Exercise 1.6.7 (Convolution)

We write $g_x:{ℝ}^d\to ℂ$, $g_x(y):=g(x-y)$. Then obviously ${(g_x)}_x$ have the same bound as $g$, i.e., for almost all $y\in {ℝ}^d:|g_x(y)|,|g(y)|\le M$.

Before we start, we establish another useful property of convolution: commutativity. Notice that the function $g{f}_x$ is simply translated function $f{g}_x$ since $g{f}_x(y)=g(y)f(x-y)=g(x-(x-y))f(x-y)=f{g}_x(x-y)$. Thus, by Exercise 1.3.15 / Exercise 1.3.20 the integrals of both functions must be equal.

(i)

By monotonicity we see that the integral in question is indeed finite $$\forall <!--\; FUNCTION\; APPLICATION\; -->x\in {ℝ}^d:{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\left|f\right|\cdot \left|g_x\right|dm\le {\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\left|f\right|Mdm=M{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\left|f\right|dm<\infty .$$

(ii)

Consider the bound functions $G(y):=\sup <!--\; FUNCTION\; APPLICATION\; -->\{g_x(y):x\in {ℝ}^d\}$. Then its is easy to verify that $|G(y)|=\sup <!--\; FUNCTION\; APPLICATION\; -->\{|g_x(y)|:x\in {ℝ}^d\}\le M$. We then have $$\forall <!--\; FUNCTION\; APPLICATION\; -->x\in {ℝ}^d:\left|{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f{g}_{x}dm\right|\le {\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\left|f{g}_x\right|dm\le {\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\left|\mathit{fG}\right|dm\le {\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\left|f\right|\cdot \left|M\right|dm=M{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\left|f\right|dm<\infty$$

as desired.

(iii)

To prove continuity, we analyse the following quantity. $$\begin{array}{llll}\hfill (f\ast g)(z)-(f\ast g)(x)& =(g\ast f)(z)-(g\ast f)(x)={\int \; <!--\; nolimits\; -->}_{{ℝ}^d}(g{f}_z-g{f}_x)dm& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_{{ℝ}^d}g(f_z-f_x)dm\le M{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}(f_z-f_x)dm& \hfill & \end{array}$$

By Proposition 1.6.13 (Translation is continuous in $L^1$), the integral on the right-hand side converges to zero, and the result follows.