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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 $a$

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    $\begingroup$ Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know. $\endgroup$ – Kavi Rama Murthy Dec 6 '18 at 10:29
  • $\begingroup$ Thank you ! I see the point now. $\endgroup$ – J.Guo Dec 6 '18 at 10:46
  • $\begingroup$ By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $\lim_{|x|\to \infty}f(x)=0$. $\endgroup$ – MaoWao Dec 6 '18 at 12:45
  • $\begingroup$ @ MaoWao If not , then there exist $a \gt0 \,, \delta \gt 0$ such that for every $M \ge 0$ ,there exist $x_0 \gt M$ , $\vert f(x) \vert \gt a$ whenever $\vert x-x_0 \vert \lt \delta$ , so we have $\int _{R_d} \vert f(x) \vert ^p \, dx \to \infty$ $\endgroup$ – J.Guo Dec 6 '18 at 13:19

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