# Prove that if $f \in L^p$ and $g \in L^q$ where $p$ and $q$ are conjugate exponents , then $\lim_{\vert x \vert \to \infty}(f*g)(x)=0$

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 $$a$$

• Hint: approximate $f$ and $g$ by smooth functions with compact support and use the inequalities you already know. – Kavi Rama Murthy Dec 6 '18 at 10:29
• Thank you ! I see the point now. – J.Guo Dec 6 '18 at 10:46
• By the way, a function can be uniformly continuous, in $p$-integrable and still not satisfy $\lim_{|x|\to \infty}f(x)=0$. – MaoWao Dec 6 '18 at 12:45
• @ 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 $\int _{R_d} \vert f(x) \vert ^p \, dx \to \infty$ – J.Guo Dec 6 '18 at 13:19