Families of Compact Operators on Hilbert Spaces and Fundamental Properties

Abstract

The aim of this chapter, from the point of view of quantum mechanical applications, is to introduce certain types of operators used to define quantum states in the standard formulation of the theory. These operators, known in the literature as operators of trace class, or nuclear operators, are bounded operators on a Hilbert space that admit a trace. In order to introduce them it is necessary to define first compact operators, also known as completely continuous operators, which play an important role in several branches of mathematics and physical applications irrespective of quantum theories.

Measure what can be measured, and make measurable what can’t be.

Galilei

Exercises

4.1

Prove that if ${\mathsf X}$ is a normed space and $T: {\mathsf X}\rightarrow {\mathsf X}$ is compact and injective, then any linear operator $S: Ran(T) \rightarrow {\mathsf X}$ that inverts T on the left cannot be bounded if $\dim {\mathsf X}=\infty$.

Solution. If S were bounded, Proposition 2.47 would allow to extend it to a bounded operator $\tilde{S}: {\mathsf Y}\rightarrow {\mathsf X}$, where ${\mathsf Y}:= \overline{Ran(T)}$, so that $\tilde{S}T = I$. Precisely as in the proof of Proposition 4.9(b), we can prove that $\tilde{S}T$ is compact if $T\in {\mathfrak B}({\mathsf X},{\mathsf Y})$ is compact and $\tilde{S}\in {\mathfrak B}({\mathsf Y},{\mathsf X})$. Then $I: {\mathsf X}\rightarrow {\mathsf X}$ would be compact, and thus the unit ball in ${\mathsf X}$ would have compact closure, breaching Proposition 4.5.

4.2

Using Banach’s Lemma 4.12 prove that in an infinite-dimensional normed space the closed unit ball is not compact.

Outline. Let $x_1, x_2, \ldots$ be an infinite sequence of linearly independent vectors with $||x_n||=1$ (hence all belonging to the closure of the unit ball). Banach’s lemma constructs a sequence of vectors $y_1,y_2,\ldots$ such that $||y_n||=1$ and $||y_{n-1}- y_{n}||>1/2$. This sequence cannot contain converging subsequences.

4.3

Prove that if $A^*= A \in {\mathfrak B}_\infty ({\mathsf H})$ on a Hilbert space ${\mathsf H}$, then

$$\sigma _p(|A|) = \{|\lambda |\,\, | \lambda \in \sigma _p(A)\}\,.$$

Conclude that if $A^*= A\in {\mathfrak B}_\infty ({\mathsf H})$,

$$sing(A) = \{|\lambda |\,\, | \lambda \in \sigma _p(A)\setminus \{0\}\}\,.$$

Solution. Expand the compact, self-adjoint operators A and |A| according to Theorem 4.20:

$$A = \sum _{\lambda \in \sigma _p(A)} \lambda P_\lambda \,\,\,\, \text { and }\,\,\,\,|A| = \sum _{\mu \in \sigma _p(|A|)} \mu P'_\mu \,,$$

with the obvious notation. By squaring A and |A| and using their continuity (this allows to consider all series as finite sums), using the idempotency and orthogonality of projectors relative to distinct eigenvectors, and recalling $|A|^2 = A^*A = A^2$, we have

$$\begin{aligned} \sum _{\lambda \in \sigma _p(A)} \lambda ^2 P_\lambda = \sum _{\mu \in \sigma _p(|A|)} \mu ^2 P'_\mu \,. \end{aligned}$$

(4.68)

Now keep in mind $P_\lambda P_{\lambda _0} = 0$ if $\lambda \ne \lambda _0$ and $P_\lambda P_{\lambda _0} = P_{\lambda _0}$ otherwise, and the same holds for the projectors in the decomposition of |A|. Composing with $P_{\lambda _0}$ on the right in (4.68), taking adjoints and eventually right-composing with $P'_{\mu _0}$ produces $\lambda _0^2 P_{\lambda _0}P'_{\mu _0} = \mu _0^2P_{\lambda _0}P'_{\mu _0}$, i.e.

$$\begin{aligned} (\lambda _0^2 - \mu _0^2) P_{\lambda _0}P'_{\mu _0} = 0\,,\end{aligned}$$

