§ 1. Convolution

1. Definition and examples

Recall (Ch. V, §6, Nos. 1 and 4; Ch. VI, §2, No. 10) that, if X and Y are locally compact spaces, $\mu$ a measure on X, and $\varphi$ a mapping of X into Y, $\varphi$ is said to be $\mu$ -proper if: a) $\varphi$ is $\mu$ -measurable; b) for every compact subset K of Y, $\varphi^{-1}(K)$ is essentially $\mu$ -integrable. Then the image measure $\nu = \varphi(\mu)$ on Y exists and has the following property: for a function f on Y, with values in a Banach space or in $\overline{\mathbf{R}}$ , to be essentially integrable for $\nu$ , it is necessary and sufficient that $f \circ \varphi$ be so for $\mu$ , in which case,

\int_Y f(y)\, d\nu(y) = \int_X f(\varphi(x)) d\mu(x).

Definition 1

Let $X_1, \ldots, X_n$ be locally compact spaces, $\mu_i$ a measure on $X_i$ ( $1 \leq i \leq n$ ); let X be the product of the $X_i$ , $\mu$ that of the $\mu_i$ . Let $\varphi$ be a mapping of X into a locally compact space Y. One says that the sequence $(\mu_i)$ is $\varphi$ -convolvable, or that $\mu_1, \ldots, \mu_n$ are $\varphi$ -convolvable, if $\varphi$ is $\mu$ -proper; in this case, the image $\nu = \varphi(\mu)$ of $\mu$ under $\varphi$ is called the convolution product of the $\mu_i$ for $\varphi$ , and is denoted $*_{\varphi}(\mu_i)_{1 \leq i \leq n}$ , or $*_{i=1}^n \mu_i$ , or $\mu_1 * \mu_2 * \cdots * \mu_n$ .

The last two notations are of course used only when there can be no doubt as to $\varphi$ .

Let f be a function on Y, with values in a Banach space or in $\overline{\mathbf{R}}$ . In order that f be essentially integrable for $\mu_1 * \cdots * \mu_n$ , it is necessary and sufficient that the function

(x_1, \ldots, x_n) \mapsto f(\varphi(x_1, \ldots, x_n))

be essentially integrable for $\mu_1 \otimes \mu_2 \otimes \cdots \otimes \mu_n$ , in which case

(1) \quad \int f\ d(\mu_1 * \cdots * \mu_n) = \int f(\varphi(x_1, \ldots, x_n)) d\mu_1(x_1) \ldots d\mu_n(x_n),

a formula that may be regarded as defining $\mu_1 * \cdots * \mu_n$ when one takes $f \in \mathscr{K}(Y)$ .

The definitions imply at once that the $\mu_i$ are convolvable if and only if the $|\mu_i|$ are. When this is the case,

|\varphi(\mu_1 \otimes \cdots \otimes \mu_n)| \leq \varphi(|\mu_1 \otimes \cdots \otimes \mu_n|) = \varphi(|\mu_1| \otimes \cdots \otimes |\mu_n|)

(Ch. VI, §2, No. 10), that is,

(2) \quad |*_i \mu_i| \leq *_i |\mu_i|.

If the $\mu_i$ are convolvable and positive, and if $\nu_i$ is a measure on $X_i$ such that $0 \leq \nu_i \leq \mu_i$ , then the $\nu_i$ are convolvable and

*_i \nu_i \leq *_i \mu_i.

Suppose $\mu_1, \mu_2, \ldots, \mu_n$ are convolvable, and that $\mu'_1, \mu_2, \ldots, \mu_n$ are convolvable ( $\mu'_1$ being a measure on $X_1$ ). By Ch. V, §6, No. 3, Cor. 1 of Prop. 6, $\mu_1 + \mu'_1, \mu_2, \ldots, \mu_n$ are convolvable and

(\mu_1 + \mu'_1) * \mu_2 * \cdots * \mu_n = \mu_1 * \mu_2 * \cdots * \mu_n + \mu'_1 * \mu_2 * \cdots * \mu_n.

