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

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

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

\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,

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

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