$L^p$ norm representation via duality

Question

I know that in a $\sigma$-finite measure space if $1\le p < \infty$ then $(L^p)'\cong L^{p'}$ isometrically via the isomorphism

$$L^{p'} \ni g\mapsto \int \cdot\, g\ \mathrm{d}\mu \in (L^p)'.$$

As a consequence one can write

$$\|g\|_{L^{p'}} = \sup_{\|f\|_{L^p}=1} \left| \int fg\ \mathrm{d}\mu \right|$$

for such $p$. While reading L. Grafakos, Classical Fourier Analysis (p. 2) I stumbled upon the statement that this norm represention also holds for $p = \infty$, i.e. $p' = 1$. How is this possible?

Answer 1

If $g$ is not zero almost everywhere, you can take $f(x)$ to be the signum of $g(x)$.

Answer 2

When $g \in L^{p'}$, the norm representation $$ \|g\|_{L^{p'}} = \sup_{\|f\|_{L^p} =1} \left| \int fg \,d\mu \right|$$ does not rely on the fact that the dual of $L^p$ is $L^{p'}$; it only relies on the fact that $L^{p'}$ is contained in the dual of $L^p$.

You've made a good point that the dual of $L^{\infty}$ is not $L^1$. However, $L^1$ is nonetheless contained in the dual of $L^{\infty}$, acting on elements of $L^{\infty}$ in the natural way, and hence for $g \in L^1$, the relevant norm representation still holds.