Example

— 1) For any $\varphi$ , the measures $\varepsilon_{x_i}$ , where $x_i \in X_i$ for $1 \leq i \leq n$ , are always convolvable and have convolution product $\varepsilon_y$ , with $y = \varphi(x_1, x_2, \ldots, x_n)$ . Consequently, if each of the $\mu_i$ has finite support, then the $\mu_i$ are convolvable and $\mu_1 * \cdots * \mu_n$ has finite support. In particular, let $M$ be a monoid $^1$ equipped with a locally compact topology; if one takes $\varphi$ to be the law of composition in $M$ then the measures on $M$ with finite support form, for convolution, an algebra that is none other than the algebra of the monoid $M$ (over $\mathbf{R}$ or over $\mathbf{C}$ , according as one considers real or complex measures) (A, III, §2, No. 6).

2) Let $M$ be a monoid equipped with the discrete topology; assume that for each $m \in M$ , there are only finitely many pairs $(m', m'') \in M \times M$ such that $m'm'' = m$ ; this amounts to saying that the law of composition in $M$ is a proper mapping of $M \times M$ into $M$ ; the measures on $M$ then form an algebra for convolution, an algebra that is none other than the *total

(1) Monoïde, in the sense of Exer. 17 of Ch. VII, §1.

algebra of the monoid $M$ (A, III, §2, No. 10); we note the following two special cases:

a) $M = N$ , the law of composition being addition. To every measure $\mu$ on $N$ , let us associate the formal series

S(\mu) = \sum_{n=0}^{\infty} \mu(\{n\}) t^n

in an indeterminate $t$ . Then $S(\mu * \mu') = S(\mu)S(\mu')$ . An analogous remark holds for formal series in any number of indeterminates.

$*b)$ $M = N^*$ , the law of composition being multiplication. To every measure $\mu$ on $N^*$ , let us associate the formal Dirichlet series

D(\mu) = \sum_{n=1}^{\infty} \mu(\{n\}) n^{-s}.

Then $D(\mu * \mu') = D(\mu)D(\mu')$ .

3) Let $X, Y, Z$ be locally compact spaces, $\varphi$ a continuous mapping of $X \times Y$ into $Z$ . If $x \in X$ and $\mu$ is a measure on $Y$ , to say that $\varepsilon_x$ and $\mu$ are $\varphi$ -convolvable comes to saying that the mapping $\varphi(x, \cdot)$ of $Y$ into $Z$ is $\mu$ -proper. One then has $\varepsilon_x * \mu = \varphi(x, \cdot)(\mu)$ .

2. Associativity

The following lemma completes Prop. 11 of Ch. V, §8, No. 5:²

Lemma 1

For $1 \leq i \leq n$ , let $X_i, Y_i$ be two locally compact spaces, $\mu_i$ a measure on $X_i$ , and $\varphi_i$ a continuous mapping of $X_i$ into $Y_i$ . Let $X = \prod_i X_i$ , $Y = \prod_i Y_i$ , $\mu = \bigotimes_i \mu_i$ , and $\varphi$ the mapping of $X$ into $Y$ that is the product of the $\varphi_i$ . If $\varphi$ is $\mu$ -proper and $\mu_i \neq 0$ for each $i$ , then the $\varphi_i$ are $\mu_i$ -proper and $\varphi(\mu) = \bigotimes_i \varphi_i(\mu_i)$ .

We can suppose that the $\mu_i$ are positive and $n = 2$ . Let $f_1 \in \mathscr{K}_+(Y_1)$ . Since $\mu_2 \neq 0$ , there exists an $f_2 \in \mathscr{K}_+(Y_2)$ such that $f_2 \circ \varphi_2$ is not $\mu_2$ -negligible. The function $(x_1, x_2) \mapsto f_1(\varphi_1(x_1)) f_2(\varphi_2(x_2))$ is essentially $\mu$ -integrable and continuous, hence $\mu$ -integrable. Therefore there exists an $x_2 \in X_2$ such that $f_2(\varphi_2(x_2)) \neq 0$ and such that the function $x_1 \mapsto f_1(\varphi_1(x_1)) f_2(\varphi_2(x_2))$ is $\mu_1$ -integrable. Therefore $f_1 \circ \varphi_1$ is $\mu_1$ -integrable,

