Particular property of $p$-norm with $1<p\le 2$

Question

Let $1<p\le 2$ and $q=p/(p-1)$. Let $u\in L_p(X,\mathcal{A},\mu).$ I must prove that $$\|u\|_p^q=\||u|^q\|_{p-1},$$ where $$\|u\|_p:=\left(\int_X |u|^p\,d\mu\right)^\frac{1}{p}.$$

Question. After several attempts I was unable to show this. Could someone show me the way??

Answer 1

The following should be obvious:$$ \|u\|_{L^p}^p = \| |u|^p \|_{L^1}.$$ Using this twice, $$ \|u\|_{L^p}^p = \| |u|^{q(p/q)} \|_{L^1}=\| |u|^q \|^{p/q}_{L^{p/q}}.$$ Raise both sides to the power $q/p$ and we get $$ \|u\|_{L^p}^q = \| |u|^q\|_{L^{p/q}} $$ Now we just use $$ p/q = p-1$$ to see that $$ \|u\|_{L^p}^q = \| |u|^q \|_{L^{p-1}}.$$ (BTW its obvious where $p\neq 1$ is needed, but I don't see where $p<2$ is needed. Indeed, if we instead impose $p\ge 2$ then we obtain actual norms on the right, since $p-1\ge 1$.)

Answer 2

Hint: note that: $$||u|^q|^{p-1} = |u|^{q(p-1)} = |u|^p$$