# A non-empty, closed subset of $L^2([0,1])$ that does not contain a vector of smallest norm

hilbert-spaceslp-spaces

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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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.