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I'm trying to solve this previous qual problem from my univeristy:

Let $L_n$ be the continuous linear functions on $L^\infty(\mathbb{R})$ given by

$$L_n(\phi)=\frac{1}{n!}\int_0^\infty x^ne^{-x}\phi(x)\,dx.$$ Prove that $L_n$ has no subsequence that converges in the weak* topology of $L^\infty(\mathbb{R})^\ast$.

So far, I managed to see that $\frac{1}{n!}x^ne^{-x}$ converges uniformly on compact sets to 0, hence the weak* limit of any subsequence of $L_n$ must take compactly supported $L^\infty$ functions to 0.

I've also tried to construct a test function directly for any subsequence but so far I have yet to come up with such a construction.

Can anyone pleas help me? Some hints would be very much appreciated! Thanks!

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  • $\begingroup$ Did $\phi$ become $f$ in your definition of $L_n$? $\endgroup$
    – Reveillark
    Sep 9 '20 at 4:20
  • $\begingroup$ First, I think you meant to have $\phi$ instead of $f$ inside the integral. Secondly, you can't use the Hahn-Banach theorem to separate in the manner you desire, because the Hahn-Banach theorem refers to the dual, not the pre-dual. $\endgroup$ Sep 9 '20 at 4:21
  • $\begingroup$ So if it converges, it has to converge to some function in $L^1$. You are close to showing that if this function exists, it must be $0$. Now try to find a function $\phi \in L^\infty$ such that $L_n(\phi)$ does not converge to $0$. $\endgroup$ Sep 9 '20 at 4:23
  • $\begingroup$ @Reveillark sorry, it's corrected now. $\endgroup$ Sep 9 '20 at 4:25
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    $\begingroup$ Yes, but those functionals are not in the predual of $L^\infty$. They are in the dual. $\endgroup$ Sep 9 '20 at 4:30
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A serious issue here is that there is a weak-* convergent sub-net, by Banach-Alaoglu. So we cannot make a topological argument, and must instead make a sequential one.

This now follows from the fact that $L^1(\mathbb R_{\ge 0})$ is weakly sequentially complete. See the attached StackExchange link: Weak limit of an $L^1$ sequence. If a subsequence $L_{n_k}$ converges $w^*$, then the sequence $\frac{1}{n_k!} x^{n_k}e^{-x}$ is weakly Cauchy. By weak sequential completeness, this means that the limit is actually represented by an $L^1$ function. In $L^1$, if a sequence converges, then a subsequence converges pointwise almost-everywhere. But the pointwise limit is always zero; as $n_k$ becomes large, $n_k!\gg x^{n_k}$. So the limiting $L^1$ function would have to be zero, but it is clearly not, because the $L_{n_k}(1)$ are all $1$, whereas integrating the $0$ function against the constant $1$ function yields $0$.

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