²The lemma follows by induction on part b) of the cited Prop. 11, which is the case $n = 2$ ; the corresponding result in the first edition of Ch. V (§8, No. 3, Prop. 7) did not include the result of part b).

which proves that $\varphi_1$ is $\mu_1$ -proper. One argues similarly for $\varphi_2$ . Then $\varphi(\mu) = \bigotimes_i \varphi_i(\mu_i)$ by Prop. 11 of Ch. V, §8, No. 5.

The following lemma completes Prop. 4 of Ch. V, §6, No. 3.\footnote{The assertion of this lemma is in fact part b) of the cited Prop. 4, a part that was not included in the first edition of Ch. V.}

Lemma 2

Let T, T', T'' be three locally compact spaces, $\mu$ a measure on T, $\pi$ a $\mu$ -measurable mapping of T into T', $\pi'$ a continuous mapping of T' into T'', and $\pi'' = \pi' \circ \pi$ . If $\pi''$ is $\mu$ -proper, then $\pi$ is $\mu$ -proper, $\pi'$ is $\pi(\mu)$ -proper, and $\pi''(\mu) = \pi'(\pi(\mu))$ .

Let K' be a compact subset of T'. Then $K'' = \pi'(K')$ is compact, therefore $\pi''(K'')$ is essentially $\mu$ -integrable, therefore $\pi^{-1}(K') \subset \pi''(K'')$ is essentially $\mu$ -integrable, thus $\pi$ is $\mu$ -proper. Then $\pi'$ is $\pi(\mu)$ -proper and $\pi''(\mu) = \pi'(\pi(\mu))$ by Ch. V, §6, No. 3, Prop. 4.

Proposition 1

Let $X_{ij}$ ( $1 \leq i \leq m, 1 \leq j \leq n_i$ ), $Y_i$ ( $1 \leq i \leq m$ ), and Z be locally compact spaces; for each i, let $\varphi_i$ be a mapping of $X_i = \prod_j X_{ij}$ into $Y_i$ ; let $\varphi$ be the product of the $\varphi_i$ , mapping $X = \prod_i X_i$ into $Y = \prod_i Y_i$ ; let $\psi$ be a mapping of Y into Z.

(i) *Let $\mu_{ij}$ be measures given, respectively, on the $X_{ij}$ , such that, for each i, the $\mu_{ij}$ ( $1 \leq j \leq n_i$ ) are $\varphi_i$ -convolvable, and such that the measures $*| \mu_{ij} |$ are $\psi$ -convolvable; then the $\mu_{ij}$ , for $1 \leq i \leq m, 1 \leq j \leq n_i$ , are $(\psi \circ \varphi)$ -convolvable and

*_{i,j} \mu_{ij} = *_i \left( *_j \mu_{ij} \right).

(ii) Assume $\psi$ and the $\varphi_i$ continuous, and let $\mu_{ij}$ be measures $\neq 0$ given, respectively, on the $X_{ij}$ and $(\psi \circ \varphi)$ -convolvable; then, for each i, the $\mu_{ij}$ ( $1 \leq j \leq n_i$ ) are $\varphi_i$ -convolvable, the measures $*_j |\mu_{ij}|$ are $\psi$ -convolvable, and the formula (3) holds.

It suffices to consider the case that all of the measures in question are $\geq 0$ .

Let us place ourselves under the hypotheses of (i). The mapping $\varphi$ is proper for $\bigotimes_{i,j} \mu_{ij}$ , and

\varphi \left( \bigotimes_{i,j} \mu_{ij} \right) = \bigotimes_i \varphi_i \left( \bigotimes_j \mu_{ij} \right) = \bigotimes_i \left( *_j \mu_{ij} \right)

(Ch. V, §8, No. 5, Prop. 11). The mapping $\psi \circ \varphi$ is proper for $\bigotimes_{i,j} \mu_{ij}$ , and

(\psi \circ \varphi)\left( \bigotimes_{i,j} \mu_{ij} \right) = \psi\left( \bigotimes_i (*_j \mu_{ij}) \right) = *_i (*_j \mu_{ij})

(Ch. V, §6, Prop. 4). Therefore the

\mu_{ij}

(

1 \leq i \leq m,\ 1 \leq j \leq n_i

) are

(\psi \circ \varphi)

-convolvable and formula (3) holds.

