Prove that $\frac{1}{n!}x^ne^{-x}$ has no weak* convergent subsequence.

Question

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!

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. — Stephen Montgomery-Smith, Sep 09 '20 at 04:21
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 $0$. — Stephen Montgomery-Smith, Sep 09 '20 at 04:23
@StephenMontgomery-Smith oh, I meant the Hahn Banach extension theorem. — Simplyorange, Sep 09 '20 at 04:26
@StephenMontgomery-Smith The thing is, by Hahn Banach extension theorem, there are nonzero linear functionals that takes compactly supported test functions to 0. Hence it is not enough to find $\phi\in L^\infty$ such that $L_n(\phi)$ does not converge to 0. — Simplyorange, Sep 09 '20 at 04:29
Yes, but those functionals are not in the predual of $L^\infty$. They are in the dual. — Stephen Montgomery-Smith, Sep 09 '20 at 04:30
@StephenMontgomery-Smith why doesn't this contradict the Banach-Alaoglu's theorem though? — Simplyorange, Sep 09 '20 at 05:08
Compactness is not necessarily the same as sequential compactness if the topology isn't metrizable. — Stephen Montgomery-Smith, Sep 09 '20 at 05:11
@StephenMontgomery-Smith But in this case, L1 is separable, hence the weak*topology on its dual should be metrizable? — Simplyorange, Sep 09 '20 at 05:15
$L_n \in (L^\infty)^*$. So $L^\infty$ is the pre-dual. I made a mistake in one of my comments. — Stephen Montgomery-Smith, Sep 09 '20 at 05:21
I should have said those functionals are not in the predual of $(L^\infty)^*$. — Stephen Montgomery-Smith, Sep 09 '20 at 05:22

score 2 · Accepted Answer · answered Sep 09 '20 at 06:12

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 $0$.

Prove that $\frac{1}{n!}x^ne^{-x}$ has no weak* convergent subsequence.

1 Answers1

Linked

Related