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I am super desperate with the following exercise:

Let $H$ be a seperable Hilbert space with othogononal basis $ {e_n} $ and let ${y_n}$ be a sequence in H such that $\sum ||y_n||² < \infty$.

1) Show that there exists a unique continuous linear map $T: H \rightarrow H$ such that $Te_n = y_n$ for all $n\ge 1$.

2) Give an upper bound for the operator norm $||T||$ in terms of the sequence $y_n$.

I am very thankful for help.

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An element $x\in H$ can be written $x=lim\sum_{i=1}^{i=n}x_ie_i$. Consider $u_x(n)=\sum_{i=1}^{i=n}x_iy_i$. You have $\|u_x(n+m)-u_x(n)\|\leq \sum_{i=n+1}^{i=n+m}|x_i|\|y_i\|\leq \sum_{i=n+1}^{i=n+m}(\sum_{i=1}^{I=n}|x_i|^2)(\sum_{i=1}^{i=n}\|y_i|^2)$ By Cauchy-Swcharz in $C^{m}$. Since $\sum_{i\geq 1}\|y_i\|^2$ is finite, for every $c>0$ There exists $N$ such that for $n>N$ and any integer $m$, $\sum_{i=1}^{i=n+m}\|y_i|^2<c$. This implies that $u_x(n)$ is a Cauchy sequence since $H$ is complete, it converges. Set $T(x)=lim_nu_x(n)$.

You have $\|T(x)\|=\|\sum x_iy_i\leq \|x\|\|y\|$ by using the argument above.

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  • $\begingroup$ Why does this show that $u_x(n) $ is a Cauchy sequence? $\endgroup$
    – Yuhe
    Commented Nov 3, 2016 at 14:58

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