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We already known that $$ (L^p(\Omega))^* = L^q(\Omega), $$ for all $1\le p < \infty $ and $q$ is the exponent conjugate to $p$. So that, $L^p(\Omega)$ is reflexive with $1<p<\infty$. However, $L^1(\Omega)$ is not, due to $$ (L^\infty(\Omega))^* \supset L^1(\Omega), $$ but I don't know why. Can we find a element $f\in (L^\infty(\Omega))^*$ and $f \notin L^1(\Omega)$?


marked as duplicate by Norbert, user147263, Namaste, Dirk, Hakim Oct 24 '14 at 12:01

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Yes. Here are two examples.

1) Let $\Omega=\mathbb{N}$, so that $L^\infty(\Omega)=\ell^\infty$. Let $\mathcal{c}$ be the subspace of convergent sequences. The linear functional $f(\{x_n\})=\lim_{n\to\infty}x_n$ is bounded on $\mathcal{c}$. By the Hann-Banach theorem it can me extended to $\ell^\infty$. This extension is not represented by any $\ell^1$ sequence.

2) $\Omega\subset\mathbb{R}^n$ open. Fix $x_0\in\Omega$ and define on $C_c(\Omega)$, the set of continuous functions with compact support, the linear functional $T\colon C_c(\Omega\to\mathbb{R}$ as $T(f)=f(x_0)$. Clearly $|T(f)|\le\|f\|_\infty$. Again by the Hann-Banach theorem, $T$ can be extended to a bounded linear functional on $L^\infty(\Omega)$. It is not represented by any $L^1(\Omega)$ function.

You can find a description of the dual of $L^\infty$ here.

  • $\begingroup$ Dear Julián Aguirre, why the extension of $T$ on $L^\infty (\Omega)$ is not represented by any $L^1 (\Omega)$ function? Thanks! $\endgroup$ – Akai Shuichi Oct 24 '14 at 18:49
  • 1
    $\begingroup$ Even on continuous functions it is not represented by an $L^1$ function, but by a measure, Dirac's measure. $\endgroup$ – Julián Aguirre Oct 25 '14 at 16:33

I assume $\Omega\subset\mathbb{R}^n$ with the usual Lebesgue measure. Fix $x\in\Omega$ and take for example $f(g)=g(x)$ for $g\in L^\infty(\Omega)\cap C(\Omega)$. Then $|f(g)|\le \|g\|_\infty$. Extend $f$ to $L^\infty(\Omega)$ by Hahn-Banach, so that $f\in (L^\infty(\Omega))^*$, but of course $f\not\in L^1(\Omega)$ ($f$ is the Dirac measure at $x$).


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