a) Let K be a compact subset of U and V an open neighbourhood of K in U. There exists a function $\varphi \in \mathscr{C}^{\infty}(U)$ with compact support contained in V such that $0\leqslant \varphi \leqslant 1$ and such that $\varphi (x) = 1$ for every $x\in K$;

b) For every locally finite open covering of U, there exists a partition of unity formed of functions of class $C^{\infty}$ which is subordinate to it.

This results from VAR, R1, p. 40, 5.3.6.

The mapping $h$ is the composite of the mapping $(f, g) : U\rightarrow E\times F$, which is of class $C^k$, and of the mapping $b: E\times F\rightarrow G$ which is of class $C^{\infty}$. It is therefore of class $C^k$. The expression for its partial derivatives results from Leibniz's formula (FVR, I, p. 28, prop. 2, cf. A, III, p. 122, corollary).

We recall that a separated locally convex topological vector space is a Montel space if it is barrelled and if every bounded subset is relatively compact (EVT, IV, p. 18, def. 4).

We also recall that if E and F are locally convex spaces, and if E is bornological (EVT, III, p. 12, def. 1), a linear mapping $u: E\rightarrow F$ is continuous if and only if the image by $u$ of every bounded subset of E is bounded in F (EVT, III, p. 11, prop. 1, (iiibis), when F is semi-normed, the general case resulting formally from this, cf. EVT, II, p. 7, prop. 5, b)).

(i) For every $t\in I$, the mapping $x\mapsto f(x, t)$ of X into E is $\nu$-integrable;

(ii) For every $x\in X$, the mapping $t\mapsto f(x, t)$ of I into E admits a derivative, denoted by $t\mapsto f'(x, t)$;

(iii) There exists a positive $\nu$-integrable function $g$ on X such that $\|f'(x, t)\|\leqslant g(x)$ for every $(x, t)\in X\times I$.

Then the mapping F of I into E defined by

is differentiable in I, and for every $t\in I$, one has

Let $t_0\in I$ and let J be an interval of $\mathbf{R}$ contained in I which is a neighbourhood of $t_0$ in I. Let $h: X\times J\rightarrow E$ be the mapping defined by $(x, t)\mapsto (f(x, t)-f(x, t_0))/(t-t_0)$ for $(x, t)\in X\times (J-\{t_0\})$ and $(x, t_0)\mapsto f'(x, t_0)$ for $x\in X$. Let $x\in X$. One has $\|h(x, t_0)\|\leqslant g(x)$ and, for every $t\not =t_0$, one has $\|h(x, t)\|\leqslant g(x)$ by FVR, I, p. 23, th. 2. By definition of the derivative, the proposition follows from corollary 1 of INT, IV, p. 144, § 4, n$^o3$, applied to the mapping $h$.

Let us resume the notation of VAR, R1, 2.4, p. 19 concerning partial derivatives.

(i) For every $t\in U$, the mapping $x\mapsto f(x, t)$ of X into E is $\nu$-integrable;

(ii) For every $x\in X$, the mapping $t\mapsto f(x, t)$ of U into E is of class $C^k$, with partial derivatives denoted by $\partial^{\alpha}f(x, t)$ for every $\alpha \in \mathbf{N}^n$ such that $|\alpha |\leqslant k$;

(iii) There exists a $\nu$-integrable function $g$ on X such that for every $\alpha \in \mathbf{N}^n$ satisfying $|\alpha |\leqslant k$ and for every $(x, t)\in X\times U$, one has $\|\partial^{\alpha}f(x, t)\|\leqslant g(x)$.

Then the mapping F of U into E defined by

is of class $C^k$ in U, and for every $\alpha \in \mathbf{N}^n$ satisfying $|\alpha |\leqslant k$ and every $t\in U$, one has

This follows from the proposition by induction on $k$.

