Analysis Problem of the Day 3

The following problem appeared as Problem 1 on Stanford’s Fall 2019 Analysis Qual:

Problem 1: Recall that a linear operator $T$ on a Hilbert space $H$ is bounded below if and only if $\|Tx\| \geq c \|x\|$ for some $c>0.$ Show that a bounded linear operator $T: H \to H$ is invertible if and only if $T$ and $T^*$ is bounded below.

Solution: The following problem is made simple by the following three facts:

The orthogonality relation
$(\ker T^*)^\perp = \overline{\text{im } T},$
which follows from the definition of the adjoint
$\langle Tx,y\rangle = \langle x, T^*y\rangle.$

2. The following characterization of boundedness below:

Lemma: An operator $T: H \to H$ is bounded below if and only if it is injective and has closed range.

Proof: If $T$ is injective and has closed range, then $T: H/\ker T \to \text{im }T$ is a bijective map, so by the open mapping theorem, the inverse is bounded, i.e. $\|T^{-1} x\| \leq C\|x\| \implies \frac{1}{C} \|x\| \leq \|Tx\|.$ Conversely, setting $x=0$ shows that a bounded below map is injective. To show the range is closed, let $Tx_n \to y \in H$ be a Cauchy sequence. Then, $\|x_n-x_m\| \leq \|Tx_n-Tx_m\| < \epsilon$ for large $n,m,$ so $x_n$ is also Cauchy and therefore converges to some $x \in H,$ and since $T$ is continuous, $Tx_n \to y=Tx.$ Thus, $y \in \text{im }T,$ so the range is closed.

3. The so-called “closed range theorem:” $T$ has closed range if and only if $T^*$ has closed range.

Proof: By symmetry, it suffices to prove only one direction. If $T: H \to H$ has closed range, then taking the dual of the sequence of maps

$H \overset{\pi}{\twoheadrightarrow} H/\ker T \overset{T}{\to} \text{im }T \overset{i}{\hookrightarrow} H$

yields the sequence

$H \to (\text{im }T)^* \overset{T^*}{\to} \text{im }T^* \hookrightarrow H.$

Since $\text{im }T$ is closed, by the open mapping theorem, the middle map is a Hilbert space isomorphism, so the image of $T^*$ is also closed.

Now, if $T$ is invertible, so is $T^*,$ with inverse $(T^{-1})^*,$ and so $T^*$ is injective and has closed range, i.e. is bounded below by the above lemma. Conversely, if $T^*$ is bounded below, then $T^*$ is injective, i.e. $\ker T^* =0,$ so by the orthogonality relation, $\text{im }T$ is dense in $H.$ By the closed range theorem, since the image of $T$ is dense and closed, it must be all of $H,$ i.e. $T$ is bijective. By the open mapping theorem, this implies that $T^{-1}$ exists and is bounded, i.e. $T$ is invertible.

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