Unbounded Operators

The associative law $\left(T_{1} T_{2}\right) T_{3} = T_{1}\left(T_{2} T_{3}\right)$ has been used freely throughout this chapter. Prove it. Prove also that $T_{1} \subset T_{2}$ implies $S T_{1} \subset S T_{2}$ and $T_{1} S \subset T_{2} S$.

Let $T$ be a densely defined operator in $H$. Prove that $T$ has a closed extension if and only if $\mathscr {D}\left(T^{*}\right)$ is dense in $H$. In that case, prove that $T^{* *}$ is an extension of $T$.

By Theorem $13.8, \mathscr {D}\left(T^{*}\right) = \left\{ 0\right\}$ for a densely defined operator $T$ in $H$ if and only if $\mathscr {G}(T)$ is dense in $H \times H$. Show that this can actually happen.

Suggestion: Let $\left\{ e_{n}: n = 1,2,3, \ldots \right\}$ be an orthonormal basis of $H$; let $\left\{ x_{n}\right\}$ be a dense subset of $H$; define $T e_{n} = x_{n}$; and extend $T$ linearly to $\mathscr {D}(T)$, the set of all finite linear combinations of the basis vectors $e_{n}$. Show that the graph of this $T$ is dense in $H \times H$.

Suppose $T$ is a densely defined, closed operator in $H$, and $T^{*} T \subset T T^{*}$. Does it follow that $T$ is normal?

Suppose $T$ is a densely defined operator in $H$, and $(T x, x) = 0$ for every $x \in \mathscr {D}(T)$. Does it follow that $T x = 0$ for every $x \in \mathscr {D}(T)$?

If $T$ is an operator in $H$, define

If $\mathscr {D}(T)$ is dense, prove that

If $T$ is also closed, prove that

This generalizes Theorem 12.10.

Consider the following three boundary value problems. The differential equation is

where $g \in L^{2}([0,1])$ is given. The choices of boundary conditions are

(i) $f(0) = f(1) = 0$.

(ii) $f^{\prime }(0) = f^{\prime }(1) = 0$.

(iii) $f(0) = f(1)$ and $f^{\prime }(0) = f^{\prime }(1)$.

Show that each of these problems has a unique solution $f$ such that $f^{\prime }$ is absolutely continuous and $f^{\prime \prime } \in L^{2}([0,1])$. Hint: Combine Example 13.4 with Theorem 13.13.

Do this also by solving the problems explicitly.

Let $H^{2}$ be the space of all holomorphic functions $f(z) = \sum c_{n} z^{n}$ in the open unit disc that satisfy

Show that $H^{2}$ is a Hilbert space which is isomorphic to $\ell^{2}$ via the one-to-one correspondence $f \leftrightarrow \left\{ c_{n}\right\}$.

Define $V \in \mathscr {B}\left(H^{2}\right)$ by $(V f)(z) = z f(z)$. Show that $V$ is the Cayley transform of the symmetric operator $T$ in $H^{2}$, given by

Find the ranges of $T+i I$ and of $T-i I$; show that one is $H^{2}$ and one has codimension 1. (Compare with Example 13.21).

With $H^{2}$ as in Exercise 9, define $V$ now by

Show that $V$ is an isometry which is the Cayley transform of a closed symmetric operator $T$ in $H^{2}$, whose deficiency indices are 0 and $\infty$.

Prove part (c) of Lemma 13.18.

Is the operator $C$ that occurs in the proof of Theorem 13.33 normal?

Prove that every normal operator $N$ in $H$, bounded or not, has a polar decomposition

where $U$ is unitary, $P$ is self-adjoint, $P \geq 0$. Moreover, $\mathscr {D}(P) = \mathscr {D}(N)$.

Prove the following extension of Theorem 12.16: If $T \in \mathscr {B}(H)$, if $M$ and $N$ are normal operators in $H$, and if $T M \subset N T$, then also $T M^{*} \subset N^{*} T$.

