image of an operator is closed iff its image of adjoint is closed

Question

Here's what I am trying to show:

Let $T : H \to K$ be a bounded operator between two Hilbert spaces $H,K$. Show that image of $T$ is closed iff image of $T^*$ is closed.

What I know is that image of $T$ is closed iff $T$ is bounded below on $(\ker T)^{\perp}$. I am trying to show that $T^*$ is bounded below on $(\ker T^*)^{\perp}$ assuming image of $T$ is closed.

Let $y \in (\ker T^*)^{\perp}$. Then we have that $\lVert T^*y \rVert =\sup_{\lVert x \rVert =1} \lvert \langle T^* y,x \rangle \rvert = \sup_{\lVert x \rVert =1} \lvert \langle y,Tx \rangle \rvert$. I am trying to get an lower bound from here but haven't been successful so far. Hints will be appreciated!

Answer 1

There is likely an easier way to do this, but here is one take based on the spirit of the polar decomposition.

Since $T^{**}=T$, we only need to prove one implication. So assume that $\def\ran{\operatorname{ran}}\ran T$ is closed, so $\|Tx\|\geq c\|x\|$ for $x\in\overline\ran T^*=(\ker T)^\perp$. With $|T|=(T^*T)^{1/2}$, we also have $$ \|\,|T|x\,\|^2=\langle |T|x,|T|x \rangle=\langle T^*Tx,x\rangle=\|Tx \|^2, $$ so $\|\,|T|x\,\|\geq c\|x \|$. Then the map $|T|:\overline\ran T^*\to\overline\ran T^*$ is injective with closed range. We also have $$ \ran |T|=(\ker |T|)^\perp=(\ker T^*T)^\perp=(\ker T)^\perp=\overline\ran T^*, $$ so $|T|$ is invertible. Let $V:\overline\ran T^*\to\ran T$ be given by $Vx=T|T|^{-1}x $. Let us see that $V$ is an isometry. We have \begin{align} \|V\,T^*x\|^2 &=\langle TV^*VT^*x,x\rangle=\langle T|T|^{-1}T^*T|T|^{-1}T^*x,x\rangle\\[0.2cm] &=\langle T|T|^{-1}|T|^2|T|^{-1}T^*x,x\rangle=\langle TT^*x,x\rangle\\[0.2cm] &=\|T^*x \|. \end{align} We also have that $V$ is onto: this follows from $\ran T=\ran TT^*$ and the invertibility of $|T|$. For if $Tx=TT^*y$, then $$ Tx=TT^*y=T|T|^{-1}|T|T^*y=V|T|T^*y. $$

Now if $\{T^*x _n\}$ is Cauchy, so is $\{VT^*x_n\}\subset\ran T$. As $\ran T$ is closed, there exists $y$ such that $VT^*x _n\to Ty $. From $\ran T=\ran TT^*$ there exists $z$ with $Ty=TT^*z $. As $V$ is a unitary, $$T^*x _n\to V^*TT^*z=|T|^{-1}T^*TT^*z=|T|T^*z=T^*(TT^*)^{1/2}z\in \ran T^*$$ and so $\ran T^*$ is closed.

The last equality above is due to $p(T^*T)T^*=T^*p (TT^*)$ for all polynomials $p$, which then gives $(T^*T)^{1/2}T^*=T^*(TT^*)^{1/2}$.

Answer 2

The idea is to factor through the quotient space: Suppose that the range of $T$, $R(T)$, is closed. Factoring through the quotient, we have that $$ \hat T \colon H / \ker T \to R(T) , \qquad x+\ker H \mapsto Tx$$ is a bijection between Hilbert spaces and thus an isomorphism. Hence, its adjoint $$ (\hat T)^* \colon R(T) \to H/\ker A $$ is also an isomorphism. So, there exists $c>0$ such that $$ \|(\hat T)^* u\| \ge c \|u\|, \quad \forall u \in R(T) $$ Notice however that for $u \in R(T)$ and any $v \in H$ $$ ( (\hat T)^*u , v) = (u,\hat T v) = (u,Av) = (T^*u,v) $$ i.e. $(\hat T)^* u = T^*u$ for $u \in R(T)$. Thus, $$ \|T^* u \| \ge c \|u\|, \quad \forall u \in R(T) = (\ker T^*)^\bot $$ i.e. $T^*$ is bounded from below on $(\ker T^*)^\bot$. Hence, $T^*$ has closed range.

---

The same logic of factoring though the quotient can be applied to general Banach spaces:

Theorem. For a bounded operator $A \colon X \to Y$ between Banach spaces the following are equivalent:

1. The range of $A$ is closed.

2. The range of $A^* \colon Y^* \to X^*$ is closed.

Of course, for Hilbert spaces (or more general, reflexive spaces) you have that $A^{**}=A$ so you only need to prove that $(1) \implies (2)$.

---

The proof is essentially due to the following:

Claim: By factoring through the quotient space, we may assume that $A$ is injective and has a dense range.

