Question about finding the norm of a bounded linear operator

Question

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?

Answer 1

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.

Answer 2

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 $||T|| = \sup|a_k|$.