(4.69)

for any $\lambda _0 \in \sigma (A)$ and $\mu _0 \in \sigma _p(|A|)$. The fact that A admits a basis of eigenvectors (Theorem 4.20) is known to be equivalent to

$$I = s\text{- }\sum _{\lambda _0\in \sigma (A)} P_{\lambda _0}\,.$$

Fix $\mu _0 \in \sigma _p(|A|)$. If $P_{\lambda _0}P'_{\mu _0}=0$ for any $\lambda _0 \in \sigma (A)$, from the above identity we would have $P'_{\mu _0}=0$, absurd by definition of eigenspace. Therefore, (4.69) withstanding, there must exist $\lambda _0 \in \sigma (A)$ such that $\lambda ^2_0 = \mu ^2_0$, i.e. $\mu _0 = |\lambda _0|$. If $\lambda _0\in \sigma _p(A)$, swapping A and |A| and using a similar argument would produce $\mu _0 \in \sigma _p(|A|)$ such that $\lambda ^2_0 = \mu ^2_0$, i.e. $\mu _0 = |\lambda _0|$. The first assertion is thus proved. The second one is evident by the definition of singular value.

4.4

Consider a separable Hilbert space ${\mathsf H}$ with basis $\{f_n\}_{n\in {\mathbb N}}\subset {\mathsf H}$ and the sequence $\{s_n\}_{n\in {\mathbb N}}\subset {\mathbb C}$ where $|s_{n}| \ge |s_{n+1}|$ and $|s_n| \rightarrow 0$ as $n\rightarrow +\infty$. Using the uniform topology prove that

$$T:= \sum _{n=0}^{+\infty } s_n f_n (f_n|\,\,)$$

is well defined and $T\in {\mathfrak B}_\infty ({\mathsf H})$. Show that if every $s_n$ is real, T is further self-adjoint and every $s_n$ is an eigenvalue of T.

Hint. Under the assumptions made, the operator $T_N:= \sum _{n=0}^{N} s_n f_n (f_n|\,\,)$ satisfies:

$$||T_Nx-T_Mx||^2 \le |s_M|^2 \sum _{n=M}^{N-1}|(f_n|x)|^2 \le |s_M|^2 ||x||^2\,$$

for $N\ge M$. Taking the least upper bound over unit vectors $x\in {\mathsf H}$ gives:

$$||T_N-T_M|| \le |s_M|^2 \rightarrow 0\quad \text { as }N, M \rightarrow +\infty ,$$

whence the first part. The rest follows by direct inspection.

4.5

Prove that if $T\in {\mathfrak B}_\infty ({\mathsf H})$ and if ${\mathsf H}\ni x_n \rightarrow x \in {\mathsf H}$ weakly, i.e.

$$(g|x_n) \rightarrow (g|x) \quad \text { as }n\rightarrow +\infty , \text { for any given }g\in {\mathsf H},$$

then $||T(x_n)-T(x)|| \rightarrow 0$ as $n\rightarrow +\infty$. Put otherwise, compact operators map weakly convergent sequences to sequences converging in norm. Extend the result to the case $T \in {\mathfrak B}_\infty ({\mathsf X},{\mathsf Y})$, ${\mathsf X}$ and ${\mathsf Y}$ normed.

Solution. Suppose $x_n \rightarrow x$ weakly. If we bear in mind Riesz’s theorem, the set $\{x_n\}_{n\in {\mathbb N}}$ is immediately weakly bounded in the sense of Corollary 2.64. According to this corollary, $||x_n|| \le K$ for any $n\in {\mathbb N}$ and for some $K>0$. So define $y_n:= Tx_n$, $y:=Tx$ and note that for any $h\in {\mathsf H}$,

$$ (h|y_n)-(h|y) = (T^*h| x_n) - (T^*h| x) \rightarrow 0 \quad \text { as } n\rightarrow +\infty , $$

