# Isometry between $L^\infty$ and $(L^1)^*$

I know that for $$p \in (1,\infty] \,$$, $$L^q(\Omega,\mathcal{M},\mu)$$ is isometric to $$(L^p(\Omega,\mathcal{M},\mu))^*$$ with $$\frac{1}{p}+\frac{1}{q}=1$$, namely that the operator: $$T:L^q \to (L^p)^*, \;\; T: g \mapsto L_{g}, \;\; L_{g}f = \int_{\Omega}{fg \:d\mu} \;\; \forall f\in L^p$$ is an isometry.

Indeed: $$||L_{g}||_{*} \leq ||g||_{L^q}$$ and we can find $$f_{0} \in L^p \,$$ s.t. $$|L_{g}f_{0}| = ||g||_{L^q}$$

So $$||Tg||_* = ||L_{g}||_{*} = ||g||_{L^q}$$ and we have an isometry.

Moreover, if $$p \in (1,\infty)$$, $$T$$ is surjective and thus it is an isomorphism. This is also the case for $$p=1$$ if $$\mu$$ is $$\sigma$$-finite.

I 'm trying to understand if $$L^\infty$$ is isometric to $$(L^1)^*$$ even if $$\mu$$ is not $$\sigma$$-finite.

Clearly: $$||L_{g}||_{*} \leq ||g||_{L^\infty}$$, I'm struggling in finding $$f_{0} \in L^1$$ s.t. $$|L_{g}f_{0}| \geq ||g||_{L^\infty}$$, and I'm questioning if such $$f_{0}$$ even exists.

• it can be shown that $\|fg\|_1=\|f\|_1\|g\|_\infty$ if and only if $|g(x)|=\|g\|_\infty$ a.e. when $f(x)\neq 0$, thus such $f_0$ doesn't exists (at least in the general case). However it can be shown also that $$\sup\left\{\left|\int fg\,\mathrm d \lambda \right|:f\in L^1 \,\land\, \|f\|_1\leqslant 1\right\}=\|g\|_{\infty }$$ holds when $\lambda$ is a $\sigma$-finite measure for any chosen $g\in L^{\infty }$, so it seems that it is an isometry. Dec 10, 2019 at 9:00

Let $$\Omega=\{a,b\}$$, with $$\mu(\{a\})=1$$, $$\mu(\{b\})=\infty$$. As every $$f:\Omega\to\mathbb C$$ is bounded, we have $$L^\infty(\Omega)\simeq\mathbb C^2$$. On the other hand, for $$f:\Omega\to\mathbb C$$ we have $$\int_\Omega |f|\,d\mu=\begin{cases} |f(a)|,&\ f(b)=0\\ \ \\ \infty,&\ f(b)\ne0\end{cases}$$ So $$L^1(\Omega)\simeq\mathbb C$$. Then the spaces are not isomorphic, and the map $$T$$ is not isometric (not even injective): for instance, $$T(0,1)=0$$, since $$T(0,1)f=f(a)\times 0+ f(b)\times 1=f(a)\times 0+0\times 1=0$$ for any $$f\in L^1(\Omega)$$.