I have some problems, help:

Let $(X,{\mathcal{A}}_{\mathcal{1}},\nu )$ 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

$\Vert Af{\Vert}_{p}=\underset{g\ge 0,\Vert g{\Vert}_{q}=1}{sup}{\int}_{X}(Af\cdot g),$

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.

Let $(X,{\mathcal{A}}_{\mathcal{1}},\nu )$ 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

$\Vert Af{\Vert}_{p}=\underset{g\ge 0,\Vert g{\Vert}_{q}=1}{sup}{\int}_{X}(Af\cdot g),$

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.