From Rudin, Real and Complex Analysis, 1st edition, Chapter 7, Problem 4
Suppose $1\le p\le \infty$, $f\in L^{1}(\mathbb{R}^{1})$, $g\in L^{p}(\mathbb{R}^{1})$. Show that the the integral defining $f*g$ exists for almost all $x$, that $f*g\in L^{p}$, and that $$\|f*g\|_{p}\le \|f\|_{1}\|g\|_{p}.$$
Show that the equality can hold if $p=1$ and if $p=\infty$, and find conditions under which this happens. Assume $1<p<\infty$, and equality holds, then either $f=0$ a.e or $g=0$ a.e.
My thoughts:
By Fubini's theorem we can prove $$\|f*g\|_{p}^{p}\le \|f\|_1^{p} \|g\|_p^{p}$$ So the first claim is okay. But I do not know how to show when the equality holds.
So it suffice to prove $$\int\left(\int f(x-y)g(y)dy\right)^{p}dx<\left(\int |f|dx\right)^{p}\left(\int |g|^{p}dy\right)$$
By Fubini's theorem we have \begin{align}\int\left(\int f(x-y)g(y)dy\right)^{p}dx&\le \int (\int |f(x-y)|^{p}|g(y)|^{p}dy)dx\\&=\left(\int |g(y)|^p dy)(\int |f(x-y)| dx\right)^{p-1}\int |f(x-y)|dy\\&=\int |g(y)^p|dy (\int |f(t)|dt)^{p} \end{align} using Lebesgue measure's translation invariance. Since the first inequality is just from absolute value, we may assume $f,g$ are non-negative without losing any generality. But I do not see which one of the last few equalities is just an inequality.