Let us place ourselves under the hypotheses of (ii). First of all, Lemma 2 proves that $\varphi$ is proper for $\bigotimes_{i,j} \mu_{ij}$ . Lemma 1 then proves that for every $i$ , $\varphi_i$ is proper for $\bigotimes_j \mu_{ij}$ , and that

\varphi\left( \bigotimes_{i,j} \mu_{ij} \right) = \bigotimes_i (*_j \mu_{ij}).

By Lemma 2,

\psi

is proper for

\bigotimes_i (*_j \mu_{ij})

. Whence the proposition.

Corollary

— Let $X_i, X'_i$ ( $1 \leq i \leq n$ ), $Y, Y'$ be locally compact spaces; let $\varphi, \varphi'$ be continuous mappings of $X = \prod_i X_i$ into $Y$ and of $X' = \prod_i X'_i$ into $Y'$ , respectively; let $f_i$ be continuous mappings of $X_i$ into $X'_i$ ( $1 \leq i \leq n$ ) and $g$ a continuous mapping of $Y$ into $Y'$ , such that $\varphi' \circ f = g \circ \varphi$ , $f$ being the mapping of $X$ into $X'$ that is the product of the $f_i$ . Let $\mu_i$ be measures given respectively on the $X_i$ , all $\neq 0$ . Then the following two assertions are equivalent: (i) $f_i$ is $\mu_i$ -proper for all $i$ , and the measures $f_i(|\mu_i|)$ are $\varphi'$ -convolvable; (ii) the $\mu_i$ are $\varphi$ -convolvable, and $g$ is proper for $*_{\varphi}(|\mu_i|)$ .

Moreover, when these assertions are verified,

*_{\varphi'}(f_i(\mu_i)) = g(*_{\varphi}(\mu_i)) = *_{g \circ \varphi}(\mu_i).

For, let $h = \varphi' \circ f = g \circ \varphi$ . By Prop. 1, the conditions (i) and (ii) are each equivalent to the following condition: (iii) the $\mu_i$ are $h$ -convolvable. When this is so,

*_{\varphi'}(f_i(\mu_i)) = *_h(\mu_i) = g(*_{\varphi}(\mu_i)).

3. The case of bounded measures

Proposition 2

— Let $X_1, \ldots, X_n, Y$ be locally compact spaces, $\mu_i$ a bounded measure on $X_i$ ( $1 \leq i \leq n$ ), $\mu$ the product of the $\mu_i$ , $\varphi$ a $\mu$ -measurable mapping of $\prod_{i=1}^n X_i$ into $Y$ . Then the $\mu_i$ are $\varphi$ -convolvable and

\left\| *_{i=1}^n \mu_i \right\| \leq \prod_{i=1}^n \| \mu_i \|.

If the

\mu_i

are moreover positive, then

\left\| *_{i=1}^n \mu_i \right\| = \prod_{i=1}^n \| \mu_i \|.