One can apply corollary 1 to the space $X =\mathbf{R}^n$, to Lebesgue measure and to the mapping $h$ defined by $(x, t)\mapsto f(x)g(t-x)$ from $\mathbf{R}^n\times \mathbf{R}^n$ into $\mathbf{C}$; indeed, for every $\alpha \in \mathbf{N}^n$ such that $|\alpha |\leqslant k$, one has the inequality

$|\partial^{\alpha}h(x, t)|\leqslant ($sup$_{|\beta|\leqslant k}$ sup$_{y\in\mathbf{R}^n}|\partial^{\beta}g(y)|)|f(x)|$

whose right-hand side is an integrable function on $\mathbf{R}^n$.

a) The function $(1 + N)^r$ belong to $\mathscr{L}^p(\mathbf{R}^n, \mu)$ if and only if $rp <-n$;

a$')$ The restriction to $\mathbf{Z}^n$ of the function $(1 + N)^r$ belong to $\ell^p(\mathbf{Z}^n)$ if and only if $rp <-n$;

b) Let V be a bounded measurable neighbourhood of 0 in $\mathbf{R}^n$. The function $N^r$ belong to $\mathscr{L}^p(\mathbf{R}^n-V, \mu)$ if and only if $rp <-n$.

By virtue of the equivalence of norms on $\mathbf{R}^n$ (EVT I, p. 14, th. 2 and TG, IX, p. 32, prop. 8), we may suppose that the norm N is given by $N(x) =$ sup$|x_i|$ for $x= (x_i)\in \mathbf{R}^n$. Let B denote the unit ball of $\mathbf{R}^n$ for this norm.

Let V be a bounded measurable neighbourhood of 0 in $\mathbf{R}^n$. The function $N^r$ is continuous and bounded on (B- $V$)$\cup (V$- B), which shows that assertion b) is valid if and only if it is so when V = B, which we shall henceforth suppose. Since the function $(1 + N)^r$ is continuous and bounded on B, and satisfies the inequalities $N^r\leqslant (1 + N)^r\leqslant 2^rN^r$ on $\mathbf{R}^n-$ B, we see that assertions a) and b) are equivalent.

Let us prove b) when V = B. We may suppose that $p= 1$. Since the case where $r >0$ is elementary, suppose that $r\leqslant 0$. For every integer $j\geqslant 1$, the set

is of measure $2^{nj}(2^n-1)$. The sets $(C_j)_{j\geqslant 1}$ form a partition of $\mathbf{R}^n-$ B. According to INT, V, p. 4, § 1, No.$^o1$, cor., one therefore has

$\mathbf{R}^{n-}Bj\geqslant 1C_j$

$\leqslant \sum_{j\geqslant 1}2^{n+nj}2^{r(j-1)}= 2^{n-r}\sum_{j\geqslant 1}2^{(n+r)j}$ which is finite if $r <-n$. On the other hand, one has (loc. cit.)

$\mathbf{R}^{n-}Bj\geqslant 1j\geqslant 1$

which is infinite if $r\geqslant -n$.

One proves a$')$ similarly by considering the sets $C_j\cap \mathbf{Z}^n$ which cover $\mathbf{Z}^n-\{0\}$ and satisfy

Card(C$_j\cap \mathbf{Z}^n$) $= (2^{j+1}-1)^n-(2^j-1)^n$,

which belongs to the interval $[(1-2^{-n})(2^{j+1}-1)^n,2^{n(j+1)}]$.

Since U is locally compact, there exists a covering $\mathscr{V}$ of U finer than $\mathscr{U}$ consisting of open sets relatively compact in U. Since U is paracompact (TG, IX, p. 51, th. 4), there exists a locally finite open covering $\mathscr{W}$ finer than $\mathscr{V}$. Then $\mathscr{W}$ is finer than $\mathscr{U}$ and consists of open sets relatively compact.

a) One has $f_m\in \mathscr{D}(\mathbf{R}^n)$ and the support of $f_m$ is contained in V ;

b) Let $p\in [1,+\infty ]$. The sequence $(f_m)_{m>m_0}$ converges to $f$ in $L^p(\mathbf{R}^n)$.

