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 $y=\{y_k\}\in \ell_2$..

  • $\begingroup$ What is your question? $\endgroup$ Commented Dec 10, 2023 at 23:49

2 Answers 2


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].

  • $\begingroup$ 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? $\endgroup$
    – fwd
    Commented Dec 11, 2023 at 0:02
  • 1
    $\begingroup$ First choose $k$ such that $\sum\limits_{i=k+1}^{\infty}y_i^{2}<\epsilon$. Let $n \to \infty$ at the end. @fwd $\endgroup$ Commented Dec 11, 2023 at 0:11
  • $\begingroup$ 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 $k$. $\endgroup$
    – copper.hat
    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 $N\in \mathbb{N}$, $\lim_n \|P_Nb_n\|=0$.


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