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Let H be a Hilbert space. Suppose $(i_k)_1^\infty$ is a complete orthonormal sequence in H. Let $a_k \in \mathbb{C}$ for $k \in \mathbb{N}$. Assume there is a bounded linear operator $T:H \rightarrow H$ such that $T(i_k) = a_ki_k$ given that ($a_k$), $\forall k \in \mathbb{N}$ is bounded. We want to find $||T||$.

We know that $||T|| = \sup\{||Tx||: x \in H, ||x|| \le 1\}$. For any given $x \in H$ we can write $x = \sum_{n=1}^\infty (x, i_n)i_n $, therefore $||x||^2 = \sum_{n=1}^\infty |(x,i_n)|^2$ and by Cauchy-Schwarz we have $|(x,i_n)| \le ||x||$. I suspect that $||T|| = sup_k |a_k|$ thinking about it intuitively but not sure how to show that. Since we know T is linear then: $Tx = \sum_{n=1}^\infty (x, i_n)Ti_n = \sum_{n=1}^\infty (x, i_n)a_ni_n.$ From this $||Tx||^2 = \sum_{n=1}^\infty |(x, i_n)|^2|a_ni_n|^2 \le (sup|a_n|)^2||x||^2$ by Bessel inequality, then $||T|| \le sup|a_n|.$ If this is correct, how could we show equality?

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    $\begingroup$ $\sum_{n=1}^\infty \|x\|^2$ isn't a helpful expression, it's infinity (you are adding the same number to itself infinitely many times). $\endgroup$ Aug 19 '13 at 4:22
  • $\begingroup$ Ye i just noticed that, let me edit it out. $\endgroup$
    – user77404
    Aug 19 '13 at 4:23
  • $\begingroup$ Your guess that $\|T\| = \sup |a_k|$ sounds good. Here's a hint to prove it: using the linearity and continuity of $T$ with the identity $x = \sum_{n=1}^\infty (x, i_n) i_n$, we have $Tx = \sum_{n=1}^\infty (x, i_n) T i_n$. What can you do with this? $\endgroup$ Aug 19 '13 at 4:24
  • $\begingroup$ @NateEldredge I edited your hint into my question, one problem I am having here is we don't know what exact norm is defined on this hilbert space. Right now we are trying to show $||Tx|| \le sup|a_k|$ $\endgroup$
    – user77404
    Aug 19 '13 at 4:32
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    $\begingroup$ To show equality, note that $\|T\| = \|T\|\|i_n\| \geq \|Ti_n\| = |a_n|$, and hence $\|T\| \geq \sup |a_n|$ $\endgroup$ Aug 19 '13 at 5:37
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As $\Arrowvert Ti_k \Arrowvert = |a_k|$, you have $\Arrowvert T \Arrowvert \geq |a_k|$ for all $k$.

You did prove $\Arrowvert T \Arrowvert \leq \sup_k|a_k|$

Thus we have equality.

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So we are trying to show $||T|| = \sup|a_k|$ and I have shown in the attempt of question above that $||T|| \ge \sup|a_k|$. To show that the other direction we use the fact that any linear operator $T:A \rightarrow A$, $||T|| \ge ||Tx||$ for all $x \in A$ such that $||x|| = 1$. So in our case chose $x = i_n$ then $||T|| \ge ||Ti_n|| = |a_n|$ for all $n \in \mathbb{N}$ so $||T|| \ge sup|a_n|$. Therefore $||T|| = \sup|a_k|$.

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  • $\begingroup$ You mean $\|T\| \geq \|Tx\|$ for all $x \in A$ such that $\|x\| = 1$ $\endgroup$ Aug 19 '13 at 12:30

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