The sequence $T_n(f)=\int_0^\infty \frac{x^n}{n!}e^{-x}f(x)dx$ in ${L^\infty}^$ is not weak-$$ seq. compact

Question

The following problem is from the spring $2020$ analysis qualifying exam given at UCLA:

Define $T_n\in L^{\infty}(\mathbb{R})^*$ as $$T_n(f)=\int_0^\infty \frac{x^n}{n!}e^{-x}f(x)dx$$

Prove that no subsequence of this sequence converges weak-∗.
Explain why this does not contradict the Banach–Alaoglu Theorem.

The second point is not too hard: $||T_n||=1$ and it does not contradict Banach-Alaoglu because Banach-Alaoglu gives compactness of the unit ball, not sequential compactness, in general. We have sequential compactness if the banach space is separable, and $L^\infty(\mathbb{R})$ is not so there's no contradiction.

However, I am stumped by the first point: my first idea was to try and find, given a sequence $T_{n_k}$, a function $f$ such that $T_{n_k}(f)$ does not converge (perhaps by some diagonalization process) but I was not able to construct such a function.

Any hint is appreciated

This could be helpful: https://math.stackexchange.com/questions/40920/weak-limit-of-an-l1-sequence — PhoemueX, May 02 '21 at 20:15
@PhoemueX Thanks! Actually, the question you linked is itself linked to another version of my question, https://math.stackexchange.com/questions/3819441/prove-that-frac1nxne-x-has-no-weak-convergent-subsequence?noredirect=1&lq=1, which I had not found before. — , May 03 '21 at 06:09

The sequence $T_n(f)=\int_0^\infty \frac{x^n}{n!}e^{-x}f(x)dx$ in ${L^\infty}^*$ is not weak-$*$ seq. compact

0 Answers0

The sequence $T_n(f)=\int_0^\infty \frac{x^n}{n!}e^{-x}f(x)dx$ in ${L^\infty}^$ is not weak-$$ seq. compact