The statement of the equality case of Minkowski is: $\|f+g\|_p=\|f\|_p+\|g\|_p$ if and only if there exist real non-negative $\alpha$, $\beta$ (not both zero) such that $\alpha f(x)=\beta g(x)$ for almost every $x$.
I'm trying to follow the proof given in this answer.
Let
$A=\{x\, st\, f(x)=0, g(x)\neq 0\}$
$B=\{x\, st\, f(x)\neq0, g(x)= 0\}$
I cannot see how we exclude the case in which both $A$ and $B$ have non-zero measure.
What the statement seems to say is that if $A$ is non negligible, then $\alpha f(x)=\beta g(x)$ implies $\beta=0$ and $\alpha\neq 0$. Then $f=0$ almost everywhere.
So it seems that in the equality case, the set on which $f$ vanishes must be the same as the set on which $g$ vanishes (up to "almost").
How do one prove that ?
Probably related : Some questions about the Minkowski inequalities But there is no convincing answer.
EDIT: computations to show where I fail to understand the previous answers. I'm trying to make explicit the proof given there : https://math.stackexchange.com/a/814294/294061
I drop the "almost everywhere" precaution.
There are reals (not both vanishing) $s$ and $t$ such that
$s|f(x)|^p=t|g(x)|^p$ (1)
And there are functions $a$ and $b$ taking values in the positive reals
$a(x)f(x)=b(x)g(x)$ (2)
I have to prove that there exist constants $\alpha$ and $\beta$ non both vanishing such that $\alpha f(x)=\beta g(x)$ for every $x$.
Case $s=0$
In that case $g(x)=0$ and we have $\alpha f(x)=\beta g(x)$ with $\alpha=0$ and $\beta=1$.
Case $s\neq 0$
Using the $p$-th root, we have
$|f(x)|=\lambda |g(x)|$
where $\lambda=\sqrt[p]{t/s}$. In particular, $a(x)$ and $b(x)$ are both non vanishing.
Case $s\neq 0$, $g(x)\neq 0$
Taking the module in (2), $a(x)\lambda|g(x)|=b(x)|g(x)|$, so that
$a(x)f(x)=\lambda a(x)g(x)$
and thus $f(x)=\lambda g(x)$.
I conclude that, if there are points in which $g(x)\neq 0$ we have to choose the constants $(\alpha,\beta)$ under the form (\alpha, \lambda\alpha).
Now we pass to the last case:
Case $s\neq 0$, $g(x)=0$
I don't know how to handle this case. If the claim is true, I must have $f(x)=\lambda g(x)$ and then $f(x)=0$.
So I have to prove that $g(x)=0\,\Rightarrow f(x)=0$.
This is the very point where I stuck.
EDIT : answer here.