# Proving a variant of closed range theorem on Hilbert space

I've been working on closed range theorem. There are a lot of materials on general Banach spaces, but not much on Hilbert spaces, so I was wondering if I could get some help. I'm trying to prove the following claim:

If a bounded linear map $T:X\to Y$ between Hilbert spaces $X$ and $Y$ has closed range if and only if there exists a constant $C>0$ so that $\|f\| \leq C\|T^*f\|$

This statement seems like the statement is a usual closed range theorem, but a bit different, especially with adjoint of the operator. Can someone help me proving this claim? Thanks!

The "ïf" direction is correct, because then you get that ($$T^*$$ has a closed range and) $$T$$ is surjective (see Theorem 2.20 from Brezis' Functional analysis book). To make it correct in both directions you need to state it as:

A bounded linear map $$T:X\to Y$$ between Hilbert spaces $$X$$ and $$Y$$ has closed range if and only if there exists a constant $$C>0$$ so that $$(*)\quad\quad\|f\|\leq C\|T^*f\|,\,\forall f\in N(T^*)^\perp$$

To prove this proposition, we use Theorem 2.19 from Brezis'book which states that $$R(T)$$ is closed iff $$R(T^*)$$ is closed and the fact that $$R(T^*)$$ is closed iff $$(*)$$ holds.

To see why $$(*)$$ is equivalent to that $$R(T^*)$$ is closed, consider the map $$\overline T^*:Y/N(T^*)\to X$$ defined by $$\overline T^*[f]=T^*f$$. This map is now injective and has the same range as $$T^*$$.

Note: The norm in $$Y/N(T^*)$$ is given by $$\|[f]\|=\inf\limits_{z\in N(T^*)}{\|f-z\|}$$ and because $$N(T^*)$$ is a closed subspace it follows that the space $$Y/N(T^*)$$ with this norm is Banach. Also because $$Y$$ is reflexive it follows that the infimum in the definition of the norm is always achieved for some $$z_0\in N(T^*)$$.

Showing the equivalence of $$R(T^*)$$ closed $$\Leftrightarrow$$ $$(*)$$ is satisfied:

Because $$R(T^*)=R(\overline T^*)$$, we need to show $$R(\overline T^*)$$ closed $$\Leftrightarrow (*)$$ is satisfied:

$$C\|\overline T^*[f]\|=C\|T^*f\|\ge \|f\|=\|f-0\|\ge \inf\limits_{z\in N(T^*)}{\|f-z\|}=\|[f]\|$$ and by the Lemma below it follows $$R(\overline T^*)$$ is closed.

Conversly, if $$R(\overline T^*)$$ is closed, then by the Lemma below it follows $$\exists C>0:\,C\|\overline T^*[f]\|\ge \|[f]\|,\,\forall [f]\in Y/N(T^*)$$ which is the same as $$C\|T^*f\|=C\|\overline T^*[f]\|\ge \|[f]\|=\inf\limits_{z\in N(T^*)}{\|f-z\|}=\|f-z_0\|$$ $$=\sqrt{(f-z_0,f-z_0)}=\sqrt{(f,f)-2(f,z_0)+(z_0,z_0)}\ge \sqrt{(f,f)}=\|f\|$$ because $$f\perp N(T^*)\ni z_0$$

Proving the equivalence above just uses the fact that if $$f\in N(T^*)^\perp$$ then $$dist(f,N(T^*))=\inf\limits_{z\in N(T^*)}{\|f-z\|}=\|[f]\|$$.

Lemma: If $$A:X\to Y$$ is a bdd linear operator which is injective, then $$R(A)\subset Y$$ is closed iff $$\exists C>0: C\|Ax\|\ge \|x\|,\,\forall x\in X$$ (i.e $$A^{-1}$$ is bounded)

The statement is wrong: $T=0$ has closed range but the inequality is obviously false.