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

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!

• Did $\phi$ become $f$ in your definition of $L_n$? Sep 9 '20 at 4:20
• 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. Sep 9 '20 at 4:21
• 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 $0$. Sep 9 '20 at 4:23
• @Reveillark sorry, it's corrected now. Sep 9 '20 at 4:25
• Yes, but those functionals are not in the predual of $L^\infty$. They are in the dual. Sep 9 '20 at 4:30

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 $$0$$.