hence also the $y_n$ converge weakly to y. Suppose, by contradiction, $||y_n - y|| \not \rightarrow 0$ as $n\rightarrow +\infty$. Then there exist $\varepsilon >0$ and a subsequence $\{y_{n_k}\}_{k\in {\mathbb N}}$ with $||y-y_{n_k}|| \ge \varepsilon$ for any $k\in {\mathbb N}$. Since $\{x_{n_k}\}_{k\in {\mathbb N}}$ is bounded by K, and T is compact, there must be a subsequence $\{y_{n_{k_r}}\}_{r\in {\mathbb N}}$ converging to some $y' \ne y$. This subsequence $\{y_{n_{k_r}}\}_{r\in {\mathbb N}}$ has to converge to $y'$ also weakly. But this cannot be, for $\{y_n\}_{n\in {\mathbb N}}$ converges weakly to $y\ne y'$. Therefore $y_n \rightarrow y$ in the norm of ${\mathsf H}$. The argument works in the more general case where $T \in {\mathfrak B}_\infty ({\mathsf X},{\mathsf Y})$, ${\mathsf X}$ and ${\mathsf Y}$ normed spaces, by interpreting ${\mathsf X}\ni x_n \rightarrow x \in {\mathsf X}$ in weak sense:

$$g(x_n) \rightarrow g(x) \quad \text { as }n\rightarrow +\infty , \text { for any given } g\in {\mathsf X}',$$

because Corollary 2.64 still holds. In the proof one uses the fact that $h\in {\mathsf Y}'$ $\Rightarrow$ $h\circ T \in {\mathsf X}'$ ($h\circ T$ is a composite of continuous linear mappings).

4.6

Referring to Example 4.27, take $T_K, T_{K'}\in {\mathfrak B}_2(L^2({\mathsf X},\mu ))$ ($\mu$ is assumed separable) with integral kernels K, $K'$. Prove that the HS operator $aT_{K}+ bT_{K'}$, $a, b \in {\mathbb C}$, has kernel $aK+bK'$.

4.7

Given $T_K \in {\mathfrak B}_2(L^2({\mathsf X},\mu ))$ with integral kernel K, prove the Hilbert–Schmidt operator $T_K^*$ has integral kernel $K^*(x,y) = \overline{K(y, x)}$.

4.8

With the same hypotheses as Exercise 4.6, show that the integral kernel of $T_{K}T_{K'}$ is

$$K''(x,z):= \int _{{\mathsf X}} K(x,y)K'(y, z)\, d\mu (y)\,.$$

4.9

Let $L^2({\mathsf X},\mu )$ be separable. Prove that the mapping $L^2({\mathsf X}\times {\mathsf X}, \mu \otimes \mu ) \ni K \mapsto T_K \in {\mathfrak B}_2(L^2({\mathsf X},\mu ))$ is an isomorphism of Hilbert spaces. Discuss whether one can view this map as an isometry of normed spaces, taking ${\mathfrak B}(L^2({\mathsf X},\mu ))$ as codomain. Discuss whether it is continuous if viewed as a homeomorphisms only.

4.10

With reference to Exercise 4.27(3), prove that if $g \in C_0([0,1])$ then

$$\left( (I-\rho T)^{-1} g\right) (x) = g(x) + \rho \int _0^x e^{\rho (x-y)} g(y) dy\,.$$

Hint. Use the operator $I-\rho T$, recalling the integral expression of T and noticing $\rho e^{\rho (x-y)} = \frac{\partial }{\partial x} e^{\rho (x-y)}$.

4.11

Let ${\mathfrak B}_D(L^2({\mathsf X},\mu ))$ be the set of degenerate operators on $L^2({\mathsf X},\mu )$ (cf. Example 4.27(4)), with $\mu$ separable. Prove the following are equivalent statements.

(a) $T\in {\mathfrak B}_D(L^2({\mathsf X},\mu ))$.

(b) Ran(T) has finite dimension.

(c) $T \in {\mathfrak B}_2(L^2({\mathsf X},\mu ))$ (hence T is an integral operator) with kernel $K(x, y) = \sum _{k=1}^N p_k(x)q_k(y)$, where $p_1,\ldots , p_N \in L^2({\mathsf X},\mu )$, $q_1,\ldots , q_N \in L^2({\mathsf X},\mu )$ are linearly independent.

