# Show that $a_n \to 0$ in weak topology

Let $$\ell_2$$ be the Hilbert space of square-summable sequences and consider $$a_n=\frac{1}{\sqrt{n}}(1,1,\dots, 1, 0,0,\dots)$$ where the first $$n$$ coordinates are one. I try to show that $$a_n \to 0$$ in weak topology.

I can check that if $$a_n=\{a_n(k)\}_{k\ge 1}$$ converges weakly to some $$a=\{a(k)\}$$, then $$a=0$$ as follows. Consider the linear functional $$f_n\in \ell_2^*=\ell_2$$ by $$f_n(x_1,x_2,\dots)=x_n$$. Then for $$m>n$$, $$f_n(a_m)=\frac{1}{\sqrt{m}}\to 0=f_n(a)=a(n)$$

So we have $$a=(0,0,\dots)$$. But we need to show that holds for any linear functional $$f\in \ell_2^*$$. I know that we can represent linear functional by for $$x\in \ell_2$$ $$f(x)=\sum_{k} x_k f(e_k)=\sum_k x_k y_k$$ where $$y=\{y_k\}\in \ell_2$$..

• What is your question? Commented Dec 10, 2023 at 23:49

Let $$(y_k) \in \ell^{2}$$. You have to show that $$\frac {y_1+y_2+...+y_n} {\sqrt n} \to 0$$ as $$n \to \infty$$. Fix $$k$$ and note that $$\frac {y_1+y_2+...+y_k} {\sqrt n} \to 0$$ as $$n \to \infty$$. Now use the fact that $$|y_{k+1}+y_{k+2}+...+y_n| \leq (\sum\limits_{i=k+1}^{\infty}y_i^{2})^{1/2} \sqrt {n-k}$$ and $$\sum\limits_{i=k+1}^{\infty}y_i^{2} \to 0$$ as $$k \to \infty$$.

[First choose $$k$$ such that $$\sum\limits_{i=k+1}^{\infty}y_i^{2}<\epsilon$$. Let $$n \to \infty$$ at the end].

• What is the order in which you send $n, k$ to infinity? It seems if you send $n$ to infinity first, the upper bound blows up. You can't send $k$ to infinity first if $n$ remains fixed, since $k < n$. Can you clarify please?
– fwd
Commented Dec 11, 2023 at 0:02
• First choose $k$ such that $\sum\limits_{i=k+1}^{\infty}y_i^{2}<\epsilon$. Let $n \to \infty$ at the end. @fwd Commented Dec 11, 2023 at 0:11
• The above written marginally differently: Write $y$ as $y=y^k + y^{\bar{k}}$, where $y^k$ is the first $k$ elements of $y$. Then $\langle y^k, a_n \rangle \to 0$, $y^{\bar{k}} \to 0$ strongly, and $|\langle y^{\bar{k}}, a_n \rangle | \le \|y^{\bar{k}}\|$. Hence $\limsup_n |\langle y, a_n \rangle | \le \|y^{\bar{k}}\|$ for all $k$. Commented Dec 11, 2023 at 0:23

We have the Cauchy-Schwarz inequality, which says that $$|\langle x,y\rangle|\leqslant \|x\|\|y\|$$.

For $$x=\sum_{k=1}^\infty x_ke_k$$ and $$y=\sum_{k=1}^\infty y_ke_k$$, we let $$\langle x,y\rangle = \sum_{k=1}^\infty \overline{x}_ky_k,$$ where $$\overline{x}_k$$ is the complex conjugate of $$x_k$$.

We also note that for $$y=\sum_{k=1}^\infty y_ke_k$$, $$\lim_N \bigl(\sum_{k=N+1}^\infty |y_k|^2\bigr)^{1/2}=0$$.

For $$N\in\mathbb{N}$$, let $$P_N\sum_{k=1}^\infty x_ke_k=\sum_{k=1}^N x_ke_k$$ and let $$Q_N\sum_{k=1}^\infty x_ke_k=\sum_{k=N+1}^\infty x_ke_k$$. Note that for any $$x,y\in \ell_2$$ and $$N\in \mathbb{N}$$, $$\langle x,y\rangle = \langle P_Nx,P_Ny\rangle+\langle Q_Nx,Q_Ny\rangle.$$ Note also that $$\|P_Nx\|^2+\|Q_Nx\|^2=\|x\|^2$$ for all $$x\in\ell_2$$, so $$\|P_Nx\|\leqslant \|x\|$$ and $$\|Q_Nx\|\leqslant \|x\|$$.

Fix $$x=\sum_{k=1}^\infty x_ke_k$$ and let $$a_n=\sum_{k=1}^n \frac{1}{\sqrt{n}}e_k$$. Fix $$\varepsilon>0$$ and choose $$N\in\mathbb{N}$$ so large that $$\|Q_Nx\|<\varepsilon/2$$. Note that for $$n>4N/\varepsilon^2$$, \begin{align*} |\langle a_n,x\rangle| & = |\langle P_Na_n,P_Nx\rangle+\langle Q_Na_n,Q_Nx\rangle| \leqslant |\langle P_Na_n,P_Nx\rangle|+|\langle Q_Na_n,x\rangle| \\ & = |\langle P_Na_n,x\rangle| +\Bigl|\sum_{k=N+1}^\infty \frac{x_k}{\sqrt{n}}\Bigr| \\ & = |\langle P_Na_n,x\rangle| + |\langle a_n,Q_Nx\rangle| \leqslant \|P_Na_n\|\|x\|+\|a_n\|\|Q_Nx\|<\varepsilon/2\cdot 1 + 1\cdot \varepsilon2/=\varepsilon.\end{align*} Since $$\varepsilon>0$$ was arbitrary, $$\lim_n \langle a_n,x\rangle=0$$. Since $$x\in \ell_2$$ was arbitrary, we are done.

Note that this argument can essentially be used to show that for any bounded sequence $$(b_n)_{n=1}^\infty\subset \ell_2$$ is weakly null if and only if for all $$N\in \mathbb{N}$$, $$\lim_n \|P_Nb_n\|=0$$.