For,

\mu'_i = |\mu_i|

is bounded and

\| \mu'_i \| = \| \mu_i \|

(Ch. III, §1, No. 8, Cor. 1 of Prop. 10). One has

|\mu_1 \otimes \cdots \otimes \mu_n| = \mu'_1 \otimes \cdots \otimes \mu'_n

(Ch. III, §4, Nos. 2, 4), therefore

\mu_1 \otimes \cdots \otimes \mu_n

is bounded and

\| \mu_1 \otimes \cdots \otimes \mu_n \| = \| \mu_1 \| \cdots \| \mu_n \|

(ibid., Prop. 4). Therefore

\varphi

\mu

-proper (Ch. V, §6, No. 1, Remark 1), that is, the

\mu_i

are

\varphi

-convolvable. One has

\| *_{i=1}^n \mu'_i \| = \| \mu'_1 \otimes \cdots \otimes \mu'_n \|

(Ch. V, §6, No. 2, Th. 1), consequently

\| *_{i=1}^n \mu'_i \| = \| \mu'_1 \| \cdots \| \mu'_n \|

. Finally,

| *_{i} \mu_i | \leq *_{i} \mu'_i

(No. 1, formula (2)), therefore

\left\| *_{i} \mu_i \right\| \leq \left\| *_{i} \mu'_i \right\| = \prod_{i=1}^n \| \mu_i \|.

Proposition 3

— Let $X_1, \ldots, X_n, Y$ be locally compact spaces, $\varphi$ a continuous mapping of $\prod_{i=1}^n X_i$ into $Y$ . Then the mapping

(\mu, \ldots, \mu_n) \mapsto *_{\varphi} (\mu_i)

\prod_{i=1}^n \mathscr{M}^1(X_i)

into

\mathscr{M}^1(Y)

is a continuous multilinear mapping. This follows from Prop. 2 and what has been said in No. 1.

4. Properties concerning supports

Proposition 4

— Let $X_1, \ldots, X_n, Y$ be locally compact spaces, $\mu_i$ a measure on $X_i$ ( $1 \leq i \leq n$ ), $S_i$ its support, and $\varphi$ a continuous mapping of $\prod_i X_i$ into $Y$ such that the restriction of $\varphi$ to $\prod_i S_i$ is proper. Then the $\mu_i$ are $\varphi$ -convolvable.

For, let $K$ be a compact subset of $Y$ . The support of $\mu = \mu_1 \otimes \cdots \otimes \mu_n$ is $S = \prod_i S_i$ (Ch. III, §4, No. 2, Prop. 2). Therefore $\overline{\varphi}(K) \cap \left( \prod_i X_i - S \right)$ is $\mu$ -negligible. On the other hand, $\overline{\varphi}(K) \cap S$ is compact. Therefore $\overline{\varphi}(K)$ is $\mu$ -integrable.

Proposition 5

Let $X_1, \ldots, X_n, Y$ be locally compact spaces, $\mu_i$ a measure on $X_i$ ( $1 \leq i \leq n$ ), $\mu$ the product of the $\mu_i$ , $\varphi$ a $\mu$ -proper mapping of $\prod_i X_i$ into $Y$ , and $S_i$ the support of $\mu_i$ .

a) *The support of $*_{i} \mu_i$ is contained in the closure of $\varphi \left( \prod_i S_i \right)$ .

b) If $\varphi$ is continuous and the $\mu_i$ are positive, then the support of $*_{i} \mu_i$ is the closure of $\varphi \left( \prod_i S_i \right)$ .

Let $S = \prod_i S_i$ be the support of $\mu$ . The support of $*_{i} \mu_i$ is contained in $\overline{\varphi(S)}$ by Ch. V, §6, No. 2, Cor. 3 of Prop. 2. If $\varphi$ is continuous and the $\mu_i$ are positive, then the support of $*_{i} \mu_i$ is $\overline{\varphi(S)}$ (loc. cit., Cor. 4 of Prop. 2).

Corollary

If $\varphi$ is continuous and the $\mu_i$ have compact support, then the $\mu_i$ are convolvable and $*_{i} \mu_i$ has compact support.

5. Vectorial expression of the convolution product

Proposition 6

Let $X, Y, Z$ be locally compact spaces, $\varphi$ a continuous mapping of $X \times Y$ into $Z$ , and $\lambda, \mu$ measures on $X, Y$ . For $\lambda$ and $\mu$ to be $\varphi$ -convolvable, it is necessary and sufficient that the mapping $(x, y) \mapsto \varepsilon_{\varphi(x,y)} = \varepsilon_x * \varepsilon_y$ of $X \times Y$ into $\mathscr{M}(Z)$ be scalarly $(\lambda \otimes \mu)$ -integrable for the topology $\sigma(\mathscr{M}(Z), \mathscr{H}(Z))$ , in which case

\lambda * \mu = \int_{X \times Y} (\varepsilon_x * \varepsilon_y) \, d\lambda(x) \, d\mu(y).