4.12

Take the set ${\mathfrak B}_D(L^2({\mathsf X},\mu ))$ of degenerate operators (cf. Example  4.27(4)) on $L^2({\mathsf X},\mu )$, with $\mu$ separable. Show ${\mathfrak B}_D(L^2({\mathsf X},\mu ))$ is a two-sided $^*$-ideal in ${\mathfrak B}(L^2({\mathsf X},\mu ))$ and a subspace in ${\mathfrak B}_2(L^2({\mathsf X},\mu ))$. In other words, prove that ${\mathfrak B}_D(L^2({\mathsf X},\mu )\subset {\mathfrak B}_2(L^2({\mathsf X},\mu ))$, that it is a closed subspace under Hermitian conjugation, and that $AD, DA \in {\mathfrak B}_D(L^2({\mathsf X},\mu )$ if $A\in {\mathfrak B}(L^2({\mathsf X},\mu ))$ and $D\in {\mathfrak B}_D(L^2({\mathsf X},\mu ))$.

4.13

Consider ${\mathfrak B}_D(L^2({\mathsf X},\mu ))$ (cf. Example 4.27(4)) with $\mu$ separable. Does the closure of ${\mathfrak B}_D(L^2({\mathsf X},\mu ))$ in ${\mathfrak B}_2(L^2({\mathsf X},\mu ))$ in the norm topology of ${\mathfrak B}(L^2({\mathsf X},\mu ))$ coincide with ${\mathfrak B}_2(L^2({\mathsf X},\mu ))$?

Hint. Consider the operator

$$T:= \sum _{n=0}^{+\infty } \frac{1}{\sqrt{n}} T_{K_n}\,,$$

where $K_n(x, y) = \phi _n(x)\phi _n(y)$, $\{\phi _n\}_{n\in {\mathbb N}}$ is a basis of $L^2({\mathsf X},\mu )$ and the convergence is uniform. Prove $T\in {\mathfrak B}(L^2({\mathsf X},\mu ))$ is well defined, but $T\not \in {\mathfrak B}_2(L^2({\mathsf X},\mu ))$ since $||T\phi _n||^2 = 1/n$.

4.14

Under the assumptions of Mercer’s Theorem 4.29, prove that if $T_K \in {\mathfrak B}_1 (L^2({\mathsf X}, d\mu ))$ then $tr(T_K) = \int _{{\mathsf X}} K(x, x) d\mu (x)$.

Hint. Expand the trace in the basis of eigenvectors given by the continuous maps in Mercer’s statement. Since the series that defines K converges uniformly on the compact set ${\mathsf X}$ of finite measure, a clever use of dominated convergence allows to show

$$\int _{\mathsf X}K(x,x) d\mu (x) = \int _X \sum _{\lambda ,i} \lambda u_{\lambda ,i}(x) \overline{u_{\lambda ,i}(x)} d\mu (x) = \sum _{\lambda ,i} \lambda \int _X u_{\lambda ,i}(x) \overline{u_{\lambda , i}(x)} d\mu (x)$$

$$= \sum _{\lambda , i} \lambda = tr(T_K)\,.$$

4.15

Consider an integral operator $T_K$ on $L^2([0,2\pi ], dx)$ with integral kernel:

$$K(x, y) = \frac{1}{2\pi }\sum _{n\in {\mathbb Z}\setminus \{0\}} \frac{1}{n^2} e^{in(x-y)}\,.$$

Prove $T_K$ is a compact Hilbert–Schmidt operator of trace class.

4.16

Consider the operator $T_K$ of Exercise 4.15 and the differential operator

$$A:= -\frac{d^2}{dx^2}\,,$$

defined on smooth maps over $[0,2\pi ]$ that satisfy periodicity conditions (together with all derivatives). What is $T_K A$?

Hint. Let $\mathbf{1}$ be the constant map 1 on $[0,2\pi ]$, and $P_0: f \mapsto (\frac{1}{2\pi }\int _0^{2\pi } f(x) dx)\mathbf{1}$ the orthogonal projector onto the space of constant maps in $L^2([0,2\pi ], dx)$. Then $T_KA = I - P_0$.

4.17

Consider an integral operator $T_s$ on $L^2([0,2\pi ], dx)$ with kernel:

$$K_s(x, y) = \frac{1}{2\pi }\sum _{n\in {\mathbb N}\setminus \{0\}} \frac{1}{n^{2s}} e^{in(x-y)}\,.$$

Prove that if $Re \, s$ is sufficiently large (how large?) the following identity makes sense:

$$tr(T_s) = \zeta _R(2s)\,,$$

where $\zeta _R= \zeta _R(s)$ is Riemann’s zeta function.

