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!


2 Answers 2


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.


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