Is boundedness required in equivalence between $\frac1n\sum_{k=1}^na_k\to0$ and $\frac1n\sum_{k=1}^na_k^2\to0$?

Question

Suppose $a_n$ is a sequence of non-negative real numbers. If $a_n$ are un-bounded, then I want to know if $\dfrac{1}{n}\sum_{k=1}^na_k\to0$ as $n\to\infty$ is equivalent to $\dfrac{1}{n}\sum_{k=1}^na_k^2\to0$ as $n\to\infty$.

I believe it is not true, but cannot figure out an argument. Some sort of $a_n$ taking very large values after very long distances...

Answer 1

How about the sequence

$ a_n = \sqrt[3]{n}$, if $n$ is the cube of an integer and $0$, otherwise.

I think this shows that the claim is not true, as you suspect.

Proof

Whenever $n=m^3$ then $\frac{1}{n}\sum_{k=1}^n a_k = \frac{m(m+1)}{2m^3}$. These are the maximum values of the sequence on the range $[m^3, (m+1)^3-1]$ so this tells us the $\limsup_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^n a_k \rightarrow 0$, and so the $\lim \rightarrow 0$

On the other hand, when $n=m^3$ then $\frac{1}{n}\sum_{k=1}^n a_k^2 = \frac{m(m+1)(2m+1)}{6m^3}$. Hence the sequence $\frac{1}{n}\sum_{k=1}^n a_k^2$ has a subsequence which tends to $\frac{1}{3}$ and thus, cannot tend to $0$.

Thanks to Bungo and DrMV for the useful discussion.

Answer 2

We've seen it's not hard to construct explicit counterexamples. But it's at least a little fussy, things have to add up just right. Just for fun we might note that it's easy to use a little functional analysis to avoid that fussiness.

Say $X$ is the space of all sequences $a=(a_1,\dots)$ such that $$\lim_{n\to\infty}\frac1n\sum_{j=1}^n|a_j|=0,$$with norm $$||a||_X=\sup_n\frac1n\sum_{j=1}^n|a_j|.$$ Then $X$ is a Banach space. (Hint: $|a_n|\le n||a||_X$, so a Cauchy sequence in $X$ converges pointwise.)

Similarly let $Y$ be the space of sequences such that $$\lim_{n\to\infty}\frac1n\sum_{j=1}^n|a_j|^2=0,$$with norm $$||a||_Y=\sup_n\left(\frac1n\sum_{j=1}^n|a_j|^2\right)^{1/2}.$$Your question is equivalent to asking whether $X\subset Y$. But if $X\subset Y$ the Closed Graph Theorem shows that $$||a||_Y\le c||a||_X$$(hint: again, $|a_n|\le n||a||_X$, also $|a_n|\le n^{1/2}||a||_Y$.) Simply calculating $|||e_n||_X$ and $||e_n||_Y$ shows that this is not so.

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Ok, I'll admit that writing down the details of why $X$ and $Y$ are complete would maybe take some space. But the point is it's straightforward; there's no point where the reader wonders how you came up with that step. As opposed to letting $a_n=n^{1/3}$ if $n$ is a perfect cube and $0$ otherwise...