4.18

Take $B\in {\mathfrak B}({\mathsf H})$ on a Hilbert space ${\mathsf H}$, and a basis N such that $\sum _{u\in N}||Bu|| <+\infty$. Prove $B\in {\mathfrak B}_1({\mathsf H})$.

Hint. Observe $|||B|\psi || = ||B\psi ||$ and $|(\psi ||B|\psi )| \le ||\psi ||\,|||B|\psi ||$.

4.19

Consider the integral operator $T_K : L^2([a,b], dx) \rightarrow L^2([a,b], dx)$

$$(T_K\varphi )(x) = \int _a^x K(x, t) \varphi (t) dt\, $$

where K is a measurable map such that $|K(x, t)|\le M$ for some $M\in {\mathbb R}$, for all $x,t \in [a, b]$, $t\le x$. Prove

$$||T^n_K|| \le \frac{M^n (b-a)^{n}}{\sqrt{(n+1)!}}\,$$

and conclude that there exists a positive integer n rendering $T_K^n$ a contraction.

Solution. In the ensuing computations $\varphi \in L^2([a,b], dx)$ implies $\varphi \in L^1([a,b], dx)$ by the Cauchy–Schwarz inequality, because the constant map 1 is in $L^2([a,b], dx)$. Define $\theta (z)=1$ for $z\ge 0$ and $\theta (z)=0$ otherwise. By construction,

$$|(T_K^n\varphi )(x)| = \int _a^b dx_1 \int ^b_a dx_2 \cdots \int _a^b dx_n \theta (x-x_1) \theta (x_1-x_2) \cdots \theta (x_{n-1}-x_n) $$

$$\times K(x, x_1)K(x_1-x_2)\cdots K(x_{n-1}, x_n)\varphi (x_n) \,.$$

Hence

$$|(T_K^n\varphi )(x)| \le M^n \int _{[a, b]^n} dx_1 \cdots dx_n |\theta (x-x_1) \cdots \theta (x_{n-1}-x_n)|\, |\varphi (x_n)| \,.$$

Using Cauchy–Schwarz on $L^2([a, b]^n, dx_1\cdots dx_n)$, and $\theta (z)^2= \theta (z) = |\theta (z)|$, we have:

$$|(T_K^n\varphi )(x)| \le M^n \sqrt{\int _{[a, b]^n} dx_1 \cdots dx_n \theta (x-x_1) \cdots \theta (x_{n-1}-x_n)}\, [b-a]^{(n-1)/2} ||\varphi ||_2\,,$$

i.e.

$$|(T_K^n\varphi )(x)| \le M^n \frac{(x-a)^{n/2}}{\sqrt{n!}} [b-a]^{(n-1)/2} ||\varphi ||_2\,.$$

Consequently

$$||T^n_K\varphi ||_2 \le \frac{M^n (b-a)^{n}}{\sqrt{(n+1)!}} ||\varphi ||_2 \,,$$

and so

$$||T^n_K|| \le \frac{M^n (b-a)^{n}}{\sqrt{(n+1)!}}\,.$$

But since:

$$\lim _{n\rightarrow +\infty } \frac{M^n (b-a)^{n}}{\sqrt{(n+1)!}} \rightarrow 0 \quad \text { as }n\rightarrow +\infty , $$

for n large enough there must exist $0<\lambda <1$ such that

$$||T^{n}_K \varphi - T^{n}_K \varphi ' ||_2 \le \lambda ||\varphi -\varphi '||_2\,, $$

making $T^n_K$ a contraction operator.

4.20

Denote by ${\mathfrak B}_D({\mathsf H})$ the family of degenerate operators on a general Hilbert space ${\mathsf H}$. This is the collection of operators $A \in {\mathfrak B}({\mathsf H})$ such that Ran(A) is finite-dimensional. Prove that:

$${\mathfrak B}_D({\mathsf H}) \subset {\mathfrak B}_1({\mathsf H}) \subset {\mathfrak B}_2({\mathsf H}) \subset {\mathfrak B}_\infty ({\mathsf H}) \subset {\mathfrak B}({\mathsf H})\,,$$

where ${\mathfrak B}_D({\mathsf H})$ is dense in ${\mathfrak B}_i({\mathsf H})$ with respect to $||\,\,||_i$, for $i=1,2, \infty$.