Suppose $T$ is a closed operator in $H, \mathscr {D}(T) = \mathscr {D}\left(T^{*}\right)$, and $\left\| T x\right\| = \left\| T^{*} x\right\|$ for every $x \in \mathscr {D}(T)$. Prove that $T$ is normal. Hint: Begin by proving that

Prove that the spectrum $\sigma (T)$ of any operator $T$ in $H$ is a closed subset of $C$. (See Definition 13.26). Hint: If $S T \subset T S = I$, and $S \in \mathscr {B}(H)$, then $S(I-\lambda S)^{-1}$ is a bounded inverse of $T-M$, for small $\left|\lambda \right|$.

Put $\phi (t) = \exp \left(-t^{2}\right)$. Define $S \in \mathscr {B}\left(L^{2}\right)$, where $L^{2} = L^{2}(R)$, by

so that $\left(S^{2} f\right)(t) = \phi (t) \phi (t-1) f(t-2)$, etc. (Note that $S$ is presented in its polar decomposition $S = P U$).

Find $S^{*}$. Compute that

Conclude that $S$ is one-to-one, that $\mathscr {R}(S)$ is dense in $L^{2}$, and that $\sigma (S) = \left\{ 0\right\}$. Define $T$, with domain $\mathscr {D}(T) = \mathscr {R}(S)$, by

Prove that $\sigma (T)$ is empty.

Let $T_{1}, T_{2}, T_{3}$ be as in Example 13.4, put

and define $T_{4} f = i f^{\prime }$ for all $f \in \mathscr {D}\left(T_{4}\right)$.

Prove the following assertions:

Show that every nonempty closed subset of $C$ is the spectrum of some normal operator in $H$ (if $\operatorname {dim} H = \infty$).

Define $Q(t) \in \mathscr {B}\left(L^{2}\right)$, where $L^{2} = L^{2}(R)$, by

If $f \in H^{2}$ (see Exercise 9) and $f(z) = \sum c_{n} z^{n}$, define

Show that each $Q(t)$ is self-adjoint (and positive). Find the infinitesimal generator $A$ of the semigroup $\left\{ Q(t)\right\}$. Is $A$ self-adjoint? Show that $A$ has pure point spectrum, at the points $\log 1, \log \frac{1}{2}, \log \frac{1}{3}, \ldots$.

For $f \in L^{2}(R), x \in R, 0 < y < \infty$, define

and put $Q(0) f = f$. Show that $\left\{ Q(y): 0 \leq y < \infty \right\}$ satisfies the conditions imposed in Definition 13.34 and that $\left\| Q(y)\right\| = 1$ for all $y$.

Find the domain of the infinitesimal generator $A$ of $\left\{ Q(y)\right\}$, and prove that

where $H$ is the Hilbert transform (Chapter 7, Exercise 24). Prove that $-A$ is positive and self-adjoint.

Show that every isometric operator in $H$ has a closed isometric extension.

On the other hand, show that some symmetric operators in $H$ have no closed symmetric extension by completing the following outline.

Let $\left\{ e_{1}, e_{2}, e_{3}, \ldots \right\}$ be an orthonormal basis of $H$. Let $X$ be the set of all finite sums $\sum \alpha_{i} e_{i}$, subject to the condition $\sum \alpha_{i} = 0$. Prove that $X$ is a dense subspace of $H$. Define $U \in \mathscr {B}(H)$ by

and let $V$ be the restriction of $U$ to $X$. Then $V$ is an isometry, with $\mathscr {D}(V) = X$, and $I-V$ is one-to-one on $X$. Hence $V$ is the Cayley transform of a symmetric operator $T$. Any closed symmetric extension of $T$ would correspond to a closed isometric extension $V_{1}$ of $V$, with $I-V_{1}$ one-to-one. But $\mathscr {D}(V)$ is dense in $H$; hence $V$ has only one closed isometric extension, namely $U$, and $I-U$ is not one-to-one.