# I have some problems, help: Let ( X , <mrow class="MJX-TeXAtom-ORD"> <mi

I have some problems, help:
Let $\left(X,{\mathcal{A}}_{\mathcal{1}},\nu \right)$ be a probability space. Let $A:{L}_{1}\to {L}_{1}$ be a contraction such that $A:{L}_{\mathrm{\infty }}\to {L}_{\mathrm{\infty }}$ is also a contraction. Suppose $A$ preserves positivity.
In a proof of why $A:{L}_{p}\to {L}_{p}$ is a contraction for every $1\le p<\mathrm{\infty }$, my lecture notes state that for all $f\ge 0$, we have
$‖Af{‖}_{p}=\underset{g\ge 0,‖g{‖}_{q}=1}{sup}{\int }_{X}\left(Af\cdot g\right),$
where $\frac{1}{p}+\frac{1}{q}=1$.
Why does the above equality hold? The strangest part to me is the fact that the LHS has a power of $\frac{1}{p}$ outside of the integral ${\int }_{X}|Af{|}^{p}$, but in the RHS the integral is not being raised to any power. I've tried to raise both sides to $p$ but I could not do much with this. If it helps, this came up in the study of conditional expectations.
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sniokd
We have by Holder's that $|{\int }_{X}\left(Af\cdot g\right)|\le ‖Af{‖}_{p}‖g{‖}_{q}=‖Af{‖}_{p}$ for all $g\in {L}^{q}$ with $‖g{‖}_{q}=1$. The question then becomes whether or not $g$ can be chosen to obtain equality.
Note that $1/p+1/q=1$ implies that $1+p/q=p$ and $1+q/p=q$. Hence, choose $g=\frac{\left(Af{\right)}^{p/q}}{‖Af{‖}_{p}^{p/q}}$, which is nonnegative because $f$ is and $A$ preserves nonnegativity. Moreover, $‖g{‖}_{q}=1$.
But then $Af\cdot g=\frac{\left(Af{\right)}^{1+p/q}}{‖Af{‖}_{p}^{p/q}}=\frac{\left(Af{\right)}^{p}}{‖Af{‖}_{p}^{p/q}}$. Integrating, shows that
${\int }_{X}Af\cdot g=\frac{‖Af{‖}_{p}^{p}}{‖Af{‖}_{p}^{p/q}}=‖Af{‖}_{p}.$