# Variant of Parseval equality in Hilbert space

We know that if $$(x_n)_{n\in\mathbb{N}}$$ is a sequence of element in a Hilbert space $$(\mathcal{H}, \langle\cdot,\cdot\rangle)$$ such that for $$n\neq m$$, $$x_n\perp x_m$$. Then the series $$\sum_{n\in\mathbb{N}} x_n$$ converges if and only if the series $$\sum_{n\in\mathbb{N}}\Vert x_n\Vert^2$$ converges, and in that case, we have $$\Bigl\Vert \sum_{n\in\mathbb{N}}x_n\Bigr\Vert^2 = \sum_{n\in\mathbb{N}}\Vert x_n\Vert^2$$

Now, if we take another sequence $$(x_n)_{n\in\mathbb{N}}$$ in $$\mathcal{H}$$ such that there exists $$N_0\geqslant 1$$ such that if $$\vert n-m\vert\geqslant N_0$$, then $$x_n\perp x_m$$. We also suppose that $$\sum_{n\in\mathbb{N}}\Vert x_n\Vert^2$$ exists (i.e. is finite). Prove that the series $$\sum_{n\in\mathbb{N}}x_n$$ is convergent and that there exists a constant $$C$$ which only depends on $$N_0$$, such that $$\Bigl\Vert \sum_{n\in\mathbb{N}}x_n\Bigr\Vert^2 \leqslant C\sum_{n\in\mathbb{N}}\Vert x_n\Vert^2$$

My try: Like in the Parseval equality proof, I try to prove that $$S_q=\sum_{n\leqslant q}x_n$$ is a Cauchy sequence: \begin{align} \Vert S_{q+q'}-S_q\Vert^2 &=\sum_{n=q+1}^{q'}\Vert x_n\Vert^2+\sum_{n=q+1}^{q'}\sum_{j=q+1}^{q'}\langle x_n,x_j\rangle + \overline{\langle x_n,x_j\rangle}\\ &=\sum_{n=q+1}^{q'}\Vert x_n\Vert^2+\sum_{n=q+1}^{q'}\sum_{j\in I_n}\langle x_n,x_j\rangle + \overline{\langle x_n,x_j\rangle}\\ &\leqslant \sum_{n=q+1}^{+\infty}\Vert x_n\Vert^2+2\sum_{n=q+1}^{q'}\sum_{j\in I_n}\vert\langle x_n,x_j\rangle\vert\\ &\leqslant \sum_{n=q+1}^{+\infty}\Vert x_n\Vert^2+2\sum_{n=q+1}^{q'}\Vert x_n\Vert\sum_{j\in I_n}\Vert x_j\Vert \end{align} where $$I_n=\{j:\vert n-j\vert\leqslant N_0\}$$. I know that $$\vert I_n\vert\leqslant N_0$$, but I can only seem to have $$\sum_{j\in I_n}\Vert x_j\Vert\leqslant N_0\sup_{j\in I_n}\Vert x_j\Vert$$

I cannot seem to go further than that. I am able to bound $$\sup_{j\in I_n}\Vert x_j\Vert$$ by $$C_j\Vert x_q\Vert^2$$ if $$x_q\neq 0$$ with $$C_j\in\mathbb{R}$$ but this does not seem to be sharp enough, as the sequence $$(C_j)_{j\in\mathbb{N}}$$ is not easily bounded.

The same happens if I consider $$\Vert S_q\Vert$$ when trying to prove the inequality. Note that $$N_0=1$$ is the same case as in the Parseval equality.

Any help would be appreciated, thanks!

## 2 Answers

In fact, you have nearly worked out it. If we choose $$q=0$$ and $$m=q+q'=q'$$ in your step, We will have $$\Vert S_{m}\Vert^2 \leqslant \sum_{n=1}^{+\infty}\Vert x_n\Vert^2+2\sum_{n=1}^{m}\Vert x_n\Vert\sum_{j\in I_n}\Vert x_j\Vert$$ And we can estimate the second term: \begin{align} 2\sum_{n=1}^{m}\Vert x_n\Vert\sum_{j\in I_n}\Vert x_j\Vert&=\sum_{n=1}^{m}\sum_{j\in I_n}2\Vert x_j\Vert\Vert x_n\Vert\\&\le \sum_{n=1}^{m}\sum_{j\in I_n}(\Vert x_j\Vert^2+\Vert x_n\Vert^2)\\&=\sum_{n=1}^{m}\sum_{n-N_0\le j\le n+N_0}(\Vert x_j\Vert^2+\Vert x_n\Vert^2), \end{align} here $$x_j=0$$ if $$j\le 0$$. and we have \begin{align} &\sum_{n=1}^{m}\sum_{n-N_0\le j\le n+N_0}(\Vert x_j\Vert^2+\Vert x_n\Vert^2)\\ &\le \sum_{n=1}^{m}(2N_0 \Vert x_n\Vert^2+\sum_{n-N_0\le j\le n+N_0}\Vert x_j\Vert^2)\\ &=2N_0\sum_{n=1}^{m}\Vert x_n\Vert^2+\sum_{n=1}^{m}\sum_{n-N_0\le j\le n+N_0}\Vert x_j\Vert^2\\ &=2N_0\sum_{n=1}^{m}\Vert x_n\Vert^2+\sum_{n=1}^{m}\sum_{|n-j|\le N_0}\Vert x_j\Vert^2\\ &\le 2N_0\sum_{n=1}^{\infty}\Vert x_n\Vert^2+\sum_{n=1}^{\infty}\sum_{j=1}^{\infty}\Vert x_j\Vert^2 I_{\{|n-j|\le N_0\}}\\ &=2N_0\sum_{n=1}^{\infty}\Vert x_n\Vert^2+\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}\Vert x_j\Vert^2 I_{\{|n-j|\le N_0\}} \\ &\le 2N_0\sum_{n=1}^{\infty}\Vert x_n\Vert^2+\sum_{j=1}^{\infty}2N_0\Vert x_j\Vert^2 \\&\le 4N_0\sum_{n=1}^{\infty}||x_n||^2 \end{align} Where $$I_{\{|n-j|\le N_0\}}=1$$ if $$|n-j|\le N_0$$,and $$0$$ otherwise.So we have $$||S_m||^2\le (4N_0+1)\sum_{n=1}^{\infty}||x_n||^2$$ for all $$m>0$$. Hence $$C=4N_0+1$$ is the desired constant.

• Hi, Nice answer ! I forgot I could actually flip the two sums in this way, thank you! – Flewer47 May 8 '20 at 15:25

$$||\sum_{i=0}^\infty x_i||^2 = ||\sum_{r = 0}^{N_0 - 1} \sum_{k = 0}^\infty x_{N_0k+r}||^2 \leq \left(\sum_{r=0}^{N_0-1}||\sum_{k=0}^\infty x_{N_0k+r}||\right)^2 \leq N_0\sum_{r=0}^{N_0-1}||\sum_{k=0}^\infty x_{N_0k+r}||^2$$ $$= N_0 \sum_{i =0}^\infty ||x_i||^2$$

• Hi, Your last inequality seems obscure to me. Can you elaborate a bit so that anyone who finds this thread can understand ? Thanks ! – Flewer47 May 8 '20 at 15:24
• The inequality is $(a_1 + \cdots a_n)^2 \leq n(a_1^2 + \cdots + a_n^2)$ and it follows from applying the Cauchy-Swartz Inequality to the vectors $v = (1, \cdots, 1)$ and $u = (a_1, \cdots, a_n)$. – cha21 May 8 '20 at 15:32
• Thank you very much for your answer ! – Flewer47 May 8 '20 at 16:58