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Let $A,B$ be complex Hilbert spaces with $\{a_n\}_{n=1}^\infty,\{b_n\}_{n=1}^\infty$ their respective orthonormal bases. Further let $\{k_n\}_{n=1}^\infty$ be a bounded complex sequence.
Define the operator $S:A\rightarrow B$ with $S(a)=\sum_{n=1}^\infty k_n(a,a_n)b_n$.

  1. How do I show that $\sum_{n=1}^\infty k_n(a,a_n)b_n$ converges in $B$?
  2. What is the adjoint $S^*$ of $S$?

What I thought:
1. From Bessel's inequality we know that $\sum_{n=1}^\infty |(a,a_n)|^2$ converges. Further I know that $$||\sum_{n=1}^\infty k_n(a,a_n)b_n||^2=\sum_{n=1}^\infty |k_n(a,a_n)|^2\leq\sup|k_n|\sum_{n=1}^\infty |(a,a_n)|^2.$$ Is this enough for convergence?
2. I thought that $S^*(b_n)=\overline{k_n}a_n$. Does that mean that $S(b)=\sum_{n=1}^\infty \overline{k_n}(b,b_n)a_n$?

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    Minor error with part $1:$ you should have $\sup |k_n|^2$. Other than tat it looks fine. – 2017-01-08
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    In 2, your analysis works because you know that $S$ is bounded. So $S^*$ exists and is bounded. – 2017-01-08
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    @Aweygan But $\sup|k_n|^2\sum|(a,a_n)|^2\in \mathbb{C}$, and not in $B$. Why does this give convergence in $B$? – 2017-01-09

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