Definition 1

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 $(x, y)\in E_1\times E_2$.