## Convolution of integrable functions

Suppose that $f, g$ are Lebesgue-integrable functions in a d–dimensional Euclidean space $\mathbb{R}^d$. The convolution of $f$ and $g$ is defined by

 $\big( f \ast g \big) (x) = \displaystyle{ \int_{\mathbb{R}^d} f(y) g(x-y)\, dy. }$

The operation is closed in the space of integrable functions:

If $f, g \in L_1(\mathbb{R}^d)$, then $\displaystyle{ \int_{\mathbb{R}^d} \lvert f(y)\rvert\, \lvert g(x-y) \rvert\, dy < \infty}$ for a.e. $x \in \mathbb{R}^d$.

Indeed; notice that the function $h\colon \mathbb{R}^d \times \mathbb{R}^d \to \mathbb{R}$ defined by $h(x,y) = \lvert f(y) g(x-y) \rvert$ is measurable and non-negative. By Tonelli’s Theorem, the iterated integrals are equal and thus

 $\displaystyle{ \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \lvert f(y)\rvert\, \lvert g(x-y)\rvert \, dy\, dx}$ $= \displaystyle{\int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \lvert f(y) \rvert\, \lvert g(x-y) \rvert\, dx\, dy}$ $= \displaystyle{\int_{\mathbb{R}^d} \bigg( \int_{\mathbb{R}^d} \lvert g(x-y) \rvert\, dx \bigg) \lvert f(y) \rvert\, dy}$ $= \lVert g \rVert_1 \lVert f \rVert_1$

Notice that the operation is commutative (by virtue of the translation-invariance of the Lebesque measure): For all $x \in \mathbb{R}^d$ where the convolution is well-defined,

 $\big( g \ast f \big) (x) = \displaystyle{ \int_{\mathbb{R}^d} f(x-y)\, g(y)\, dy = \int_{\mathbb{R}^d} f(y)\, g(x-y)\, dy = \big( f \ast g \big) (x) }$.

Unfortunately, there is no unit for this operation, and thus $\big( L_1(\mathbb{R}^d), + , \ast \big)$ does not have the structure of an algebraic ring.

The convolution is also well-defined in a general $L_p$ space for $1 \leq p \leq \infty$, with the following fundamental result:

Suppose $g \in L_1(\mathbb{R}^d)$, $f \in L_p(\mathbb{R}^d)$ for $1 \leq p \leq \infty$. Then $f \ast g$ is a well-defined $L_p(\mathbb{R}^d)$ function satisfying $\lVert f \ast g \rVert_p \leq \lVert g \rVert_1\, \lVert f \rVert_p$.

This is direct using Tonelli’s Theorem and Hölder’s Inequality: Assume $f$ and $g$ are non-negative, and let $q>0$ such that $\frac{1}{p} + \frac{1}{q} =1$.

 $\hfill\lVert f \ast g \rVert_p^p$ $= \displaystyle{\int_{\mathbb{R}^d} \bigg( \int_{\mathbb{R}^d} g(y)^{1/p} g(y)^{1-1/p} f(x-y)\, dy \bigg)^p dx}$ $\leq \displaystyle{\int_{\mathbb{R}^d} \bigg( \int_{\mathbb{R}^d} g(y)^{p \cdot 1/p} f(x-y)^p dy \bigg) \bigg( \int_{\mathbb{R}^d} g(y)^{q \cdot 1/q} dy \bigg)^{p/q}\, dx}$ $= \lVert g \rVert_1 \lVert f \rVert_p^p \lVert g \rVert_1^{p/q} = \lVert g \rVert_1^p \lVert f \rVert_p^p.$

Another interesting property of the convolution is that it preserves the “best” smoothness, provided the smooth function has compact support:

Suppose $f \in L_p(\mathbb{R}^d)$ for $1 \leq p \leq \infty$, and $\phi \in C_c^m(\mathbb{R}^d)$ for some $m \geq 1$. Then $f \ast \phi \in L_p(\mathbb{R}^d) \cap C^m(\mathbb{R}^d)$.

The continuity follows easily, as it does the fact that the convolution is in $L_p(\mathbb{R}^d)$. To prove the statement concerning the smoothness, it will be enough to show that for the partial differential operator $D=\frac{\partial}{\partial x_1}$, it is $D( f \ast \phi) = f \ast D\phi$.

1. September 18, 2013 at 2:58 pm

please give some reference also … and thanks for the young’s inequality