# Does $\sum\limits_{k=1}^{\infty} (a_{k+1} - a_k)^2 <\infty$ imply $\frac{1}{n^2} \sum\limits_{k=1}^n a_k^2 \to 0$?

I'm currently working on some problem regarding Dirichlet forms and, as the title states, I'm trying to figure out if $$\sum\limits_{k=1}^{\infty} (a_{k+1} - a_k)^2 <\infty \Rightarrow \frac{1}{n^2} \sum\limits_{k=1}^n a_k^2 \to 0$$ holds. I've already tried it with some examples (mostly 'mainstream' sequences) and it seems to always work. However, I don't seem to be able to proof it and thus I tend to think the statement is false...

We have $a_n = a_1 + \sum\limits_{k=1}^{n-1} a_{k+1} - a_k$. Using $(a+b)^2 \leq 2(a^2+b^2)$ and the Cauchy-Schwarz inequality we get

$$a_n^2 = \left( a_1 + \sum\limits_{k=1}^{n-1} (a_{k+1} - a_k) \right)^2 \leq 2 a_1^2 + 2 \left( \sum\limits_{k=1}^{n-1} (a_{k+1}-a_k) \right)^2$$ $$\leq 2a_1^2 + 2 (n-1) \cdot \sum\limits_{k=1}^n (a_{k+1}-a_k)^2 \leq 2a^2_1 + C \cdot (n-1)$$

But this only implies that $\frac{1}{n^2} \sum\limits_{k=1}^n a_k^2$ is bounded. Any hints/ideas or a counterexample would be much appreciated.

Here is a sledgehammer solution: Using the Hardy's inequality, we get

$$\sum_{n=1}^{\infty} \frac{a_{n}^{2}}{n^{2}} < \infty.$$

Now you can apply the Kronecker's lemma to obtain the desired conclusion.

This at least shows that your guess is correct. I believe that there is a much simpler solution, and I will update my solution when I find it.

• Thanks for the timely answer. It indeed feels like one of those things were one thinks "there has to be an easy way" :) – GenericNickname Oct 24 '14 at 16:47

Here is a solution using Cauchy's inequality and Stolz–Cesàro theorem.

By Stolz–Cesàro theorem, we need to prove

$$\lim\dfrac{1}{n^2}\sum_{k=1}^n a_k^2 = \lim \dfrac{a^2_{n+1}}{2n+1} =0$$ provided the second limit exists.

Let $b_k = a_{k+1}- a_k$ with $b_0 = a_1$, we have $\sum_k b_k^2 < \infty$ and $a_{n+1} = \sum_{k=0}^{n}b_k$.

\begin{align} a_{n+1}^2 = (\sum_{k=0}^{n}b_k)^2 \leq 2(\sum_{k=0}^{N}b_k)^2 + 2(\sum_{k=N+1}^{n}b_k)^2 \leq 2(\sum_{k=0}^{N}b_k)^2 + 2(n-N)(\sum_{k=N+1}^{n}b_k^2) \end{align}

Thus we have \begin{align} \limsup_n\dfrac{a_{n+1}^2}{n} &\leq \limsup_n \dfrac{2(\sum_{k=0}^{N}b_k)^2 + 2(n-N)(\sum_{k=N+1}^{n}b_k^2)}{n}\\ &\leq 2(\sum_{k=N+1}^{\infty}b_k^2) \end{align}

Since it's true for all $N$ and $\sum_{N}^\infty b_k^2 \to 0$ as $N\to \infty$, we have the conclusion.

• Indeed we have a simple solution! (+1) – Sangchul Lee Oct 24 '14 at 19:18