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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Masacroso
    Dec 10, 2019 at 9:00

1 Answer 1


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 $f\in L^1(\Omega)$.


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