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Suppose $x_t$ is a nonnegative sequence satisfying $$ \sum_{t=1}^{+\infty} x_t^2 < \infty.$$ I am trying to get a precise estimate for how fast $\sum_{t=1}^T x_t$ can grow as a function of $T$. Application of Cauchy-Schwarz gives that $$\sum_{t=1}^T x_t \leq \sqrt{T} \sqrt{\sum_{t=1}^{+\infty} x_t^2},$$ so $O(\sqrt{T})$ is one upper bound. My question is whether in fact $$ \lim_{T \rightarrow +\infty} \frac{1}{\sqrt{T}} \sum_{t=1}^T x_t = 0.$$

Here is why one might hope that such a thing is true. First, Cauchy-Schwarz is tight when the two vectors are multiples of each other, and since $x_t \rightarrow 0$, the vector $(x_1, \ldots, x_T)$ is very far from being a multiple of $(1,...,1)$. Second, if we try to come up with a tight example, the natural guess might be $x_t = 1/(\sqrt{t} \log^c(t))$ for some $c>0$, since its square is close to being the slowest decaying summable sequence. But in that case $\sum_{t=1}^T x_t = O(\sqrt{T}/\log(T))$, and the limit is indeed zero.

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  • $\begingroup$ I think you need $c\gt\frac12$? $\endgroup$
    – joriki
    Feb 4, 2020 at 19:01

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Notice that the limit is trivially zero, for any $\mathscr{l}^{1}$-sequence. W.log we may assume that the sequence $\left\{a_{n}\right\}_{n\geq1}$ is positive. Notice that for any $\mathscr{l}^{1}$-sequence $\left\{b_{n}\right\}$, we have that $$ \frac{1}{\sqrt{N}}\sum_{n=1}^{N}a_{n} = \frac{1}{\sqrt{N}}\sum_{n=1}^{N}(a_{n}-b_{n}) + \frac{1}{\sqrt{N}}\sum_{n=1}^{N}b_{n} \leq \left( \sum_{n=1}^{\infty}(a_{n}-b_{n})^{2} \right)^{1/2} + \frac{1}{\sqrt{N}} \sum_{n=1}^{N}b_{n} $$ Now letting $N\rightarrow \infty$, we get that $$ \limsup_{N\rightarrow \infty} \frac{1}{\sqrt{N}}\sum_{n=1}^{N}a_{n} \leq \left(\sum_{n=1}^{\infty}(a_{n}-b_{n})^{2} \right)^{1/2}. $$ Choosing $\left\{b_{n}\right\}$ to be an approximate of $\left\{a_{n}\right\}$ in $\mathscr{l}^{2}$-norm (This is possible since $\mathscr{l}^{1}$ forms a dense subspace of $\mathscr{l}^{2}$), proves the claim.

To prove that this is sharp, suppose there exists another function $\phi:\mathbb{N} \rightarrow (0,\infty)$, with the property $$\lim_{N\rightarrow \infty} \frac{1}{\phi(N)}\sum_{n=1}^{N}a_{n} \rightarrow 0 \qquad, \, \forall \left\{a_{n}\right\}_{n\geq 1} \in \mathscr{l}^{2}. $$ This means precisely that the family of bounded linear functionals $\frac{1}{\phi(N)}L_{N}$, with $$L_{N}(\left\{a_{n}\right\}_{n\geq 1} )= \sum_{n=1}^{N}a_{n}$$ converge to $0$ in the weak-star topology of $\mathscr{l}^{2}$. By the principle of uniform boundedness, it then follows that the family $\frac{1}{\phi(N)}L_{N}$ is uniformly bounded in the dual-norm, i.e there exists a constant $C>0$, independent of $N\geq 1$, such that $$ \lvert \lvert L_{N} \rvert \rvert \leq C\, \phi(N), \qquad , \, \forall N\geq 1. $$ It is straightforward to prove that the dual-norm of $L_{N}$ is equal to $\sqrt{N}$, hence by the above we conclude that $\sqrt{N} \leq C \phi(N)$, for all $N\geq 1$.

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  • $\begingroup$ Do you mean $\ell^1$? Analysis isn't my thing so I'm scared to edit your answer directly, but you might want to use \ell for the cursive l if that's what you want. $\endgroup$ Feb 4, 2020 at 18:59
  • $\begingroup$ @HallaSurvivor: Interesting – they look the same in my browser. $\endgroup$
    – joriki
    Feb 4, 2020 at 19:01
  • $\begingroup$ Hmm... I'm on the app, so maybe mathjax is a bit busted on my end... Sorry for the false alarm ^_^ $\endgroup$ Feb 4, 2020 at 19:01

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