Prove that the convolution is continuous

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I need some help with this exercise:

Let $f \in L^p(\Bbb R)$ and let $g$ be a continuous function with compact support. Prove that:

a) If $p=1$ or $p=+\infty$, then $f*g$ is continuous.

b) If $1\lt p \lt+\infty$, then $f*g$ is continous.

I have no idea how to solve this, but I know that if $1\le p \lt +\infty$ then the set of continuous functions with compact support is dense in $L^p(\Bbb R)$ so I guess that first I can suppose that $f$ is continuous with compact support and then I'll have to use this density property.

Also, given that the exercise is separated in two parts, I imagine that the solution is different in each case, is that right?

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Here are the ideas. Try to work out the details on your own. It is enough to assume that $1\leq p <\infty$. By H\"older's inequality and translation invariance of Lebesgue in $\mathbb{R}^n$ \begin{aligned} |(f*g)(x+h)-(f*g)(x+k)|&\leq \int|(f(x+h-y)-f(x+k-y)||g(y)|\,dy\\ &\leq \|\tau_{-(k-h)}f-f\|_p\|g\|_q. \end{aligned} where $\tau_h$ is the translation operator $\tau_hf(x)=f(x-h)$. The result follows once you show that $\|\tau_hf-f\|_p\rightarrow0$ as $h\rightarrow0$. This is typically done by using approximations with continuous functions of compact support.

If $1<p<\infty$, by using continuous functions of compact support you can actually prove that $f*g\in\mathcal{C}_0(\mathbb{R}^n)$.