Understanding proof when equality holds in Minkowski's inequality I have a simple question about a fact that is constantly mentioned when we have $\|f+g\|_p = \|f\|_p + \|g\|_p$ in $L^p(\mu)$ space for $1<p<\infty$ (I hope someone find this useful).
I read the arguments given here, and here and a proof almost related here. In all of these proofs it's mentioned that when the two Hölder's inequalities below that result on the steps of Minkowski's inequality:
$$\int |f| \cdot |f+g|^{p-1} \, d\mu \leq \|f\|_p \cdot \|f+g\|_p^{p-1}$$
$$\int |g| \cdot |f+g|^{p-1} \, d\mu \leq \|g\|_p \cdot \|f+g\|_p^{p-1}$$
These two holds with equality thanks to the given hypothesis, and that's clear to me. However, then comes a fact that is not as clear to me as it should be (since I found no justification for this) that is: these two equalities implies that there is $\alpha,\beta \geq 0$ such that $|f|^p = \alpha |f+g|^p$ and $|g|^p = \beta |f+g|^p$ almost everywhere. Any hint or reference is appreciated. Thanks!
 A: I am gonna give it a try, let me know if this is what your question is about!
Holder inequality is a consequence of the Young's inequality, the latter stating that
for $p,q\in (1,\infty)$ conjugate numbers
$$AB\leq \frac{A^p}{p}+\frac{B^q}{q}$$
$A,B\geq 0$ where equality occurs only if $B=A^{p-1}$.
Then in order to obtain Holder's inequality set $$A:=\frac{|f|}{\|f\|_p}$$
$$B:=\frac{|f+g|^{p-1}}{\|(f+g)^{p-1}\|_q}$$
Then
$$AB=\frac{|f|\cdot |f+g|^{p-1}}{\|f\|_p\|(f+g)^{p-1}\|_p}\leq \frac{|f|^p}{p\|f\|^p_p}+\frac{|f+g|^{(p-1)q}}{q\|(f+g)^{p-1}\|^q_q}$$
Integrating both sides we obtain the Holder's inequality.
Now in order to have equality in the final expression we need equality on the latter, which is true if $B=A^{p-1}$.
$$\frac{|f+g|^{p-1}}{\|(f+g)^{p-1}\|_q}=\bigg(\frac{|f|}{\|f\|_p}\bigg)^{p-1}$$
$$|f|^{(p-1)}=\bigg(\frac{\|f\|_p^{(p-1)}}{\|(f+g)^{p-1}\|_q}\bigg)\cdot|f+g|^{p-1}$$
rising both sides to $p/(p-1)$
yields to the desired result
$$|f|^{p}=\bigg(\frac{\|f\|_p^{(p-1)}}{\|(f+g)^{p-1}\|_q}\bigg)^{p/p-1}\cdot|f+g|^{p}$$
calling $\alpha=\bigg(\frac{\|f\|_p^{(p-1)}}{\|(f+g)^{p-1}\|_p}\bigg)^{p/p-1}\geq 0$.
$$\alpha=\frac{\|f\|_p^p}{\bigg(\int |f+g|^{q(p-1)}d\mu\bigg)^{\frac{1}{q}\frac{p}{p-1}}}$$
notice that $q(p-1)=p$ then we obtain
$$\alpha=\frac{\|f\|_p^p}{\int |f+q|^p d\mu}=\frac{\|f\|_p^p}{\|f+g\|_p^p}$$