By TG, IX, p. 14, remark, one has $d(K,\mathbf{R}^n-V)>0$. Let $m_0\geqslant 1$ be an integer such that $m^{-1}_0< d(K,\mathbf{R}^n-$ V) ; it satisfies the required condition. For every $m > m_0$, the function $f_m$ is continuous (INT, VIII, p. 166, § 4, n$^o5$, prop. 11) and with support contained in $K + V_m$ (INT, VIII, p. 126, § 1, n$^o4$, prop. 5), hence in V. By Corollary 2 of IV, p. 198, one has $f_m\in \mathscr{D}(\mathbf{R}^n)$. If $p\not = +\infty$, then the sequence $(f_m)$ converges to $f$ in $L^p(\mathbf{R}^n)$ (INT, VIII, p. 172, § 4, n$^o7$, prop. 20).[^1]

Suppose that $p= +\infty$. Let $\varepsilon >0$. The function $f$ is uniformly continuous on $\mathbf{R}^n$, therefore there exists an integer $m_1> m_0$ such that, for every $m\geqslant m_1$ and all $x\in \mathbf{R}^n$ and $y\in V_m$, one has $|f(x)-f(x-y)|< \varepsilon$.

For every $m\geqslant m_1$, the function $\varphi_m$ vanishes outside $V_m$, is positive and of integral 1. One therefore deduces the inequality

for every $x\in \mathbf{R}^n$, whence the result.

b) For every $p\in [1,+\infty [$, the canonical injection of $\mathscr{D}(U)$ into $\mathscr{L}^p(U)$ is continuous and $\mathscr{D}(U)$ is dense in $L^p(U)$.

Every continuous seminorm on $\mathscr{K}(U)$ is continuous on $\mathscr{D}$(U), hence the canonical injection of $\mathscr{D}(U)$ into $\mathscr{K}(U)$ is continuous.

Let $f\in \mathscr{K}(U)$. There exists a sequence $(f_m)_{m\in\mathbf{N}}$ in $\mathscr{D}(U)$ which converges uniformly to $f$ on U and such that the supports of the $f_m$ are contained in a fixed relatively compact neighbourhood of the support of $f$ (Lemma 3). The sequence $(f_m)$ therefore converges to $f$ in $\mathscr{K}(U)$. This concludes the proof of a).

Let $p\in [1,+\infty [$. Let $f\in \mathscr{D}(U)$ and let K be the support of $f$. One then has the inequality $N_p(f)\leqslant \mu(K)^{1/p}$ sup$_{x\in K}|f(x)|$, hence the canonical injection of $\mathscr{D}(U)$ into $\mathscr{L}^p(U)$ is continuous. The last part of assertion b) then results from a).

Since the product of two test functions is again a test function, Leibniz's formula (FVR, I, p. 28, prop. 2) shows that the space $\mathscr{D}(U)$ is a topological algebra.

More generally, let $f\in \mathscr{C}^{\infty}(U)$. The linear mapping $\varphi \mapsto f \varphi$ of $\mathscr{D}(U)$ into $\mathscr{D}(U)$ is then continuous. In fact, for every compact subset K of U and every $\alpha \in \mathbf{N}^n$, one has

$0\leqslant \beta \leqslant \alpha$

(cf. loc. cit.).

sup$_{x\in K}|\partial^{\alpha}\varphi (x)|\leqslant M_{\alpha}$.

In fact, the space $\mathscr{D}(U)$ is bornological and every bounded subset of $\mathscr{D}(U)$ is contained in one of the bounded sets described in the statement.

In particular, a sequence $(f_m)_{m\in\mathbf{N}}$ of distributions converges to a distribution $f$ if and only if, for every $\varphi \in \mathscr{D}(U)$, the sequence $(\langle f_m, \varphi \rangle )_{m\in\mathbf{N}}$ converges to $\langle f, \varphi \rangle$.

Since $\mathscr{D}'(U)$ is a Montel space, hence barrelled, this follows from the Banach-Steinhaus theorem (EVT, III, p. 26, cor. 3 of th. 1).

