Spring 2020 - Problem 6

Solution.

1. Notice that $L_n\p{1}$ is just the integral of the Gamma function, so $L_n\p{1} = 1$ for all $n$. Let $\set{L_{n_k}}_k$ be a subsequence and $\epsilon > 0$. Let $r_1 = 0$ and $R_1 > r_1$ be so large that

   $$\frac{1}{n_1!} \int_{R_1}^\infty x^{n_1} e^{-x} \,\diff{x} < \epsilon.$$

   In general, let $r_k = R_{k-1}$ and notice that

   $$\frac{1}{n!} \int_0^{r_k} x^{n} e^{-x} \,\diff{x} \xrightarrow{n\to\infty} 0,$$

   so replace $n_k$ with an integer so large that

   $$\frac{1}{n_k!} \int_0^{r_k} x^{n_k} e^{-x} \,\diff{x} < \epsilon$$

   and pick $R_k > r_k$ large enough that

   $$\frac{1}{n_k!} \int_{R_k}^\infty x^{n_k} e^{-x} \,\diff{x} < \epsilon.$$

   These are picked so that

   $$\begin{gathered} 1 = \frac{1}{n_k!} \int_0^{r_k} x^{n_k} e^{-x} \,\diff{x} + \frac{1}{n_k!} \int_{r_k}^{R_k} x^{n_k} e^{-x} \,\diff{x} + \frac{1}{n_k!} \int_{R_k}^\infty x^{n_k} e^{-x} \,\diff{x} \\ \frac{1}{n_k!} \int_{r_k}^{R_k} x^{n_k} e^{-x} \,\diff{x} \geq 1 - 2\epsilon \end{gathered}$$

   for every $k \geq 1$. Thus, if we let

   $$f\p{x} = \sum_{k=1}^\infty \p{-1}^k \chi_{\p{r_k,R_k}},$$

   then we have $f \in L^\infty\p{\R}$. If $k$ is odd, then

   $$L_{n_k}\p{f} \leq -\p{1 - 2\epsilon}$$

   and if $k$ is even,

   $$L_{n_k}\p{f} \geq 1 - 2\epsilon.$$

   Hence, if $\epsilon$ is small enough, say, $\epsilon = \frac{1}{4}$, then $L_{n_k}\p{f}$ is essentially alternating, so this subsequence cannot converge weak-*.

2. (1) simply tells us that the unit ball in $\p{L^\infty\p{\R}}^*$ is not sequentially compact, which is not necessarily equivalent to compact when the weak-* topology is not metrizable.