Equality in convolution inequality As we know the following inequality holds: $\lVert f \ast g \rVert_{L^p}\leq \lVert f\rVert_{L^1} \lVert g \rVert_{L^p}$ for $1\leq p \leq \infty$. My question is when the inequality becomes an equality, i.e, $\lVert f \ast g \rVert_{L^p} = \lVert f\rVert_{L^1} \lVert g \rVert_{L^p}$
Thanks.
 A: Here is a sufficient condition: Assume $f\ge0$, $g\ge0$, $p=1$. Then using
Fubini, we get
$$
\|f\ast g\|_{L^1} = \int_{\mathbb R} \int_{\mathbb R} f(y) g(x-y)dy  \ dx
= \int_{\mathbb R}f(y) \int_{\mathbb R}  g(x-y)dx  \ dy
= \|g\|_{L^1} \int_{\mathbb R}f(y)  dy =  \|g\|_{L^1} \|f\|_{L^1},
$$
which is equality.
A: Assume $1<p<\infty$. By Holder's inequality
\begin{align*}
|(f\ast g)(x)|  & \leq\int|f(x-y)||g(y)|\,dy=\int(|f(x-y)||g(y)|^{1/p}
)|g(y)|^{1/p^{\prime}}\,dy\\
& \leq\left(  \int|f(x-y)|^{p}|g(y)|\,dy\right)  ^{1/p}\left(  \int
|g(y)|\,dy\right)  ^{1/p^{\prime}}
\end{align*}
and so
$$
|(f\ast g)(x)|^{p}\leq\int|f(x-y)|^{p}|g(y)|\,dy\left(  \int|g(y)|\,dy\right)
^{p/p^{\prime}}.
$$
Integrating and using Fubini's theorem
\begin{align*}
\int|(f\ast g)(x)|^{p}dx  & \leq\int\int|f(x-y)|^{p}|g(y)|\,dydx\left(
\int|g(y)|\,dy\right)  ^{p-1}\\
& =\int|g(y)|\int|f(x-y)|^{p}\,dxdy\left(  \int|g(y)|\,dy\right)  ^{p-1}\\
& =\int|g(y)|dy\left(  \int|g(y)|\,dy\right)  ^{p-1}\\
& =\int|f(z)|^{p}dz\left(  \int|g(y)|\,dy\right)  ^{p}.
\end{align*}
If you have equality then all inequalities must be equalities. It follows
that
$$
|(f\ast g)(x)|^{p}=\int|f(x-y)|^{p}|g(y)|\,dy\left(  \int|g(y)|\,dy\right)
^{p/p^{\prime}}%
$$
for a.e. $x$. Find $F$ of measure zero such that this equality holds for all
$x\in\mathbb{R}^{n}\setminus F$. But since this came from Holder's inequality,
you must have equality in Holder's inequality, that is, if you consider the
functions $h(y)=|f(x-y)||g(y)|^{1/p}$ and $s(y)=|g(y)|^{1/p^{\prime}}$ then
$\frac{(h(y))^{p}}{\Vert h\Vert_{L^{p}}}=\frac{(s(y))^{p^{\prime}}}{\Vert
s\Vert_{L^{p^{\prime}}}}$. This implies 
$$
\frac{|f(x-y)|^{p}|g(y)|}{\Vert h\Vert_{L^{p}}}=\frac{|g(y)|}{\Vert
s\Vert_{L^{p^{\prime}}}}$$
for a.e $y$. But if $|g(y)|>0$ then $|f(x-y)|^{p}=\frac{\Vert h\Vert_{L^{p}}
}{\Vert s\Vert_{L^{p^{\prime}}}}$. So if $g\neq0$ on a set of infinite
measure, then $f$ has to be a constant in that set, which is impossible since
$f\in L^{p}$, which means that $f=0$. 
I have never thought about this before. For some reason I have the feeling I am doing something wrong, but I don't see where....
PS: If instead of $q=1$ you consider Young's inequality with $1<q<\infty$, then the best constant is no longer one and the equality (with the sharp constant) is reached by Gaussian functions. You can find this in the book of Lieb and Loss Lieb and Loss. 
