Resolvent identities, analyticity and resolvent bounds

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Problem 10B.6 from Axler S. "Measure, Integration & Real Analysis" [Axl19]More problems in Section 10B "Spectrum"

(a)(b)

Suppose $T$ is a bounded operator on a complex nonzero Banach space $V$ .

(a)

Prove that the function

\alpha \mapsto \varphi \left((T-\alpha I)^{-1} f\right)

is analytic on $\mathbb {C} \backslash \operatorname {sp}(T)$ for every $f \in V$ and every $\varphi \in V^{\prime }$ .

(b)

Prove that $\operatorname {sp}(T) \neq \varnothing$ .

Problem 11.7 from Brayman V.B., Chaikovskyi A.V., Konstantinov O.Y., Kukush A.G., Mishura Y.S., Nesterenko O.N. "Functional Analysis and Operator Theory" [BCK+24]More problems in Section 11.2 "Problems to Solve"

Let $A \in \mathcal{L}(X)$ . Prove that for $\left|\lambda \right| > \left\| A\right\|$ the inequality $\left\| R_{\lambda }(A)\right\| \leq (\left|\lambda \right|- \left\| A\right\| )^{-1}$ holds.

Problem 11.8 from Brayman V.B., Chaikovskyi A.V., Konstantinov O.Y., Kukush A.G., Mishura Y.S., Nesterenko O.N. "Functional Analysis and Operator Theory" [BCK+24]More problems in Section 11.2 "Problems to Solve"

Let $X$ be a Banach space, $A \in \mathcal{L}(X), \lambda \in \rho (A)$ . Prove that $A$ and $R_{\lambda }(A)$ commute.

Problem 11.45 from Brayman V.B., Chaikovskyi A.V., Konstantinov O.Y., Kukush A.G., Mishura Y.S., Nesterenko O.N. "Functional Analysis and Operator Theory" [BCK+24]More problems in Section 11.2 "Problems to Solve"

Let $X$ be a Banach space, $A, B \in \mathcal{L}(X)$ . Prove that for all $\lambda \in \rho (A) \cap \rho (B)$ the following identity holds:

(A-\lambda I)^{-1}-(B-\lambda I)^{-1} = (A-\lambda I)^{-1}(B-A)(B-\lambda I)^{-1}.

Problem 12.18 from Brayman V.B., Chaikovskyi A.V., Konstantinov O.Y., Kukush A.G., Mishura Y.S., Nesterenko O.N. "Functional Analysis and Operator Theory" [BCK+24]More problems in Section 12.2 "Problems to Solve"

(1)(2)

Let $A$ be a self-adjoint compact operator in a Hilbert space $H, A = \sum_{n = 1}^{\infty } \lambda_{n}\left(\cdot , \varphi_{n}\right) \varphi_{n}$ be its Schmidt expansion. Prove that

(1)

$\forall k \geq 1: A^{k} x = \sum_{n = 1}^{\infty } \lambda_{n}^{k}\left(x, \varphi_{n}\right) \varphi_{n}, x \in H$ ;

(2)

$\forall \lambda \in \mathbb {C} \backslash \left\{ 0, \lambda_{n}, n \geq 1\right\}$ :

R_{\lambda }(A) y = \sum _{n = 1}^{\infty } \frac{\left(y, \varphi _{n}\right)}{\lambda _{n}-\lambda } \varphi _{n}-\frac{1}{\lambda } P y = \frac{1}{\lambda }\left(\sum _{n = 1}^{\infty } \frac{\lambda _{n}\left(y, \varphi _{n}\right)}{\lambda _{n}-\lambda } \varphi _{n}-y\right), y \in H

where $P$ is the orthogonal projector onto $\operatorname {Ker} A$ .

Problem 12.4 from Бородин П.А., Савчук А.М., Шейпак И.А. "Задачи по функциональному анализу" [БСШ17]More problems in Section 12.1 "Spectrum"

R_{\lambda }(A)-R_{\mu }(A) = (\lambda -\mu ) R_{\lambda }(A) R_{\mu }(A).

