A non-empty, closed subset of $L^2([0,1])$ that does not contain a vector of smallest norm Would anyone be so kind to tell me if my answer to a problem is correct? Thank you! Happy new year!
The problem appeared on the UW-Madison Analysis Qual in January 2016.
Give an example of a non-empty closed subset of $L^2([0,1])$ that does not contain a vector of smallest norm.
I know an example that works. It's $\frac{n+1}{n}e_n$ where $e_n$ is an orthonormal basis for the $L^2([0,1])$ space. You can choose $e_n=\sqrt{2}\sin(n\pi x)$.
What I would like to know is the following:
My initial guess was $\{\sqrt{n+1}\chi_{[0,1/n]} : n\in \mathbb{N}\}$ where $\chi_{[0,1/n]}$ is the characteristic function of $[0,1/n]$. I thought this example worked because the $L^2$ norm of each term is $\sqrt{\frac{n+1}{n}}$, which is decreasing in $n$, so the smallest norm is never achieved. Does this example work? Is the set closed?
I have done some computation and realized that the distances between arbitrary terms does go to zero, which means we can't use the same argument as the orthonormal basis example.
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 A: Well, it is closed. Assume a sequence from $\{\sqrt{n+1}\chi_{[0,1/n]}\}$ were convergent in $L^2$ and denote the limit by $g$. The sequence must have a subsequence that converges to $g$ a.e. Suppose $g$ is not in the set $\{\sqrt{n+1}\chi_{[0,1/n]}\}$. Since $\sqrt{n+1}\chi_{[0,1/n]}$ converges to zero function a.e. as $n\to\infty$, any non-constant convergent sequence must converge to the zero function a.e. So $g = 0$ a.e. However, $||\sqrt{n+1}\chi_{[0,1/n]}||_2>1$ and $||g||_2=0$, so there can't be an $L^2$ convergence. Contradiction. $g$ is in the set and the set is closed.
A: Your answer works! What you can show is that with your set denoted as $\{y_n\}_{n \ge 1}$ is that any non-constant convergent sequence $\{x_n\}_{n \ge 1}$ that is a subset of your set must eventually satisfy for all $m$, there exists an $M$ so that $\{x_n\}_{n \ge M} \subset \{y_n\}_{n \ge m}$, and further that for all $y_0$ in the set and $\epsilon>0$, $\|y_n - y_0\| > \epsilon$ for all $n \ge m$ for some $m$. These can be combined to show that any subsequence of your set that is convergent must be eventually constant.
A: Your example does work. If $K\subset\Bbb N$ is an infinite  subset along which $\{f_n\}$ converges in $L^2$ to some $g\in L^2$, then there is a further subsequence along which the convergence is a.e, and this implies that $g=0$ a.e. But this is impossible because $\|g\|_2=0\not=1=\lim_{n\in K}\|f_n\|_2$. In short, the only convergent sequences to be extracted from your set $F:=\{f_n\}_{n\in\Bbb N}$ are eventually constant, so $F$ is closed.
