# Weak convergence of partial sums

I recently came across an interesting problem on weak convergence in $\ell^2 (\Bbb N)$.

Suppose that we have canonical basis $\{e_i\}$ in $\ell^2 (\Bbb N)$. We need to prove that the sequence $v_n=\frac{1}{\sqrt n}\sum_{i=1}^ne_i$ converges weakly to zero. In other words, we need to show that $$\lim_{N\to\infty}\sum_{i=1}^N\frac{x_i}{\sqrt N}=0$$ for any sequence of real numbers $x_i$ such that $\sum_{i=1}^\infty|x_i|^2<\infty$.

The approach from functional analysis yields that there exists a weakly convergent subsequence because $\|v_n\|_{\ell^2}=1$; it easy to show that any weakly convergent subsequence converges weakly to zero.

Now let's take $W$ - set of all weakly convergent subsequences of the sequence $\{v_n\}$ and examine all $S=\{v_{n_k}\}$ - set of elements of $\{v_n\}$ such that $\forall k$ $v_{n_k}$ does not belong to any subsequence in $W$. In other words,

$$S= \{v_n\}\setminus \bigcup_{\exists h\in W:\,v\in h} \{v\}$$ If $S$ is non-empty finite, then we can safely put all its elements into any weakly convergent subsequence, which implies a contradiction.

If $S$ is non-empty countable, then this is again a bounded sequence, hence it has a weakly convergent subsequence, thus some of elements of $S$ can't belong to $S$, which implies contradiction.

Therefore, $S=\emptyset$, which gives us that any $v_n$ belongs to some weakly convergent subsequence and all those subsequences weakly converge to zero. I'd like to prove now that this implies that the whole sequence weakly converges; I think that Zorn's lemma could help, but this problem was given to students who don't even know what Zorn's lemma is, so I'd like to avoid it.

I'd be grateful for hints on how to finalise the proof of this result.

Your arguments seem to be too complicated. You can use the convergence principle:

A sequence $(x_n)$ converges weakly to $x$, iff every subsequence contains a subsequence converging weakly to $x$.

This is the case for your example.

To prove the convergence principle: Assume every subsequence has a subsequence converging weakly to $x$. Assume $(x_n)$ does not converge to $x$ weakly.

Then there is $f\in X^*$ such that $f(x_n)$ does not converge to $f(x)$. Then there is an $\epsilon>0$ such that $|f(x_n)-f(x)|>\epsilon$ for infinitely many $n$, which gives a subsequence $(x_{n'})$ with $|f(x_{n'}-f(x)|>\epsilon$. By assumption, this subsequence has a subsequence $(x_{n''})$ converging weakly to $x$, implying $f(x_{n''})\to f(x)$. A contradiction. Hence $x_n$ converges weakly to $x$.

• Indeed, I've overcomplicated the proof. Thanks! – TZakrevskiy Jun 26 '14 at 13:26

A different solution:

Let $(x_k)$ be a sequence in $\displaystyle l^2(\mathbb{N})$. Put $\displaystyle S_n=\sum_{k=1}^n x_k^2$, with $S_0=0$. The sequence $S_n$ converge to a limit $S$.

We may suppose that $x_k\geq 0$ for all $k$; hence we have $x_k=\sqrt{S_k-S_{k-1}}$.

Put: $$T_n=\frac{1}{\sqrt{n}}\sum_{k=1}^n x_k=\frac{1}{\sqrt{n}}\sum_{k=1}^n \sqrt{S_k-S_{k-1}}$$

We have to show that $T_n$ has limit $0$.

Let $\varepsilon>0$. There exists a $N>0$ such that if $m\geq k\geq N$, we have $0\leq S_m-S_{k}\leq \varepsilon^2$.

Let $n>N$, we use the Cauchy-Schwarz inequality:

$$(\sum_{k=N+1}\sqrt{S_k-S_{k-1}})^2\leq (n-N)(\sum_{k=N+1}(S_k-S_{k-1}))=(n-N)(S_n-S_N)\leq n \varepsilon^2$$

Hence, for $n\geq N$:

$$\frac{1}{\sqrt{n}}\sum_{k=N+1}^n x_k=\frac{1}{\sqrt{n}}\sum_{k=N+1}^n \sqrt{S_k-S_{k-1}}\leq \varepsilon$$

$$T_n=\frac{1}{\sqrt{n}}(\sum_{k=1}^N x_k)+\frac{1}{\sqrt{n}}(\sum_{k=N+1}^n x_k)\leq \frac{M}{\sqrt{n}}+\varepsilon$$ with $\displaystyle M=\sum_{k=1}^N x_k$.

There exists a $H>N$ such that $\displaystyle \frac{M}{\sqrt{n}}<\varepsilon$ for $n\geq H$. Hence for $n\geq H$, we have $0\leq T_n\leq 2\varepsilon$, and we are done.

• It is interesting to see the proof from the point of of real analysis. – TZakrevskiy Jun 26 '14 at 14:11