Holder's inequality: $||f^*||_q =1 $.

Question

This is from Royden. I just don't understand the last statement in the proof. Why $||f^*||_q = 1$? Could you give some help?

Answer 1

Here $\infty >p>1$ so that conjugate exponent $q$ of $p$ is actually an element of $(1,\infty )$.

Note that $f\in L^p(X,\mu)$ so that $\int_X\ |f|^p\ d\mu\ <\infty$. But $f^*=||f||^{1-p}_p sgn(f) |f|^{p-1}$ so that $$\int_X\ |f^*|^q\ d\mu\ =\ \int_X ||f||^{(1-p)q}_p |sgn(f)|^q |f|^{(p-1)q}\ d\mu= ||f||^{(1-p)q}_p\ \int_X |f|^p\ d\mu =||f||_p^p||f||^{(1-p)q}_p=||f||_p^0=1$$Since $p+q-pq=0$.Therefore $||f^*||_q=(\int_X |f^*|^q\ d\mu)^{\frac{1}{q}}=1^{\frac{1}{q}}=1$

In order to define $||f||^{(1-p)q}$ I have excluded the case when $f=0\ a.e.\ \mu$.

Answer 2

For $p\gt1$, $(p-1)q=p$, $$ \begin{align} \int_E|f^\ast(x)|^q\,\mathrm{d}\mu &=\|f\|_p^{(1-p)q}\int_E|f(x)|^{(p-1)q}\,\mathrm{d}\mu\\ &=\|f\|_p^{-p}\int_E|f(x)|^p\,\mathrm{d}\mu\\[3pt] &=1 \end{align} $$