# Densely defined linear operators on Hilbert space and their ranges

1. Aug 16, 2011

### AxiomOfChoice

Suppose $T$ is an injective linear operator densely defined on a Hilbert space $\mathcal H$. Does it follow that $\mathcal R(T)$ is dense in $\mathcal H$? It seems right, but I can't make the proof work...

There is a theorem that speaks to this issue in Kreyszig, and also in the notes provided by my professor; however, my professor's notes seem to indicate that the answer to the above question is "YES", whereas Kreyszig seems to indicate that it's "NO".

Oh...and if micromass reads this, then: Thank you, so much, for helping me out over the past several days.

2. Aug 16, 2011

### Citan Uzuki

Not at all, assuming that R(T) refers to the range of T. Let $H=\ell^2$, and consider the map $(x_1, x_2, x_3, \ldots) \mapsto (0, x_1, x_2, x_3, \ldots)$. This is everywhere defined, injective, and even continuous, but the range is not dense in H.

3. Aug 16, 2011

### micromass

Staff Emeritus
It's (by definition) true for invertible operators however (that is: densely defined operators with a densely defined inverse). Maybe your professor meant that??

4. Aug 16, 2011

### AxiomOfChoice

That is indeed possible. This all revolves around a theorem which intends to prove that $T^*$ is injective and $(T^*)^{-1} = (T^{-1})^*$.

I'm interested in your definition of "invertible operator", though...this is the first time I've heard that! Does that pertain to something specific (i.e., spectral theory of linear operators on Banach spaces)?

5. Aug 16, 2011

### AxiomOfChoice

Great. Thanks a lot.

6. Aug 17, 2011

### micromass

Staff Emeritus
Well, an operator T with dense domain D(T) is invertible if there exists an operator S with dense domain D(S) such that D(T)=R(S), D(S)=R(T) and STx=x for all x.

Since your theorem deals with invertible operators, I think that this is indeed the case here...