To say that $\lambda$ and $\mu$ are $\varphi$ -convolvable signifies that, for every $f \in \mathscr{H}(Z)$ , $f \circ \varphi$ is $(\lambda \otimes \mu)$ -integrable, that is, for every $f \in \mathscr{H}(Z)$ the function $(x, y) \mapsto \langle f, \varepsilon_{\varphi(x,y)} \rangle$ is $(\lambda \otimes \mu)$ -integrable, that is, again, that the mapping $(x, y) \mapsto \varepsilon_{\varphi(x,y)}$ of $X \times Y$ into $\mathscr{M}(Z)$ is scalarly $(\lambda \otimes \mu)$ -integrable for $\sigma(\mathscr{M}(Z), \mathscr{K}(Z))$ . If this is the case, then

\langle \lambda * \mu, f \rangle = \int f(\varphi(x, y)) d\lambda(x) d\mu(y) = \int_{X \times Y} \langle \varepsilon_{\varphi(x,y)}, f \rangle d\lambda(x) d\mu(y),

whence $\lambda * \mu = \int_{X \times Y} \varepsilon_{\varphi(x,y)} d\lambda(x) d\mu(y)$ .

Proposition 7

Let X, Y, Z be locally compact spaces, $\varphi$ a continuous mapping of $X \times Y$ into Z, and $\lambda, \mu$ measures on X, Y. Assume that for every $x \in X$ , $\varepsilon_x$ and $\mu$ are $\varphi$ -convolvable. For $\lambda$ and $\mu$ to be $\varphi$ -convolvable, it is necessary and sufficient that the mapping $x \mapsto \varepsilon_x * |\mu|$ of X into $\mathscr{M}(Z)$ be scalarly $\lambda$ -integrable for the topology $\sigma(\mathscr{M}(Z), \mathscr{K}(Z))$ , in which case $\lambda * \mu = \int_X (\varepsilon_x * \mu) d\lambda(x)$ .

Suppose that $\lambda$ and $\mu$ are $\varphi$ -convolvable. For every $f \in \mathscr{K}(Z)$ , $f \circ \varphi$ is $(|\lambda| \otimes |\mu|)$ -integrable, therefore the function $x \mapsto \int_Y f(\varphi(x, y)) d|\mu|(y) = \langle f, \varepsilon_x * |\mu| \rangle$ (which by hypothesis is defined for all $x \in X$ ) is $\lambda$ -integrable; thus $x \mapsto \varepsilon_x * |\mu|$ is scalarly $\lambda$ -integrable for $\sigma(\mathscr{M}(Z), \mathscr{K}(Z))$ , and

\langle f, \lambda * \mu \rangle = \int_X d\lambda(x) \int_Y f(\varphi(x, y)) d\mu(y) = \int_X \langle f, \varepsilon_x * \mu \rangle d\lambda(x),

whence $\lambda * \mu = \int_X (\varepsilon_x * \mu) d\lambda(x)$ . Conversely, suppose that the mapping $x \mapsto \varepsilon_x * |\mu|$ of X into $\mathscr{M}(Z)$ is scalarly $\lambda$ -integrable for $\sigma(\mathscr{M}(Z), \mathscr{K}(Z))$ . Let $f \in \mathscr{K}_+(Z)$ . Then the function $(x, y) \mapsto f(\varphi(x, y))$ is continuous and (Ch. V, §8, No. 3, Prop. 5)

\iint^* f(\varphi(x, y)) d|\lambda|(x) d|\mu|(y) = \int^* d|\lambda|(x) \int^* f(\varphi(x, y)) d|\mu|(y) = \int^* \langle f, \varepsilon_x * |\mu| \rangle d|\lambda|(x) < +\infty.

Therefore $f \circ \varphi$ is $(\lambda \otimes \mu)$ -integrable, so that $\lambda$ and $\mu$ are $\varphi$ -convolvable.

Exercises

See the exercises for § 1.