The convolution of $f$ and $g$ on $R^d$ equipped with the lebsgue measure is defined by $$(f*g)(x)=\int_{R_d} f(x-y)g(y) \, dy$$ Prove that if $f \in L^p$ and $g \in L^q$ where $p$ and $q$ are conjugate exponents , then $$\lim_{\vert x \vert \to \infty}(f*g)(x)=0$$
There's something I have already proved:
(a) If $f \in L^p$ and $g \in L^1$ , then $f*g \in L^p$ with $$\vert\vert f*g \vert\vert _{L^p} \le \vert\vert f \vert\vert _{L^p} \vert\vert g \vert\vert _{L^1}$$ (b) If $f \in L^p$ and $g \in L^q$ where $p$ and $q$ are conjugate exponents , then $f*g \in L^{\infty}$ with $$\vert\vert f*g \vert\vert _{L^\infty} \le \vert\vert f \vert\vert _{L^p} \vert\vert g \vert\vert _{L^q}$$ Moreover , the convolution $f*g$ is uniformly continuous on $R^d$
I want to show that $f*g \in L^a$ for some $a \lt \infty$ , then by the uniform ontinuous I can get the desired conclution. However , can I find the desired $a$ ？
Here's a proof based on Kavi Rama Murthy's comment. The original question is from Exercise 17 in Stein and Shakarchi's Functional Analysis, Chapter 1. I will quote known results from the same book.
We know that the family of continuous functions with compact support is dense in $L^p$ (Exercise 7). For any $\epsilon > 0$, we can find $f' \in L^p$ and $M>0$ such that $\|f-f'\|_{L^p} < \frac{\epsilon}{2 \|g\|_{L^q}}$ and $f'(x) = 0$ when $|x| > M$. Note that $f' \in L^1$ since it has finite support. (Proposition 1.4)
Similarly, we can find $g' \in L^q$ and $N>0$ such that $\|g-g'\|_{L^q} < \frac{\epsilon}{2 \|f'\|_{L^p}}$ and $g'(x) = 0$ when $|x| > N$.
From (b), we have
For any $|x| > M+N$, we have $$ \begin{align} \big|f*g(x)\big| &\leq \big|[(f-f')*g](x)\big| + \big|[f'*(g-g')](x)\big| + \big|[f'*g'](x)\big| \\ & \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} + \int \big|f'(x-y)g'(y)\big| \mathrm{d}y \\ &= \epsilon + \int_{|y| \leq N} \big|f'(x-y)g'(y)\big| \mathrm{d}y + \int_{|y| > N} \big|f'(x-y)g'(y)\big| \mathrm{d}y \\ &\leq \epsilon, \end{align} $$
our desired result.