2.24

Part a.) Given the equation: $$(f*g)(x) = \int_{\mathbb{R}^d} f(x-y)g(y) \hspace{0.1cm}dy$$ If we assume $f$ is integrable, and $\exists M \geq 0$ such that $|g| \leq M \hspace{0.1cm} \forall x \in \mathbb{R}^d$, we have: $$\big|(f*g)(x) - (f*g)(z)\big| = \ldots$$ $$\ldots = \Bigg|\int_{\mathbb{R}^d} \bigg(f(x-y) - f(z - y)\bigg)g(y) \hspace{0.1cm}dy\Bigg| \leq \ldots$$ $$\ldots \leq M\int_{\mathbb{R}^d} \bigg|f(x-y) - f(z - y)\bigg| \hspace{0.1cm}dy =\ldots$$ $$\ldots = M\int_{\mathbb{R}^d} \bigg|f(-y) - f((z - x) - y)\bigg| \hspace{0.1cm}dy$$ Thus since $f$ is $L^1(\mathbb{R}^d)$, by Proposition 2.5, $\forall \epsilon > 0,\hspace{0.25cm} \exists \delta$ such that $||z - x|| < \delta \hspace{0.25cm} \Rightarrow ||f(y) - f(y-(z-x))||_{L^1} < \epsilon$. Thus, since $$\big|(f*g)(x) - (f*g)(z)\big| \leq ||f(y) - f(y-(z-x))||_{L^1}$$ The convolution $(f*g)(x)$ must be uniformly continuous.

Part b.) If $f$ and $g$ are both $L^1(\mathbb{R}^d)$, we proved in exercise 21 part d that $(f*g)(x)$ is also $L^1(\mathbb{R}^d)$. Thus, since $(f*g)(x)$ is uniformly continuous, and integrable, we have (by exercise 6 part b) that: $$\lim_{|x| \to \infty} (f*g)(x) = 0$$ ...as desired.