Expansion coefficients of an orthonormal basis must satisfy $c_n n^{1/2}\to 0$ as $n\to \infty$ for $\sum|c_n|^2$ to converge.

Question

Expansion coefficients with respect to an orthonormal basis must satisfy $c_n n^{1/2}\to 0$ as $n\to \infty$ in order that $\sum|c_n|^2$ may converge. Is this true or false? Give a proof or counter-example.

Hint: Consider the sequence defined by $n^{-1/3}$ when $n$ is a perfect square, $c_n=0$ otherwise. I'm not sure how to use this hint, any hints or solutions are greatly appreciated.

Answer 1

Let $I = \{ 4^k \}_{k=0}^\infty$. Define $c_n = {1 \over \sqrt{n}}$ if $n \in I$, and $c_n = 0$ otherwise.

Then $\sqrt{n} c_n = 1$ infinitely often, hence $\sqrt{n}c_n$ does not converge to zero.

But $\sum_{n \in \mathbb{N}} |c_n|^2 = \sum_{n \in I} |c_n|^2 = \sum_{n \in I} {1 \over n} =\sum_k {1 \over 4}^k = {1 \over 1-{1 \over 4}}$.

Answer 2

In that case $c_n \sqrt{n} = n^{1/2 - 1/3} = n^{1/6} \to \infty $ however, $\sum n^{-2/3} = \sum k^{-4/3} < \infty $.

Here we sum over $n \in \{ 1,4,9,16,\dots\}= \{ k^2 : k \in \mathbb{N}\}$ or just $n=k^2$.

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I graphed the function $f(t) = \mathrm{Re}\left[\sum k^{-1/3}e^{2\pi i k^2 t}\right]$ on $[0,1]$. It is rather chaotic.