When does $\|f*g\|_{p}=\|f\|_{1}\|g\|_{p}$? 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.
 A: Here we assume $f$ and $g$ to be positive functions. 
Here is my result:
if $$\|f*g\|_{p}=\|f\|_{1}\|g\|_{p} $$ then either 
$g=0~~a.e$ or $p =\infty$ or $f=0~~~a.e$. 
see all the details below.

Claim: for, $f\in L^1$ and $g\in L^p$ we have, for all $x\in\Bbb R,$ $$x\mapsto h(x) =\|f\|_1^{p-1} |g|^p*|f|(x)-|f*g(x)|^p \ge 0.$$ 

proof: By Holder's inequality
\begin{align}|f*g(x)|&\le \int_\Bbb R |g(y)||f(x-y)|^{1/p} |f(x-y)|^{1-1/p}dy\\
&\le \left(\int_\Bbb R |g(y)|^p|f(x-y)|dy\right)^{1/p}\left( \int_\Bbb R |f(x-y)|dy\right)^{1-1/p}\\&=\left[\|f\|_1^{p-1} |g|^p*|f|(x)\right]^{1/p}\end{align}
That is $$\color{red}{h(x)= \|f\|_1^{p-1} |g|^p*|f|(x)-|f*g(x)|^p \ge 0~~~\forall ~~x\in\Bbb R. }$$
However, by Fubini, 
\begin{align}\|f*g\|_{p}=\|f\|_{1}\|g\|_{p} &\Longleftrightarrow \int_\Bbb R |g*f(x)|^{p} dx =  \|f\|_1^{p-1}\left(\int_\Bbb R |g(y)|^p\right)\left(\int_\Bbb R |f(x-y)|dy\right)\\ &\Longleftrightarrow \int_\Bbb R |g*f(x)|^{p} dx =  \|f\|_1^{p-1}\int_\Bbb R \left(\int_\Bbb R |g(y)|^p|f(x-y)|dy\right)dx
\\ &\Longleftrightarrow \int_\Bbb R |g*f(x)|^{p} dx =  \|f\|_1^{p-1}\int_\Bbb R [|g|^p*|f|](x)dx
\\ &\Longleftrightarrow \int_\Bbb R |g*f(x)|^{p}- \|f\|_1^{p-1}[|g|^p*|f|](x)dx=0
\\ &\Longleftrightarrow \int_\Bbb R h(x)dx=0
\end{align}
But since $h\ge 0$ We conclude that 
\begin{align}\|f*g\|_{p}=\|f\|_{1}\|g\|_{p} &\Longleftrightarrow 
h(x)=0~~a.e\\ 
 &\Longleftrightarrow \color{blue}{|g*f(x)|^{p} =\|f\|_1^{p-1}[|g|^p*|f|](x)~~a.e}\\
 &\Longleftrightarrow \color{blue}{\int_\Bbb R g(y)f(x-y)dy =\left(\int_\Bbb Rf(x-y)dy\right)^{1-1/p}\left(\int_\Bbb Rg^p(y)f(x-y)dy\right)^{1/p}~~a.e}\tag{I}
\end{align}
On the other hands, in any measure space, the Holder's inequality says:
$$\int_E|f_1f_2|\leq\left(\int_E|f_1|^a\right)^{\frac{1}{a}}\left(\int_E|f_2|^a\right)^{\frac{1}{b}}$$
where $\displaystyle\frac{1}{a}+\frac{1}{b}=1$. Equality holds when $\alpha |f_1|^a=\beta |f_2|^b$ almost everywhere on $E$, where $\alpha$ and $\beta$ are constants.

Thus, let $x\in \Bbb R$  fixed such that 
  $$\color{blue}{|g*f(x)|^{p} =\|f\|_1^{p-1}[|g|^p*|f|](x)}$$ 
  we consider $f\in L^1$ we consider, the measure 
  $$\color{red}{d\mu_x(y)= |f(x-y)|dy}$$
  $(\Bbb R, \mathcal{B}, \mu_x)$ is a measure space,

It springs from (I) that, 
\begin{align}
 &\int_\Bbb R g(y)f(x-y)dy =\left(\int_\Bbb Rf(x-y)dy\right)^{1-1/p}\left(\int_\Bbb Rg^p(y)f(x-y)dy\right)^{1/p}~\\
 &\Longleftrightarrow \\
& \int_\Bbb R g(y)d\mu_x(y) =\left(\int_\Bbb R \mathbf1_\Bbb R(y)d\mu_x(y)\right)^{1-1/p}\left(\int_\Bbb Rg^p(y)d\mu_x(y)\right)^{1/p}
\end{align}
Applying Holder inequality above there exists a constant $C_x$ such that,
$$g(y) =C_x ~~\mu_x-~~a.e$$
Improvement:

Then there exists $A\subset \Bbb R. $
  such that, $$ g(y) =C_x~~~ \forall ~~y\in\Bbb R\setminus A$$ and $$\mu_x(A) = 0~~\Longleftrightarrow  \int_A  |f(x-y)|dy= 0$$
  Then, either $|A_x|= 0$ ($A_x$ is Lebesgue negligible ) or $A_x$ is not Lebesgue negligible and  $f(x-y) = 0 ~~~a.e $ on $A$.

Assume that: $|A_x|=0$ then obviously we get that
$$g(y) =C_x~~a.e~~on~~\Bbb R$$
which means that $g$ is constant almost everywhere. since the only function satisfying $f(x) = h(y)$ for almost all $x$ and $y$ are constant function.
Thus, we either have, $\color{red}{g =0~~a.e}$ or $\color{red}{p =\infty}$ since, $\color{red}{g\in L^p(\Bbb R)}$
Now Assume that:  $A_x$ is not Lebesgue negligible Then,  $f(x-y) = 0 ~~~a.e $ on $A_x$.
Then repeating this argument for all $x\in\Bbb R$ we arrive at $f= 0~~a.e$

Or use this to conclude $f(y-x)$ integrable implies $f=0$ a.e.
