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!
Best Answer
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 $0$.