Equality condition for convolution's $L^p$ norm. Suppose that $1< p< \infty$, $f\in L^1(R)$, and $g\in L^p(R)$ and that $\|f*g\|_p=\|f\|_1\|g\|_p$. Show that then either $f=0$ a.e. or $g=0$ a.e.
I have solved for $g=0$ a.e. if $\|f\|_1>0$ using the equality condition for Hölder ; $\alpha f^p=\beta g^q$ for some $\alpha, \beta \ge 0$. Is this still useful for supposing $\|g\|_p>0$ and deriving $f=0$ a.e?

In general, if you want to prove a statement of the form $A\implies (B\ \text{or}\ C)$, it is logically equivalent to show $(A\ \text{and}\ (\text{not}\ B))\implies C$. The other answer essentially shows how this works in this particular case, but also you could convince yourself of this by making a truth table with columns

*

*$A$,

*$B$,

*$C$,

*$A\implies (B\ \text{or}\ C)$,

*$(A\ \text{and}\ (\text{not}\ B))\implies C$,

and then checking that no matter what values of $0$ or $1$ ($\text{False}$ or $\text{True}$) you assign to the first three statements $A, B, C$, both of the last two statements will agree.
