9-6

Question: Prove Theorem 9.4, that is, prove that if f \in L(\R^n) and K is bounded and uniformly continuous on \R^n, then f*K is bounded and uniformly continuous on \R^n.
Answer: Since K is bounded on \R^n, there is a constant M > 0 such that K( \mathbf{x}) \le M for all \mathbf{x} \in \R^n. Then for all \mathbf{x} \in \R^n,
\begin{align}
|(f * K)(\mathbf{x})| & =\left|\int_{\R^n} f(\mathbf{x} - \mathbf{t}) K(\mathbf{t}) \ d\mathbf{t}\right| \\
& \le\left|\int_{\R^n} f(\mathbf{x} - \mathbf{t}) \ d\mathbf{t}\right| M \\
& \le\|f\|_1 M \\
& <\infty .
\end{align}
This shows that f*K is bounded on \R^n. Let \varepsilon > 0. Since K is uniformly continuous on \R^n, there exists \delta > 0 such that for any \mathbf{x}, \mathbf{y} \in \R^n, if \|\mathbf{x}-\mathbf{y}\| < \delta, then 
|K(\mathbf{x})-K(\mathbf{y})| < \dfrac{\varepsilon}{\|f\|_1+1}.
Hence, for all \|\mathbf{x}-\mathbf{y}\|<\delta, we have
\begin{align}
|(f * K)(\mathbf{x}) - (f * K)(\mathbf{y})| & =\left|\int_{\R^n} f(\mathbf{t}) K(\mathbf{x} - \mathbf{t}) \ d \mathbf{t} - \int_{\R^n} f(\mathbf{t}) K(\mathbf{y} - \mathbf{t}) \ d\mathbf{t}\right| \\
& =\left|\int_{\R^n} f(\mathbf{t})[K(\mathbf{x} - \mathbf{t}) - K(\mathbf{y} - \mathbf{t})] \ d \mathbf{t}\right| \\
& \le \frac{\varepsilon}{\|f\|_1+1} \int_{\R^n}|f(\mathbf{t})| \ d \mathbf{t} \\
& =\frac{\varepsilon}{\|f\|_1+1}\|f\|_1 \\
& <\varepsilon .
\end{align}
This proves that f*K is uniformly continuous.