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I am trying to show that the operator for Hilbert space is bounded. It is given that $\{e_n, n \in N\}$ be an orthonormal basis for Hilbert space .I have started the proof with for $x=\sum_{n\in N}(x|e_n)e_n$. We define $Tx= \sum_{n\in N}(1+\frac{1}{n})(x|e_n)e_n$ , this implies that $\| {Tx}\|^2= \sum_{n\in N}(1+\frac{1}{n})^2|(x|e_n)|^2$. Now I don't know how to proceed to show that this $T$ is bounded.

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$\| {Tx}\|^2= \sum_{n\in N}(1+\frac{1}{n})^2|(x|e_n)|^2 \le \sum_{n\in N}4|(x|e_n)|^2=4\sum_{n\in N}|(x|e_n)|^2=4||x||^2$.

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\begin{align*} \| \sum_{n = 1}^{\infty} (1 + 1/n) \left< x, e_n \right> e_n \|^2 & = \sum_{n = 1}^{\infty} (1 + 1/n)^2 |\left< x , e_n \right> |^2 \\ & \leq 2^2 \sum_{n = 1}^{\infty} |\left< x , e_n \right> |^2 \\ & = 4 \| x \|^2 \end{align*}

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  • $\begingroup$ We have $\| \sum_{n = 1}^{\infty} (1 + 1/n) \left< x, e_n \right> e_n \|^2 = \sum_{n = 1}^{\infty} (1 + 1/n)^2 |\left< x , e_n \right> |^2$ $\endgroup$
    – Fred
    Aug 15 '17 at 8:03
  • $\begingroup$ Right you are. Edited. $\endgroup$
    – AJY
    Aug 15 '17 at 16:17

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