_-$ , which is a closable partial operator (prop. 13 of IV, p. 242) and symmetric (since Δ is formally symmetric), hence $u$ is closable and symmetric (proposition 2 of IV, p. 228 and lemma 3 of IV, p. 239). By the above corollary, there therefore exists a self-adjoint extension of $u_{(\mathbf{C})}$ . It is a Laplacian on U (IV, p. 243).

Exercises

Unless otherwise stated, the Banach spaces and Hilbert spaces below are assumed to be complex.

See the exercises for § 4.

[^1]: Care should be taken not to confuse the notion of graph considered here with that introduced in TA, II, p. 155, def. 1.