The $L^p$ norm is the supremum of pairings against unit $L^q$ functions

For every $g\in L^q(\mu )$ with $\mathrm{\parallel }g{\mathrm{\parallel }}_q\le 1$, [L1] gives $\mid \int fgd\mu \mid \le \mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\parallel }g{\mathrm{\parallel }}_q\le \mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{.}$ Therefore the supremum is at most $\mathrm{\parallel }f{\mathrm{\parallel }}_p$.

If $f=0$, the displayed formula is immediate. Hence assume from now on $f\mathrm{\ne }0$. For a measurable representative $u$ of $f$, define its conjugate phase by ${\theta }_u(x):=\{\begin{array}{ll}\overline{u(x)}\mathrm{/}\mathrm{\mid }u(x)\mathrm{\mid },& u(x)\mathrm{\ne }0,\\ 0,& u(x)=0.\end{array}$ Then $\mathrm{\mid }{\theta }_u\mathrm{\mid }\le 1$ and $u{\theta }_u=\mathrm{\mid }u\mathrm{\mid }$ pointwise.

Assume $1<p<\mathrm{\infty }$ and choose a representative $u$ of $f$. Define $g(x):=\frac{\mathrm{\mid }u(x){\mathrm{\mid }}^{p-1}{\theta }_u(x)}{\mathrm{\parallel }f{\mathrm{\parallel }}_p^{p-1}}\mathrm{.}$ Since $(p-1)q=p$, $\mathrm{\mid }g{\mathrm{\mid }}^q=\frac{\mathrm{\mid }u{\mathrm{\mid }}^p}{\mathrm{\parallel }f{\mathrm{\parallel }}_p^p},$ so $\mathrm{\parallel }g{\mathrm{\parallel }}_q=1$. Also $u{\theta }_u=\mathrm{\mid }u\mathrm{\mid }$ gives $\int ugd\mu =\frac{1}{\mathrm{\parallel }f{\mathrm{\parallel }}_p^{p-1}}\int \mathrm{\mid }u{\mathrm{\mid }}^{p}d\mu =\mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{.}$ Hence the supremum is at least $\mathrm{\parallel }f{\mathrm{\parallel }}_p$.

Assume $p=1$ and choose a representative $u$ of $f$. Put $g:={\theta }_u$. Then $g\in L^{\mathrm{\infty }}(\mu )$ with $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, and $\int ugd\mu =\int \mathrm{\mid }u\mathrm{\mid }d\mu =\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{.}$ So the supremum is at least $\mathrm{\parallel }f{\mathrm{\parallel }}_1$.

Step 1.1 gives the upper bound, while step 2.1 or step 2.2 gives an $L^q$ function of norm at most one attaining it. Therefore the displayed supremum equals $\mathrm{\parallel }f{\mathrm{\parallel }}_p$.