Fall 2011 - Problem 3

Solution.

1. First, we need to show that the convolution is well-defined. By Hölder's inequality,

   $$\int_{\R^3} \abs{f\p{x - y}g\p{y}} \,\diff{y} \leq \norm{f}_{L^p}\norm{g}_{L^q} < \infty$$

   for any $x$, so $f\p{x - y}g\p{y} \in L^1\p{\R^3}$ for all $x \in \R^3$.

   For $x, y \in \R^3$, we have

   $$\begin{aligned} \abs{\p{f * g}\p{x} - \p{f * g}\p{y}} &\leq \int_{\R^3} \abs{f\p{x - z} - f\p{y - z}}\abs{g\p{z}} \,\diff{z} \\ &\leq \norm{g}_{L^q} \p{\int_{\R^3} \abs{f\p{z} - f\p{z + \p{y - x}}}^p \,\diff{z}}^{1/p}. \end{aligned}$$

   Because translation is continuous in $L^p\p{\R^3}$ (e.g., it is true for compactly supported smooth functions, which are dense in $L^p\p{\R^3}$), this quantity tends to $0$ as $\abs{x - y} \to 0$, so $f * g$ is continuous.

2. Notice that by dominated convergence,

   $$\lim_{N\to\infty} \int_{\R^3} \abs{f\p{x - y}g\p{y} - f\p{x - y}g\p{y}\chi_{\set{\abs{x}<N}}\p{x}} \,\diff{y} = 0,$$

   since $2f\p{x - y}g\p{y} \in L^1\p{\R^3}$. Let $\epsilon > 0$. If $N$ is large enough, then the integral above is smaller than $\epsilon$ and so

   $$\begin{aligned} \limsup_{\abs{x}\to\infty}\,\abs{\p{f * g}\p{x}} &\leq \limsup_{\abs{x}\to\infty}\,\p{\int_{\R^3} \abs{f\p{x - y}g\p{y} - f\p{x - y}g\p{y}\chi_{\set{\abs{x}<N}}\p{x}} \,\diff{y} + \int_{\R^3} \abs{f\p{x - y}g\p{y}\chi_{\set{\abs{x}<N}}} \,\diff{y}\p{x}} \\ &\leq \epsilon, \end{aligned}$$

   since the second term vanishes for any fixed $N \in \N$. Sending $\epsilon \to 0$, we get the claim.