Fall 2009 - Problem 1

Solution.

Let $f_n = n\chi_{\br{0, \frac{1}{n^2} + \frac{1}{n^3}}}$ and consider $A = \set{f_n \mid n \geq 1} \subseteq L^2\p{\br{0, 1}}$.

First, notice that

so $A$ does not have an element with minimal normal. Furthermore, $A$ is closed. Indeed, suppose there is a sequence $\set{f_{n_j}} \subseteq A$ which converges to $f \in L^2\p{\br{0, 1}}$. Passing to a subsequence if necessary, we may assume that $f_{n_j}\p{x}$ converges to $f\p{x}$ almost everywhere.

If $J \coloneqq \set{n_j \mid j \geq 1}$ is bounded, then $f \in A$, since each $f_n$ is isolated. Otherwise, if $J$ is unbounded, there exists a subsequence $\set{n_{j_k}}$ which increases to $\infty$. Since $\set{f_{n_j}}$ converges pointwise almost everywhere to $f$, so must every subsequence, so

almost everywhere. But this would mean that

which is impossible. Hence, $J$ must have been bounded to begin with, i.e., every convergent sequence in $A$ converges in $A$, so $A$ is closed.