Closed range of $T^*T$ for Hilbert space map.

Question

Earlier today, I came across a paper that stated (but did not prove!) the following theorem.

Theorem: Let $T:A \rightarrow B$ be a bounded linear Hilbert space map. Then the following are equivalent.

1. $T$ has closed range;
2. $T^*$ has closed range;
3. $T^*T$ has closed range.

The equivalence of (1) and (2) easily follows from the closed range theorem for Banach spaces. I have also managed to show that (1) + (2) implies 3 by using the identity $\text{image}(T) = \ker(T^*)^\perp$, which holds when $T$ has closed range. This identity allows you to show that if $T$ has closed range, then $T^*T$ and $T^*$ have the same range.

I am struggling to show that (3) implies the other conditions though. It seems like there should be a way to maybe take advantage of the fact that $\ker(T^*T) = \ker(T)$ and $\ker(T^*T)^\perp = \text{image}(T^*T)$ (since $T^*T$ has closed range). But I’m not seeing how to put the pieces together. If anyone has a reference for this result, or knows how to prove it, I’d appreciate it!

Answer 1

Suppose that $\{T^*x_n\}$ is a Cauchy sequence, $z=\lim_nT^*x_n$. Let $P$ be the orthogonal projection onto $\ker TT^*=\ker T^*$ and let $y_n=x_n-Px_n$, so $T^*y_n=T^*x_n$. The projection $I-P$ is the orthogonal projection onto $$ (\ker TT^*)^\perp=(\ker T^*)^\perp=\overline{\operatorname{ran}}T. $$ Hence $y_n=(I-P)x_n\in \overline{\operatorname{ran}}T$. By approximating the $y_n$ better and better with elements of $\overline{\operatorname{ran}}T$, we get a sequence $\{Tw_n\}\subset{\operatorname{ran}}T$ such that $$ z=\lim_nT^*x_n=\lim_nT^*y_n=\lim_nT^*Tw_n. $$ So $z\in \overline{\operatorname{ran}}T^*T$. As this range is closed, $z\in \operatorname{ran}T^*T$. Thus $\{T^*x_n\}$ converges to $z\in\operatorname{ran}T^*T\subset\operatorname{ran}T^*$.

Answer 2

First suppose that $T$ is injective and has dense range. To show that $R(T)$ is closed, suppose that $Tx_n \to y$ in $H$. Then $T^* Tx_n \to T^* y \in R(T^*T)$ so that $T^*y =T^*Tx$ for some $x \in H$. Hence, $y-Tx \in \ker T^* = R(T)^\bot =(0)$ i.e. $$y = Tx \in R(T).$$

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The idea is that, by factoring through the quotient, we can always assume that $T$ is injective and has a closed range. Indeed, one can consider $$\hat T \colon H / \ker T \to R(T).$$ Then $\hat T$ is injective, has a dense range and $R(T) = R(\hat T)$. Moreover, $R(T^*) = R((\hat T)^*)$ and $$R(T^*T) = R( (\hat T)^* \hat T).$$ You can now apply the first argument for $\hat T$ to infer that $R(\hat T)$ is closed and thus $R(T)$ is closed. See this for more details.

Answer 3

For a positive operator $S:A\to A$ the range of $S^2$ is closed iff the range of $S$ is closed. Indeed $\ker S=\ker S^2$ Hence $V:=\overline{{\rm Im}\,S}=\overline{{\rm Im}\,S^2}.$ If the range of $S^2$ is closed then $S^2:V\to V$ is invertible. Hence $S:V\to V$ is invertible. Thus its range is closed. The converse follows the same way.

For $T$ we have $$\|Tx-Ty\|=\|T(x-y)\|=\||T|(x-y)\|=\||T|x-|T|y|\|$$ Thus the range of $T$ is closed iff the range of $|T|:A\to A$ is closed (see the spoiler). Now apply the above to $S:=|T|$ to get that the range of $T$ is closed iff the range of $T^*T=|T|^2=S^2$ is closed.

Assume operators $S:X\to Y$ and $T:X\to Z$ where $X,Y,Z$ are Banach spaces, satisfy $\|Sx\|=\|Tx\|$ for any $x.$ Let the range of $S$ be closed. Assume $Tx_n\to y.$ Then $Tx_n$ is a Cauchy sequence. So is $Sx_n,$ because $\|Tx_n-Tx_m\|=\|Sx_n-Sx_m\|.$ Therefore $Sx_n$ is convergent to some $Sx.$ We have $\|Tx_n-Tx\|=\|Sx_n-Sx\|\underset{n\to\infty}{\longrightarrow}0.$ Hence $y=Tx.$

Answer 4

Upon reflection, I found a more simple argument than the answers posted. I’m still enormously grateful for the help.

As noted in the question, we know $\ker(T^*T) = \ker(T)$, and $\text{image}(T^*T) = \ker(T^*T)^\perp$. We can also use the fact that $\ker(T)= \text{image}(T^*)^\perp$ to derive $\ker(T)^\perp = \overline{\text{image}(T^*)}$. We then can check:

$$\begin{align*} \overline{\text{image}(T^*)} & = \ker(T)^\perp \\ & = \ker(T^*T)^\perp \\ & = \text{image}(T^*T) \\ & \subseteq \text{image}(T^*) \end{align*}$$ Thus the image of $T^*$ contains its own closure, and is closed.