Convolution of integrable functions

Leave a comment Go to comments

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. Anonymous
    September 18, 2013 at 2:58 pm
    Reply

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

  1. No trackbacks yet.

Leave a comment