# Use Holder inequality to show $f_n * g_n$ converges to $f*g$ uniformly

Assume that $G$ is a locally compact abelian group. Let $C_c(G)$ be a set of complex-valued continuous functions with compact support. Let $C_0(G)$ be a set of complex-valued continuous functions which vanish at infinity.

The following is extracted from 'Fourier Analysis on Groups' by Rudin, page $5:$

Proposition: If $1 < p < \infty, 1/p + 1/q = 1, f \in L^p(G)$ and $g \in L^q(G),$ then $f*g \in C_0(G).$

Proof: Choose sequences $\{f_n\}$ and $\{g_n\}$ in $C_c(G)$ such that $\|f_n-f\|_p \rightarrow 0$ and $\|g_n - g\|_q \rightarrow 0$ as $n \rightarrow \infty.$ Holder's inequality shows that $f_n * g_n \rightarrow f*g$ uniformly. If $f , g \in C_c(G),$ then $f*g \in C_c(G).$ It follows that $f_n * g_n \in C_c(G).$. Hence, $f*g \in C_0(G).$

Question: $(1)$ How to use Holder inequality to show that $f_n *g_n$ converges to $f*g$ uniformly. $(2):$ How to obtain $f*g \in C_0(G)$?

For question $(1),$ we need to show that for any $\varepsilon>0,$ there exists a natural number $N$ such that for all $n \geq N$ and all $x \in G,$ we have $$|(f_n *g_n)(x) - (f*g)(x)| < \varepsilon.$$

My attempt: By definition of convolution, we have $$|(f_n *g_n)(x) - (f*g)(x)| = \left| \int_G f_n(x-y)g_n(y) dy - \int_G f(x-y) g(y)dy\right| .$$ I do not see how to apply Holder inequality here.

I have no idea how to start question $(2).$

Any hint would be appreciated.

Hint. We have that $$|(f_n*g_n)(x) - (f*g)(x)| \leq \int_G |f_n(x-y)-f(x-y)||g_n(y) |dy\\ + \int_G |f(x-y)| |g_n(y) -g(y)|dy$$ Then use Holder's inequality, \begin{align*} |(f_n*g_n)(x) - (f*g)(x)|&\leq \left(\int_G |f_n(x-y)-f(x-y)|^pdy\right)^{1/p} \|g_n\|_q\\&\quad+ \left(\int_G |f(x-y)|^pdy\right)^{1/p} \|g_n-g\|_q\\ &= \|f_n -f \|_p\|g_n\|_q+\|f\|_p\|g_n-g\|_q. \end{align*}
• But $\|g_n\|_q$ depends on $n,$ right? How to make the term $\|f_n - f\|_p \|g_n\|_q$ small? – Idonknow Jul 11 '17 at 6:39
• @Idonknow $\|g_n\|_q\to \|g\|_q$ therefore $\|g_n\|_q$ is bounded – Robert Z Jul 11 '17 at 6:41
• If each $h_n$ has compact support and $h_n$ converges uniformly to $h$ then $h$ vanish at infinity. Proof. Let $x_n\to +\infty$ then there is $N$ such that for all $n\geq N$, $\|h_n-h\||<\epsilon$. Let $m\geq N$ such that $h_m(x_n)=0$ for all $n\geq N'\geq N$. Then $|h(x_n)|\leq |h_m(x_n)|+\|h_n-h\||\leq \epsilon$ which means that $|h(x_n)|\to 0$. – Robert Z Jul 11 '17 at 9:12