Let V be an open set contained in U. The transpose of the canonical linear mapping of $\mathscr{D}(V)$ into $\mathscr{D}(U)$ is a continuous linear mapping of $\mathscr{D}'(U)$ into $\mathscr{D}'$(V), called the restriction of distributions from U to V and denoted by $r_{VU}$, or sometimes $r_{V,U}$.

One has $r_{VV}= 1_{\mathscr{D}'(V)}$. If $W\subset V\subset U$ are open sets in U, then one has $r_{WV}\circ r_{VU}=r_{WU}$. In other words, if $\mathscr{T}$ denotes the topology on U, the projective system $\mathscr{D}'(U) = ((\mathscr{D}'(V))_{V\in\mathscr{T}},(r_{WV}))$ is a presheaf on U with values in the Species of structures of locally convex topological vector spaces (TA, I, p. 42, def. 1 and p. 66, §10).

Let us recall (TA, I, p. 43, def. 2) that this means that for every open subset V of U and every open covering $(V_i)_{i\in I}$ of V, the following conditions are satisfied:

(i) The mapping $(r_{V_iV})_{i\in I}:\mathscr{D}'(V)\rightarrow \prod_{i\in I}\mathscr{D}'(V_i)$ is injective;

(ii) For every family $(f_i)\in \prod_{i\in I}\mathscr{D}'(V_i)$ such that

for every pair $(i, i')\in I\times I$, there exists a distribution $f\in \mathscr{D}'(V)$ such that for every $i\in I$, one has $r_{V_iV}(f) =f_i$.

Let V be an open set of U and $\mathscr{V}= (V_i)_{i\in I}$ an open covering of V. Let $(W_j)_{j\in J}$ be a locally finite open covering of V, finer than $\mathscr{V}$ and formed of relatively compact subsets (lemma 2 of IV, p. 201). Let us fix a partition of unity $(\varphi_j)_{j\in J}$ subordinate to the covering $(W_j)_{j\in J}$ such that the support of $\varphi_j$ is contained in $W_j$ for every $j\in J$ (lemma 1 of IV, p. 196).

Let $\varphi \in \mathscr{D}(V)$. Since the set of $j\in J$ such that $W_j$ meets the support of $\varphi$ is finite (TG, I, §9, n$^o1$, p. 59), one has

where the sum consists of only a finite number of non-zero terms.

Let us prove (i). Suppose that $f\in \mathscr{D}'(V)$ satisfies $r_{V_iV}(f) = 0$ for every $i\in I$. This means that $\langle f, \varphi \rangle = 0$ for every test function $\varphi$ whose support is contained in one of the open sets $V_i$. This is fortiori true if the support of $\varphi$ is contained in one of the open sets $W_j$. But then, for every $\varphi \in \mathscr{D}$(V), we have

hence $f= 0$.

Let us prove (ii). Let $(f_i)_{i\in I}$ be a family such that $f_i\in \mathscr{D}'(V_i)$ for every $i\in I$ and $r_{V_i\cap V_{i'},V_i}(f_i) =r_{V_i\cap V_{i'},V_{i'}}(f_{i'})$ for all $i$ and $i'$ in I. Let $\iota : J\rightarrow I$ be a mapping such that $W_j\subset V_{\iota(j)}$ for every $j\in J$. For every $j\in J$, put $\widetilde{f}_j=r_{W_j,V_{\iota(j)}}(f_{\iota(j)})$. We then have

Exchanging the role of $j$ and $j'$ and noting that

by assumption, we deduce that

for all $(j, j')\in J^2$.

For $\varphi \in \mathscr{D}$(V), put

where the sum is finite since only a finite number of terms can be non-zero.

