range of $\lambda I - T$ and the null space of $\bar \lambda I - T^*$

Question

I have a few questions about the following exercise in Stein's Real analysis.

Exercise 29

Let $T$ be a compact operator on a Hilbert Space $\mathcal H$ and assume $\lambda \neq 0$.

(a) Show that the range of $\lambda I - T$ defined by $\{g \in \mathcal H \ : \ \exists f \in \mathcal H, \ g = (\lambda I - T)f \}$ is closed.

(b) Show, by example, that this may fail when $\lambda = 0$.

(c) Show that the range of $\lambda I - T$ is all of $\mathcal H$ if and only if the null space of $\bar \lambda I - T^*$ is trivial.

This page explains the answer of (a), and below is my attempt to solve (c).

Consider the case that $\lambda I - T$ is all of $\mathcal H$, and I assumed that $\exists f \neq 0$ such that $\bar \lambda f = T^* f$. $f$ is an eigenvector of $T^*$ with eigenvalue $\bar \lambda$. Then, $f$ is an eigenvector of $T$ with eigenvalue $\lambda$ : $Tf = \lambda f$. But, this does not give an useful conclusion, such as $\exists g \in \mathcal H$ s.t. $g \neq (\lambda I - T)f $ for all $f \in \mathcal H$...

Any help about part (c) would be appreciated. Also, some hints about $\lambda = 0$ so that the range is not closed (regarding (b)) would also be appreciated. Thank you.

(Later, I'll write down what I've done regarding (a) that the writer in the linked page had skipped, such as $\lambda I - T$ is bounded, $V_{\lambda}$ is closed, and $S$ is bijection)

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1st: $\lambda I - T$ is bounded

Consider $\{f_n\} \subset \mathcal H$ such that $\lim_{n \to \infty}\Vert f_n-f \Vert = 0$ for $f \in \mathcal H$.

$$\Vert (\lambda I - T)f_n - (\lambda I - T)f \Vert$$ $$ =\Vert \lambda (f_n - f) - T(f_n - f) \Vert $$ $$ \le |\lambda|\Vert f_n-f \Vert + \Vert T(f_n-f) \Vert$$ $$ \le |\lambda|\Vert f_n-f \Vert + \Vert T \Vert \Vert f_n-f \Vert$$

Note that $T$ is compact, and thus bounded.

Then $\lim_{n \to \infty}\Vert (\lambda I - T)f_n - (\lambda I - T)f \Vert = 0$.

Thus $\lambda I - T$ is continuous and bounded.

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2nd: $V_{\lambda}$ is closed

Choose the convergent sequence $\{f_n\}$ from $V_{\lambda}$ : $\lim_{n \to \infty}\Vert f_n-f \Vert = 0$ for $f \in \mathcal H$

$(\lambda I - T)f_n = 0$ for all $n \in \mathbb N$, so using the result above,

$$\Vert (\lambda I - T)f_n - (\lambda I - T)f \Vert = \Vert (\lambda I - T)f \Vert \le |\lambda|\Vert f_n-f \Vert + \Vert T \Vert \Vert f_n-f \Vert$$

and taking the limit of $n$ to $\infty$, $\Vert (\lambda I - T)f \Vert = 0$. This means $f \in V_{\lambda}$, so $V_{\lambda}$ is closed.

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3rd: $S: V_{\lambda}^{\bot} \to \mathcal R$ is bijective

It suffices to show that $S$ is injective. Denote the range of $\lambda I - T$ as $\mathcal R$ for simplicity.

Consider $g_1, g_2 \in \mathcal R$ such that $g_1 = (\lambda I - T)f_1$ and $g_2 = (\lambda I - T)f_2$ and $f_1, f_2 \in V_{\lambda}^{\bot}$

When $g_1 = g_2$, $T(f_1-f_2) = \lambda (f_1-f_2)$ holds.

So, $f_1 - f_2 \in V_{\lambda}$.

Since $V_{\lambda}^{\bot}$ is a subspace, $f_1-f_2 \in V_{\lambda}^{\bot}$

Using the fact $V_{\lambda} \cap V_{\lambda}^{\bot} = \{0\}$, $f_1-f_2 = 0$.

Thus, $f_1 = f_2$.

This proves that $S$ is injective, and thus $S$ is bijective.

Answer 1

Suppose $R(\lambda I-T)\neq \mathcal{H}$. Since it is closed, we know by Hahn-Banach that there exists $y \in R(\lambda I-T)^{\perp}, \ y \neq 0$. Thus $\langle y,(\lambda I-T)x\rangle=0 \ \forall x \in \mathcal{H}$. Simply using the definition of the adjoint, we thus see that $\langle (I\overline{} -T^*)y,x\rangle=0 \ \forall x \in \mathcal{H}$, so that $y \in Ker(I\overline{} -T^*)$, showing that the kernel is nontrivial, so one implication is proved.

For the other direction, if $R(\lambda I-T)= \mathcal{H}$, let $y \in Ker(I\overline{} -T^*)$. Then

$$ 0=\langle (I\overline{} -T^*)y,x\rangle =\langle y, (\lambda I-T)x\rangle $$ for all $x \in \mathcal{H}$, so $\langle y,z\rangle =0$ for any $z \in R(\lambda I-T)=\mathcal{H}$, so that $y=0$. Hence $R(\lambda I-T)= \mathcal{H} \implies Ker(I\overline{} -T^*)=\{0\}$.

EDIT

For the first part, any compact operator with infinite dimensional range will do. Indeed, suppose that $R(T)$ is infinite dimensional and closed. Then $(R(T),||\cdot||)$ is a Banach space, and $R(T)=\bigcup_{n \in \mathbb{N}}\overline{T(B_n(0))}$ where $B_n(0)$ is the ball of radius n in $\mathcal{H}$. By the Baire category theorem, at least one of these sets has non-empty interior, so there exists m, and $x \in R(T), \ \epsilon>0$ so that $\overline{B^{R(T)}_\epsilon(x)} \subset \overline{T(B_m(0))}$. But by definition of a compact operator, $\overline{T(B_m(0))}$ is compact, so that $\overline{B^{R(T)}_\epsilon(x)}$ is compact, and hence $R(T)$ must be finite dimensional.

EDIT 2

For an explicit example for part (b), let $e_n$ be a Hilbert basis for $\mathcal{H}$. Then let $Tx=\sum \frac{1}{n}\langle x,e_n \rangle e_n$. It is easy to see this is linear and bounded. Further you can check it is compact either by noting it is the limit of a sequence of finite rank operators, or by the following more direct argument:

Let $||x_m|| \leq 1$. Then $|\langle x_m, e_n \rangle| \leq 1$ for all $n,m$, so by Bolzano-Weierstrass and a diagonal argument, there is some (non-relabeled) subsequence so that $\langle x_m, e_n \rangle$ converges to some $a_n, |a_n|\leq 1$ as $ m \to \infty$ for all $n$. Then

$||Tx_m-\sum_n \frac{1}{n}a_ne_n||^2=\sum_n \frac{1}{n^2}|\langle x_m, e_n\rangle-a_n|^2$. Since $|\langle x_m,e_n\rangle-a_n|\leq 2$, the above expression converges to 0 by the dominated convergence theorem, showing compactness. However, note that we do not have closed range. Indeed, consider $y_m=\sum_{n=1}^m e_n$. Then $Ty_m \to y=\sum_n \frac{1}{n}e_n$, but this cannot be in $R(T)$, since if $Tz=y$, then $\langle z, e_n \rangle =1$ for all $n$ which cannot be.