Using this identity, prove that

R_{\lambda }(A) R_{\mu }(A) = R_{\mu }(A) R_{\lambda }(A)

Problem 12.5 from Бородин П.А., Савчук А.М., Шейпак И.А. "Задачи по функциональному анализу" [БСШ17]More problems in Section 12.1 "Spectrum"

Let $\lambda_{0}, \lambda \in \rho (A)$ . Prove that if $\left|\lambda -\lambda_{0}\right| < \frac{1}{\left\| R_{\lambda_{0}}(A)\right\| }$ , then

\left\| R_{\lambda }(A)-R_{\lambda _{0}}(A)\right\| \leqslant \frac{\left|\lambda -\lambda _{0}\right| \cdot \left\| R_{\lambda _{0}}^{2}(A)\right\| }{1-\left|\lambda -\lambda _{0}\right| \cdot \left\| R_{\lambda _{0}}(A)\right\| }

Problem 12.6 from Бородин П.А., Савчук А.М., Шейпак И.А. "Задачи по функциональному анализу" [БСШ17]More problems in Section 12.1 "Spectrum"

(a)(b)

Prove that $R_{\lambda }(A)$ is

(a)

a continuous;

(b)

an analytic function of the variable $\lambda$ on the domain $\rho (A) \cup \left\{ \infty \right\}$ with values in $\mathcal{B}(X)$ . Prove that if $\lambda_{0} \in \rho (A)$ , then for $\left|\lambda -\lambda_{0}\right| < \frac{1}{\left\| R_{\lambda_{0}}(A)\right\| }$ the resolvent admits the representation

R_{\lambda }(A) = \sum _{k = 0}^{\infty }\left(\lambda -\lambda _{0}\right)^{k}\left(R_{\lambda _{0}}(A)\right)^{k+1}

and for $\left|\lambda \right| > r(A)$

R_{\lambda }(A) = -\frac{1}{\lambda } \sum _{n = 1}^{\infty } \frac{A^{n}}{\lambda ^{n}}

Problem 12.45 from Бородин П.А., Савчук А.М., Шейпак И.А. "Задачи по функциональному анализу" [БСШ17]More problems in Section 12.1 "Spectrum"

Let $X$ be a normed space, $A, B \in \mathcal{B}(X)$ . Prove that if $A$ and $B$ commute, then $R_{\lambda }(A)$ commutes with $B$ for all $\lambda \in \rho (A)$ . Prove that if $B$ and $R_{\lambda }(A)$ commute for some $\lambda \in \rho (A)$ , then $A$ and $B$ commute.

Problem 86 from Halmos P.R. "A Hilbert Space Problem Book" [Hal82]More problems in Chapter 11 "Spectral Radius"

The resolvent of every operator is analytic at each point of its domain, and at $\infty$ ; its value at $\infty$ is (the operator) $0$ .

Remark.

Suppose that $A$ is an operator on a Hilbert space $\mathbf{H}$ . If $\lambda$ does not belong to the spectrum of $A$ , then the operator $A - \lambda$ is invertible; write $\rho (\lambda ) = (A-\lambda )^{-1}$ . (When it is necessary to indicate the dependence of the function $\rho$ on the operator $A$ , write $\rho = \rho_A$ .) The function $\rho$ is called the resolvent of $A$ . The domain of $\rho$ is the complement of $\operatorname {spec} A$ ; its values are operators on $\mathbf{H}$ .

Consider, quite generally, functions $\varphi$ whose domains are open sets in the complex plane and whose values are operators on $\mathbf{H}$ . Such a function $\varphi$ is called analytic if, for each $f$ and $g$ in $\mathbf{H}$ , the numerical function $\lambda \mapsto \left\langle \varphi (\lambda )f, g \right\rangle$ (with the same domain as $\varphi$ ) is analytic in the usual sense. (This concept is sometimes called weak analyticity.) In case the function $\psi$ defined by $\psi (\lambda ) = \varphi (1/\lambda )$ can be assigned a value at the origin so that it becomes analytic there, then (just as for numerical functions) $\varphi$ is called analytic at $\infty$ , and $\varphi$ is assigned at $\infty$ the value of $\psi$ at $0$ .