The mapping $\lambda$ is a linear form on $\mathscr{D}(V)$. Let us prove that it is a distribution. Let B be a bounded subset of $\mathscr{D}(V)$ and let K be a compact subset of V such that $B\subset \mathscr{C}_K^{\infty}(V)$. By TG, I, §9, n$^o1$, p. 59, there exists a finite subset $J'$ of J such that

for every $\varphi \in B$. Since $\widetilde{f}_j$ is a distribution for every $j\in J'$ and $\mathscr{D}(V)$ is a topological algebra, we deduce that $\lambda$ is bounded on B, whence the result (lemma 4).

Let $j\in J$ and let $\varphi \in \mathscr{D}(V)$ whose support is contained in $W_j$. We then have

by (1), since $\varphi \varphi_{j'}$ has support contained in $W_j\cap W_{j'}$. Consequently

whence $r_{W_jV}(\lambda ) =\widetilde{f}_j$ for every $j\in J$.

Let $i\in I$. Let us show finally that the restriction of $\lambda$ to $V_i$ coincides with $f_i$. By condition (i), applied to the covering of $V_i$ by the open sets $W_j$, it is enough to verify that for every $j\in J$, the restriction of $\lambda$ to $V_i\cap W_j$ coincides with that of $f_i$. By what precedes, it is a question of verifying that the restriction of $f_i$ to $V_i\cap W_j$ is that of $\widetilde{f}_j$. Now one has

where one has used the assumption concerning the family $(f_i)$. But then

which permits one to conclude.

Let K be a compact subset of U. For every function $\varphi \in \mathscr{C}_K^{\infty}$(U), one has $|\langle \nu , \varphi \rangle |\leqslant p_{0,K}(\varphi )|\nu |$(K), hence the restriction of $\nu$ to $\mathscr{D}(U)$ is continuous. Since $\mathscr{D}(U)$ is dense in $\mathscr{K}(U)$ (prop. 4, a) of IV, p. 202), the restriction of $\nu$ to $\mathscr{D}(U)$ is zero if and only if $\nu$ is zero.

We shall identify the space $\mathscr{M}(U;\mathbf{C})$ of complex measures on U with a subspace of $\mathscr{D}'(U)$. We shall also identify the space $L^1_{loc}(U)$ with a subspace of $\mathscr{D}'(U)$ by the mapping deduced by passing to quotients from the mapping which associates to $f\in \mathscr{L}_{loc}^1(U)$ the measure $f\cdot \mu$ (INT, V, p. 44, § 5, No.$^o2$, def. 2). In other words, for $f\in L^1_{loc}(U)$ and $\varphi \in \mathscr{D}$(U), one has

In particular, for $p\in [1,+\infty ]$, this permits one to identify the space $L^p(U)$ with a subspace of $\mathscr{D}'(U) ($cf. INT, V, p. 43, § 5, No.$^o1)$.

Let $f\in L^p(U)$. Let K be a compact subset of U, and denote by $\varphi_K$ its characteristic function. For every test function $\varphi$ with support contained in K, Hölder's inequality implies

where $q$ is the conjugate exponent of $p$.

In particular one can identify $\mathscr{D}(U)$ with a subspace of $\mathscr{D}'(U)$.

Let $\lambda$ be a linear form on $\mathscr{D}'(U)$ which is zero on $\mathscr{D}(U)$. Since $\mathscr{D}(U)$ is reflexive, there exists a test function $\varphi \in \mathscr{D}(U)$ such that $\lambda (f) =\langle f, \varphi \rangle$ for every $f\in \mathscr{D}'(U)$. Hence

whence $\varphi = 0$. The proposition then follows from the Hahn–Banach theorem (EVT, II, p. 49, cor. 3 (ii)).

For $h\in \mathscr{C}^{\infty}$(U), the transpose of the continuous linear mapping $\varphi \mapsto h\varphi$ of $\mathscr{D}(U)$ into itself is a continuous linear mapping of $\mathscr{D}'(U)$ into itself, which is denoted by $f\mapsto hf$. This definition is justified because if $f$ is the distribution associated with a measure $\nu$ on U, then $hf$ is associated with the measure $h\cdot \nu$. In fact, for every test function $\varphi \in \mathscr{D}$(U), one computes