# How fast can the sum of a square-summable sequence grow?

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.

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

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 $$N\geq 1$$.
• 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. Feb 4, 2020 at 18:59