Convolution product, Lp spaces

Question

I wonder how to prove the following statement,

Let p,q be real numbers s.t $1\leq p \leq\infty$, $1\leq q \leq\infty$ and $ \frac{1}{p}+ \frac{1}{q}=1$

Let $f \in L^p(\mathbb R^n)$ and $g \in L^q(\mathbb R^n)$ be given. Define$$(f*g)(x)=\int_{\mathbb R^n}f(x-y)g(y)dy$$

Show that $$f*g \in C(\mathbb R^n) \cap L^\infty(\mathbb R^n)$$ If, in adittion $1< p <\infty$, then $$\lim_{|x| \rightarrow\infty}(f*g)(x)=0$$

In order to prove the first part I've considered that If $f*g \in L^\infty(\mathbb R^n)$ then $f*g$ has to be measurable and $|f*g(x)| \leq C$ a.e in $\mathbb R^n$, where C is a constant.

I think $f*g$ satisfies this properties, $f \in L^p(\mathbb R^n)$ and $g \in L^q(\mathbb R^n)$ so f and g are measurable, we know that the product of measurable function is measurable, so $f*g$ is measurable. Furthermore, I've already proved that $f*g$ is an integrable function, thus there exists a constant C defined as above. Thus I conclude $f \in L^p(\mathbb R^n)$.

But what about $f*g \in C(\mathbb R^n)$? I saw a lemma which says that if f and g have compact support, then the convolution product has compact support, but I don't know how to prove that f and g have compact support. On the other hand, I saw another lemma which says that if $f \in C_{c}(\mathbb R^n)$ and $g \in L_{LOC}^1(\mathbb R^n)$ then $f*g$ is well define for all $x |in (\mathbb R^n)$) and more over $f*g \in C(\mathbb R^n)$. There is a th which establishes that $g \in L_{LOC}^q(\mathbb R^n)$ if $gX_k \in L^p(\mathbb R^n)$ for all compac set in $\mathbb R^n$. Well $$gX_k(x)= g$$ for all $x \in k $ , where k is a compact set. I found that $gX_k(x)$ cause it's a product of measurable functions, and$\int_{\mathbb R^n}|gX_k(t)|^{q}dt=\int_{k}|g(t)|^{q}dt+\int_{\mathbb R^n-k}0dt=\int_{k}|g(t)|^{q}dt \leq \infty$. If $g \in L_{LOC}^q(\mathbb R^n)$ then $g \in L_{LOC}^1(\mathbb R^n)$. But now I don't how to prove $f \in C_{c}(\mathbb R^n)$.

I'm not sure all the step above are right and I don't know prove the limit equals cero, may if x goes to infinity then x doesn't belong to the support thus the function equals cero?

Thank you in advance.

Answer 1

If both $f$ and $g$ are continuous with compact support, then so is their convolution. And continuous functions with compact support are dense in $L^p$ for $1<p<\infty$. So, you need to do the following:

1. Given $f\in L^p$ and $g\in L^q$, approximate them with $C_c$ functions $f_n$ and $g_n$ in the respective norm.
2. Show that $f_n*g_n$ is a Cauchy sequence in the uniform norm. This will imply that its limit, which is $f*g$, is a continuous function vanishing at infinity.

For the second step, Hölder's inequality is useful: $$\begin{split} |f_n*g_n(x)-f_m*g_m(x)|&\le |f_n*g_n(x)-f_n*g_m(x)|+|f_n*g_m(x)-f_m*g_m(x)| \\ &\le \|f_n\|_p \|g_n-g_m\|_q + \|f_n-f_m\|_p \|g_m\|_q \end{split}$$