Clarifying a step in the equality case of the Hölders inequalite

Question

Can anyone please explain to me why $f'g' = \frac{1}{p}f'^p + \frac{1}{q}g'^q$ iff $f'^p = g'^q$ in this post On the equality case of the Hölder and Minkowski inequalites

Answer 1

Suppose ${f'}^p={g'}^q$ a.e. then observe that $$\frac{1}{p}f'^p+\frac{1}{q}g'^{q}=\frac{|f|^p}{p\|f\|_p^p}+\frac{|g|^q}{q\|g\|_{q}^q}=\frac{(p+q)\|g\|_q^q|f|^p}{pq\|g\|_q^{q}\|f\|_p^p}=\left(\frac{1}{p}+\frac{1}{q} \right)\frac{|f|^p}{\|f\|_p^p}=\frac{|f|^p}{\|f\|_p^p}=\frac{(|f|^p)^{\frac{1}{p}+\frac{1}{q}}}{(\|f\|_p^p)^{\frac{1}{p}+\frac{1}{q}}}=\frac{|f| |f|^{\frac{p}{q}}}{\|f\|_p\|f\|_p^{\frac{p}{q}}}=\frac{|f||g|}{\|f\|_p\|g\|_q}=f'g'$$where the second last equality is because $|f|^p=|g|^q$ so $|f|^{\frac{p}{q}}=(|f|^p)^{\frac{1}{q}}=(|g|^q)^{\frac{1}{q}}=|g|$ and $$\|f\|_{p}^{\frac{p}{q}}=\left(\int |f|^p \right)^{\frac{1}{q}}=\left(\int|g|^q \right)^{\frac{1}{q}}=\|g\|_q$$.

You may want to do similar argument for reverse direction.

-----------------------------Addition-----------------------------------

For the converse, you may want to use the inequality $$ab\leq \frac{1}{p}a^p+\frac{1}{q}b^q$$ is true. So let $f'=x$ and $g'=y$. Then we have $$k(x,y)=\frac{1}{p}x^p+\frac{1}{q}y^q-xy\geq 0$$

We know that the equality hold by our assumption. Now, take the derivative with respect to $x$, then we have $$ \frac{\partial k}{\partial x}=x^{p-1}-y $$ And set it equal to $0$ to find the condition make the inequality equality. Then you get $$x^{\frac{p}{q}}=x^{p-1}=y\implies x^p=y^q$$ which is desired result.