I am doing exercises about Holder and Minkowsvi inequalities and I have a doubt to an exercise that seems easy, but I am stucked. It uses Holder inequality in functions $\varphi,\psi$.
Choosing $\varphi=\psi^{p-1}/||\psi||_p^{p/q}$, show that $\displaystyle{||\psi||_p=\sup_{||\varphi||_{q}=1}\int |\varphi\psi|d\mu}$.
Here, $1/p+1/q=1$.
Thanks.
The supremum on RHS is over $\phi$ (satisfying $\|\phi\|_q=1$), not over $\psi$. With this change the proof goes as follows: RHS $\leq $ LHS is immediate from Holder's in equality. To prove the reverse inequality note that with the special choice of $\phi$ we have $\int |\phi|^{q} =\int|\psi|^{(p-1)q} /(\|\psi\|_p^{p})$. From the fact that $\frac 1 p +\frac 1 q=1$ verify that $(p-1)q=p$. It follows now that $\int |\phi|^{q}= \int|\psi|^{p} /(\|\psi\|_p^{p})=1$. Hence RHS $\geq \int |\phi \psi|$. I leave it to you to verify that $\int |\phi \psi|=\|\psi\|_p=$ LHS.