More precisely, let $H$ be a subspace(need not be close) of $L^2(X,\mu)$ where $\mu$ is probablity measure. Denote $[H]$ be closure of $H$ in $L^2(\mu)$. Assume that $H\cap L^{\infty}$ is dense in $H$ in $L^{2}$-norm. i.e. closure of following three sets $H\cap L^{\infty}$, $[H]\cap L^{\infty}$ and $H$ are equal to $[H]$.
Does the closure of two set with respect to $L^2$-norm equal ?
(1) $H \cap \{ f \in L^{\infty} : ||f||_{\infty} \leq 1 \}$
and
(2) $[H] \cap \{ f \in L^{\infty} : ||f||_{\infty} \leq 1 \}$
where $||.||_{\infty}$ denote the $L^{\infty}$ norm.
It seems to me this is not true. Take $H = C[0,1]$ as a subspace of $L_2[0,1]$. Then,
$H\cap L^{\infty} = H$ is dense in $H$ in all possible senses, $[H] = L_2[0,1]$.
Consequently,
$H \cap \{ f \in L^{\infty} : ||f||_{\infty} \leq 1 \}$ is the unit ball of $H = C[0,1]$,
but $[H] \cap \{ f \in L^{\infty} : ||f||_{\infty} \leq 1 \}$ is the unit ball of $L_\infty[0,1]$,
so they are not equal.
Dear Prof. Vladimir Kadets,
I guess your example does not work for the following reason.
By your hypothesis, take any $f\in L^{\infty}$ such that $ ||f||_{\infty} \leq 1 $, then there exists a subsequence $f_{n}\in C[0,1] $ such that $f_{n}$ goes to $f$ in $L^2$-norm. By choosing a subsequence one can assume that $ f_{n} $ goes to $ f $ a.e.
Now define,
g(z)= z, if $ |z| \leq 1 $
= 1/z if $ |z|>1 $.
Thus $ g $ is continuous and $ g_{n}:= g(f_{n}) $.
Hence $g_n$ are all continuous and $ || g_{n} ||_{\infty} \leq 1$. Moreover, $g_{n}$ goes to $ f $ in $L^2$- norm by Dominated convergence theorem.
Dear Debabrata,
You have right, I misunderstood the question. By unknown reason I read
instead "Does the closure of two set with respect to $L^2$-norm equal ?"
"Does the closure of two sets with respect to $L^\infty$-norm equal ?
Best, Vladimir
Hi all,
Does the following hold. Reason for asking the above question are described in the attached file. I am looking for the answer of the following question. Assume that the measure is a non atomic probability measure.
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