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

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

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

\begin{align} \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \lvert f(y)\rvert\, \lvert g(x-y)\rvert \, dy\, dx & = \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \lvert f(y) \rvert\, \lvert g(x-y) \rvert\, dx\, dy \\ & = \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 \end{align}

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,

\begin{equation} \big( g \ast f \big) (x) = \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). \end{equation}

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 \( \frac{1}{p} + \frac{1}{q} =1 \).

\begin{align} \lVert f \ast g \rVert_p^p &= \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 \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. \end{align}

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 \( D( f \ast \phi) = f \ast D\phi \).