TITLE: Bounded linear operator on a Hilbert space
QUESTION [5 upvotes]: I am having a bit of difficulty with the following homework problem.

Let $\{x_n\}$ be an orthonormal basis in a Hilbert space $V$ over $\mathbb{C}$ and let $\{c_n\}_{n \in \mathbb{N}}$ be a fixed bounded sequence of complex numbers. Consider the bounded linear operator $T: V \to V$ defined by $T(x_n) = c_nx_n$.

There are numerous parts to the question, but below are the ones I am having trouble with


Find the adjoint operator $T^*$ and its norm $||T^*||$
If T is invertible, is its inverse continuous?
Show that any linear operator on a normed space is continuous if the unit sphere is compact.



I have managed to find $T^*$. As for the norm, I know that $||T^*|| = ||T||$. But is there an explicit  value for $||T||$ that can be found? I can't think of a way to find $||T||$ explicitly since we don't know what the norm on $V$ is.

I am not really sure how to do this one. Firstly, I know that a linear operator is continuous iff it is bounded, so I need to show that a linear operator $T: V \to V$ is bounded if the unit sphere $\{x \in V : ||x|| = 1\}$ is compact. I have been told to assume that $T$ is unbounded and try to get a contradiction. If T is unbounded then $||T|| = \sup_{||x|| = 1}\{||Tx||\} = \infty$. I don't know what to do from here.

REPLY [4 votes]: We do know the norm on $V$, because we know that $\{x_n\}$ is an orthonormal basis.  That means that each $v\in V$ can be written as $v=\sum_n a_nx_n$ with $a_n=\langle v,x_n\rangle$ and $\|v\|^2=\sum_n\|a_n\|^2$.  Using this fact, you should be able to find the norm of $\|T\|$ in terms of the sequence $\{c_n\}$.
This is typically false.  If $c_n=0$ for some $n$, the map is not injective.  If $0$ is in the closure of $\{c_n\}$, then the map is not surjective.  The sum you mention would converge if the sequence $\left\{\frac{1}{c_n}\right\}$ is bounded, so that would be a good condition to focus on.  You may also find it useful to note that a bijective bounded linear operator on a Hilbert space automatically has a bounded inverse.
You could combine the facts that “Every linear mapping on a finite dimensional space is continuous” and the Characterization of normed vector spaces of finite dimension in terms of compactness of the unit sphere.