Counter Example About $L^{\infty}$ function

dual-spaceslp-spacesreal-analysisriesz-representation-theorem

Let $(\Omega,\Sigma,\mu)$ be a measure space.

Let $1<p<\infty$ and $q={p\over p-1}.$

Let $f\in L^p(\Omega,\Sigma,\mu).$

Then by the Riesz Representation Theorem, we know that, $$||f||_p=\sup_{||g||_q\le1} |\int fg\,d\mu|. $$

My text says that, this equality does not always hold if $p=\infty.$ Intuitively I think this is true, because the dual space of $L^\infty(\Omega,\Sigma,\mu)$ is not $L^1(\Omega,\Sigma,\mu).$ But I cannot come up with a concrete counter example.

My question is, can any one help me with coming up such a counter example? Thanks.

Best Answer

I think the claim is true if the measure space has the following property: every $A\in \Sigma$ with $\mu(A)>0$ contains a subset of finite, positive measure. Define $A:=\{x:\ |f| \ge \|f\|_\infty-\epsilon\}$. Take $B\subset A$ with $0<\mu(B)<+\infty$, define $g:=\mu(B)^{-1}\chi_B\cdot sign(f)$.

Hence, a counter-example has to use some exotic measure space: Take the measure space $\Omega =\{0\}$, $\mu(\Omega)=+\infty$. Then $L^1(\mu)$ contains only the null function. So this supremum is zero. $L^\infty(\mu)$ contains all functions from $\Omega$ to $\mathbb R$.

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