This is an exercise I have encountered in my functional analysis class where we assume the operator may be unbounded and we assume nothing about the domain except that it is a vector subspace of $H$.
Let $H$ be a Hilbert space, and we define an operator $ S : D(S) \rightarrow H$ which is injective. We have the following additional statements about $S$
- $S$ is a closed operator.
- $S$ has dense range.
- $S$ has closed range.
- For some positive constant $C$ and all $\phi \in D(S)$, one has $\|S\phi\| \geq C \|\phi\|$.
We are asked to prove the following:
a. 1-3 imply 4 (as a hint, we are advised to use the closed graph theorem on $S^{-1}$).
b. 2-4 imply 1.
c. 1 and 4 imply 3.
I think I did manage parts a,b (but I am unsure of the reasoning), with the idea being that 2 and 3 together imply that the range of $S$ is all of $H$ so that $S$ is injective and surjective, and statement 4 means that $S^{-1}$ is bounded. The only thing I am stuck on is part c which I cannot figure out how to do, it has me stumped. The definition of closed operator here is the usual one where the graph is closed. Also, I think but I do not know if this is true, $S$ is closed iff $S^{-1}$ is closed for injective $S$ but I do not know if this is true (I used it on a and b, if this is incorrect, my work for a and b is flawed). For part c, we have a bijection $S : D(S) \rightarrow \text{Ran}(S)$ which we know to be closed, then its inverse $S^{-1}: \text{Ran}(S) \rightarrow D(S)$ should also be closed and its domain has to be closed, should it not? Why do we need 4? What am I doing wrong here? I mainly need help on C, but would appreciate help on A and B also. I thank all helpers who can assist.
C) is quite easy: Let $S\phi_n \to \psi$. We have to show that $\psi $ belongs to range of $S$. $\|S\phi_n-S\phi_m\| \to 0$ and hence $\|\phi_n- \phi_m\| \to 0$ By 4). Since $H$ is complete there exist $\phi$ such that $\phi_n \to \phi$. But $\phi_n \to \phi$ and $S(\phi_n) \to \psi$ implies that $\psi =S(\phi)$ which finishes the proof.