# 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 (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!

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 $$\alpha=\frac{\|f\|_p^p}{\int |f+q|^p d\mu}=\frac{\|f\|_p^p}{\|f+g\|_p^p}$$

• Yes! This answer my question. Very nice and detailed proof for the exact value of the constants. Thanks a lot! Commented Nov 6, 2019 at 13:49
• @CrisPaint I am glad to help! Commented Nov 6, 2019 at 14:17