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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.

Laurent Claessens
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  • I think this follows from the equality case of Holder's inequality, which is usually used in the proof of Minkowski's inequality. – Peter Oberly May 03 '21 at 05:17
  • @Peter Oberly This is the way explained in the given link. But I fail to fill the gap in the particular case where A and B have non vanishing measure. (or to show that this case is impossible) – Laurent Claessens May 03 '21 at 08:12
  • The equality holds iff we have $|f+g|=|f|+|g|$ AND two equalities in the Holder inequalities applied. So the first thing is to understand precisely when the equality occurs in Holder's inequality. Then this will follow easily. This post is related: https://math.stackexchange.com/questions/87636/on-the-equality-case-of-the-h%C3%B6lder-and-minkowski-inequalities – Just dropped in May 03 '21 at 08:57
  • Also, you asked the exact same question yesterday; for house-keeping reasons I would recommend deleting one of the two. – Just dropped in May 03 '21 at 08:58
  • @JustDroppedIn I add some computations to show where I'm stuck when trying to understand the linked answer. – Laurent Claessens May 03 '21 at 11:37

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