I was doing this exercise:
Suppose that $(X,\mathcal{S},\mu )$ is a measure space, $1<p<\infty $ and $f,g\in \mathcal{L}^p(\mu )$. Prove that Minkowski's s inequality is an equality if and only if there exists non-negative numebrs $a$ and $b$, not both zero, such that $af(x)=bg(x)$ almost everywhere.
My wrong proof below:
We want to show that if $\left(\int |f+g|^p\right)^{1/p}=\left(\int |f|^p\right)^{1/p}+\left(\int |g|^p\right)^{1/p}$ then there are some $a,b\geqslant 0$, not both zero, such that $af(x)=bg(x)$ a.e.
WLOG we can assume that $\|f+g\|_p=1$ and $f,g\neq 0$ a.e., therefore $\|f\|_p,\|g\|_p\in(0,1)$ and so there is some constant $b>0$ such that $b\|f\|_p=\|g\|_p$, but this means that $$ \int b^p|f|^p \,\mathrm d \mu=\int |g|^p \,\mathrm d \mu \implies |bf|=|g|\text{ a.e. }\tag1 $$ Then there is some measurable function $h:X\to \Bbb F$ such that $|h(x)|=1$ a.e. and $b fh=g$, thus we want to show that $h=1$ a.e. Now note that $$ \|f+g\|_p=\|f(1+hb)\|_p\quad \,\land\,\quad \|f\|_p+\|g\|_p=(1+b)\|f\|_p\\ \therefore\, \|f(1+bh)\|_p=\|f(1+b)\|_p\implies |1+h(x)b|=1+b \,\text{ a.e. }\tag2 $$ Because $b>0$ and $|h(x)|=1$ a.e. then is easy to conclude that $h=1$ a.e., finishing one direction of the proof.
The other direction is easy to see due to the homogeneity of $\|{\cdot}\|_p$, so we are done.$\Box$
However above I didnt used the fact that $p\in (1,\infty )$, indeed the fake proof above seems to hold for any chosen $p>0$. However it doesn't hold for $p=1$ because if $f$ and $g$ are non-negative then is easy to check that $\|f+g\|_1=\|f\|_1+\|g\|_1$, but in general $f$ and $g$ doesn't need to be proportional a.e.
And probably the statement doesn't hold either when $p\in (0,1)$. Where is my mistake in the above proof?
EDIT: I see my mistake... It is the assertion that $|bf|=|g|$ a.e. I forget that $\int h\,\mathrm d \mu =0\Rightarrow h=0$ a.e. just holds when $h=|h|$.