Proof of claim: Let $V= \overline{R(A)}$ and consider the factorization $$ \hat A \colon X / \ker A \to V $$ Then $\hat A$ is injective, has dense range and $R(A)= R(\hat A)$. In particular, $R(A) $ is closed iff $R(\hat A)$ is closed. It remains to show that $R(A^*)$ is closed if and only if $R((\hat A)^*)$ is closed. Note that $ (\hat A)^* \colon (\overline{R(A)})^* \to (X/\ker A)^*, $ where $$ (X / \ker A)^* = \ker A^\bot = \overline{R(A^*)}^{w^*} $$ and $$ (\overline{R(A)})^* = Y^*/ (\overline{R(A)})^\bot = Y^*/R(A)^\bot. $$ So, in fact, $$ (\hat A)^* \colon Y^*/R(A)^\bot \to \overline{R(A^*)}^{w^*}. $$ Now, I claim that $R(A^*) = R( (\hat A)^*)$. In fact, $$ (\hat {A})^* (y^* + R(A)^\bot) = A^* y^* , \quad \forall y^* \in Y^*. $$ Fix $y^* \in Y^*$ and let $x \in X$. We calculate \begin{align*} \langle (\hat A)^* (y^* + R(A)^\bot) ,x\rangle &= \langle y^*+R(A)^\bot,\hat A x \rangle \\ &=\langle y^* + R(A)^\bot,Ax \rangle \\ &= \langle y^* + R(A)^\bot , Ax \rangle \\ &= \langle y^*,Ax \rangle \\ &= \langle A^*y^*,x \rangle. \end{align*} The second to last equality follows from the identification of $Y^*/R(A)^\bot $ with $ \overline{R(A)}^*$: Every element $y^* + R(A)^\bot$ of $Y^*/R(A)^\bot$ can be identified with a bounded functional on $\overline{R(A)}$ such that it acts by way of $ \langle y^* + R(A)^\bot , y \rangle = \langle y^*,y \rangle $ for every $y \in \overline{R(A)}$. This finishes the proof of the claim.

---

Henceforth, we assume that $A$ is 1-1 and has a dense range.

$(1) \implies (2)$: If $A$ has a closed range, then $R(A) = Y$; hence $A$ is isomoprhic and consequently so is $A^*$. In particular, $R(A^*)$ is closed.

The inverse implication $(2) \implies (1)$ is trickier (unless, as mentioned above, $X$ is reflexive): Suppose that $R(A^*)$ is closed. We first show that $R(A^*)$ is weak-star closed. This will then imply that $R(A)$ is closed (in fact, these two are equivalent). Notice that $\ker A^* = R(A)^\bot = (0)$ since $R(A)$ is dense in $Y$ and thus $A^* \colon Y^* \to R(A^*)$ is an isomorphism between Banach spaces. This implies that $A^*$ is bounded from below i.e. there is $c>0$ such that $$ \|A^* y^* \| \ge c \|y^* \| , \quad \forall y^* \in Y^* \tag{$\star$} $$ To show that $R(A^*)$ is weak-star closed, we make use of Krein-Smulian's theorem: If $(A^* y_i^*)$ is a bounded net in $R(A^*)$ with $\|A^* y_i^*\| \le r$ that converges weak-star to some $x^* \in X^*$ then $\|y^*_i\| \le c^{-1} r$ by $(\star)$. By Alaoglu's theorem we may assume that $y_i^* \xrightarrow{w^*} y^*$ for some $y^*$. Hence $A^* y_i^* \xrightarrow{w^*} A^*y^* $ and so $x^* = A^* y^* \in R(A^*)$. This shows that $R(A^*)$ is weak-star closed.

Now, $$ R(A^*) = \overline{R(A^*)}^{w^*} = \ker A ^\bot = (0)^\bot = X^* $$ so $A^* \colon Y^* \to X^*$ is an isomorphism. Finally, this implies that $A\colon X \to Y$ is an isomorphism and consequently $R(A)$ is closed. Indeed, by the open mapping theorem, there exists $c>0$ such that $$ A^*B_{X^*} \supset c B_{x^*} $$ and consequently, for every $x^* \in B_{X^*}$, there exists $y^* \in B_{Y^*}$ such that $cx^*= A^* y^*$. Hence, $$ | \langle x^*,x \rangle | = \tfrac 1c | \langle y^*,Ax \rangle| \le \tfrac 1c \|Ax\|, \quad \forall x^* \in X^*, \ x \in X, $$ and thus $ \|Ax\| \ge c\|x\|$ for all $x \in X$. This implies that $A \colon X \to R(A)$ is an isomorphism.

Answer 3

The proof follows directly from the polar decomposition $T=U|T|$ where $U$ is the partial isometry which maps isometrically $\overline{{\rm Im}\, |T|}$ onto $\overline{{\rm Im}\, T}$ and $T$ vanishes on $\ker |T|=\ker T.$ Thus $U^*$ maps isometrically $\overline{{\rm Im}\, T}$ onto $\overline{{\rm Im}\,|T|}.$ Hence if the range of $T$ is closed, so is the range of $|T|.$ We also have $T^*=|T|U^*.$ The range of the right hand side coincides with the range of $|T|$. Hence if the range of $|T|$ is closed so is the range of $T^*.$