# Weakly convergent sequence

Consider a sequence $(x_n)_n$ in Hilbert space $H$ such that $\langle x_m,x_n\rangle=\delta_{mn}$ where $\delta_{mn}$ equals one if $m = n$ and $C$ otherwise. Prove that $(x_n)_n$ is a weakly convergent sequence.

• One could use Bessel's inequality. – David Mitra Dec 9 '12 at 14:04
• Bessel's inequality could be applied to an orthonormal sequence only. – Takasima Senko Dec 9 '12 at 14:40
• Oops, I misread that "$C$"... – David Mitra Dec 9 '12 at 14:44
• If the problem is correct, we necessarily have $C=0$, otherwise we have $\langle x_m,x_1\rangle =C$ for $m\geqslant 2$. – Davide Giraudo Dec 9 '12 at 14:56
• Could you explain this in detail? – Takasima Senko Dec 9 '12 at 15:03

## 2 Answers

Let $F:=\overline{\operatorname{Span}\{x_n,n\geqslant 1\}}$: it's a closed subspace of $H$, so we have $H=F\oplus F^\perp$.

We know, as $\{x_n\}$ is bounded, that it has a weakly converging subsequence. What we have to prove is that the limit doesn't depend on the choice of the subsequence.

Let $u_0$ a weak limit of $\{x_{k'}\}$. We can write $u_0=u'_0+u''_0$, where $u'_0\in F$ and $u''_0\in F^\perp$.

Testing the definition of weak convergence $\langle x_{k'},v\rangle\to \langle u,v\rangle$ with $v=u''$, we get that $u''=0$, and for each $j$, $\langle u',x_j\rangle=C$.

This determines $u'$, because if $u_1$ and $u_2$ are weak limits of a subsequence of $\{x_n\}$, using the preceding argument we get that $\langle u_1'-u_2',x_j\rangle=C-C=0$ for all $j$, so $\langle u'_1-u'_2,w\rangle=0$ for all $w$ in the linear span of the $x_k$, and by density for all $w\in F$, and $u''_1-u''_2=0$.

• I tried as your sketch but I have no result. Could you please give me some hints? – Takasima Senko Dec 9 '12 at 15:35
• If $C=0$ then we could use Bessel's inequality. But how can you prove that $C=0$? And I don't understand your last question. Would you please post your proof? – Takasima Senko Dec 9 '12 at 16:07
• You have written $\langle x_m,x_1\rangle =C$ for $m\geq 2$ recently but I think this is the assumptions of the problem because we have $\langle x_m,x_n\rangle =C$ when $m\neq n$. Please let me know if I misconstrued your idea. – Takasima Senko Dec 9 '12 at 16:25
• I understand your point: the convergence is not necessarily to $0$. I will think more about the problem (at least, the beginning show that we can restrict ourselves to the closure of the span of the $x_j$). – Davide Giraudo Dec 9 '12 at 16:29
• @DonDon I've rewritten the argument, now I hope it's correct. – Davide Giraudo Dec 9 '12 at 16:57

We have $y_n=\frac{1}{n}\sum\limits_{i=1}^{n}x_i$ converges to $y$ in $H$. Then $e_i=x_i-y$, $1\leq i\leq n$ is an orthonormal system. Hence, $(e_i)$, $1\leq i\leq n$ are linearly independent.

Let $F_n=\operatorname{Span}\{x_1,\ldots ,x_n\}$, it is a closed subspace of $H$. For all $x$ in $H$, $x=u_n+v_n$, $u_n\in F_n$, $v_n\in F_n^\perp$. Then $u_n=\sum\limits_{k=1}^{n}\lambda_ke_k$ and $\langle x,e_k\rangle=\lambda_k$. So $u_n=\sum\limits_{k=1}^{n}\langle x,e_k\rangle e_k$.

By Pythagorean Theorem, $||x||=||\sum\limits_{k=1}^{n}\langle x,e_k\rangle e_k+v_n||=\sum\limits_{k=1}^{n}|\langle x,e_k\rangle|^2+||v_n||^2$. Therefore, $\sum\limits_{k=1}^{n}|\langle x,e_k\rangle|^2\leq||x||$. In other words, the series $\sum\limits_{k=1}^{\infty}|\langle x,e_k\rangle|^2$ converges. So $\lim\limits_{n\to\infty}\langle x,e_k\rangle=0$. We can derive that $\lim\limits_{n\to\infty}\langle x,x_k\rangle=\lim\limits_{n\to\infty}\langle x,y\rangle$. Q.E.D