Problem 642 from Kirillov A.A., Gvishiani A.D. "Теоремы и задачи функционального анализа" [KiGv88]More problems in Subsection V.1.2 "Functions of Bounded Self-Adjoint Operators"

Compute explicitly the resolvent of the Volterra operator in Problem 641.

Problem 710 from Kirillov A.A., Gvishiani A.D. "Теоремы и задачи функционального анализа" [KiGv88]More problems in Subsection V.2.2 "The Spectral Theorem"

Suppose that $A$ is a bounded selfadjoint operator with spectral measure $\lambda$ . Prove the equality

\operatorname {s-lim}_{\varepsilon \to 0}\frac\varepsilon \pi \int _a^b[(a-\lambda )^2+\varepsilon ^21]^{-1}\, d\lambda =\frac12\lambda \left(\left\{ a\right\} \right)+\lambda \left((a,b)\right)+\frac12\lambda \left(\left\{ b\right\} \right) =\frac12\lambda \left([a,b]\right)+\frac12\lambda \left((a,b)\right).

Problem 119 from Le A. "Functional Analysis Problems with Solutions" [Le13]More problems in Chapter 8 "Bounded Operators on Banach Spaces and Their Spectra"

Let $X$ be a Banach space and $A, B \in \mathcal{B}(X)$ . Show that for any $\lambda \in \rho (A) \cap \rho (B)$ ,

R(B ; \lambda )-R(A ; \lambda ) = R(B ; \lambda )(A-B) R(A ; \lambda )

Problem 2.18.11 from Rudin W. "Real and Complex Analysis" [Rud87]More problems in Chapter 2.18 "Elementary Theory of Banach Algebras"

The resolvent $R(\lambda , x)$ of an element $x$ in a Banach algebra with unit is defined as

R(\lambda , x) = (\lambda e-x)^{-1}

for all complex $\lambda$ for which this inverse exists. Prove the identity

R(\lambda , x)-R(\mu , x) = (\mu -\lambda ) R(\lambda ) R(\mu )

and use it to give an alternative proof of Theorem 18.5.

Problem 5.1.41 from Steeb W., Hardy Y. "Problems and Solutions in Introductory and Advanced Matrix Calculus" [StHa17]More problems in Section 5.1 "Chapter Problems"

Let $H, H_{0}, V$ be $n \times n$ matrices over $\mathbb {C}$ and $H=H_{0}+V$ . Let $z \in \mathbb {C}$ and assume that $z$ is chosen so that $(H_{0}-zI_{n})^{-1}$ and $(H-zI_{n})^{-1}$ exist. Show that

(H-zI_{n})^{-1} = (H_{0}-zI_{n})^{-1} - (H_{0}-zI_{n})^{-1}V(H-zI_{n})^{-1}.

This is called the second resolvent identity.

Problem 25.1.3 from Steeb W., Hardy Y. "Problems and Solutions in Introductory and Advanced Matrix Calculus" [StHa17]More problems in Section 25.1 "Chapter Problems"

Let $A$ be an $n \times n$ matrix over $\mathbb {C}$ . If $\lambda$ is not an eigenvalue of $A$ , then the matrix $(A - \lambda I_{n})$ has an inverse, namely the resolvent

R_{\lambda } = (A - \lambda I_{n})^{-1}.

Let $\lambda_{j}$ be the eigenvalues of $A$ . For $\left|\lambda \right| \geq a$ , where $a$ is any positive constant greater than all the $\left|\lambda_{j}\right|$ the resolvent can be expanded as

R_{\lambda } = -\frac{1}{\lambda }\left(I_{n} + \frac{1}{\lambda }A + \frac{1}{\lambda ^{2}}A^{2} + \cdots \right).

Calculate

-\frac{1}{2\pi i}\oint _{\left|\lambda \right|=a} \lambda ^{m} R_{\lambda } \, d\lambda , \qquad m = 0, 1, 2, \ldots .

References

[Axl19] Axler S. Measure, Integration & Real Analysis. 3rd ed., Springer, 2019.

[BCK+24] Brayman V.B., Chaikovskyi A.V., Konstantinov O.Y., Kukush A.G., Mishura Y.S., Nesterenko O.N. Functional Analysis and Operator Theory. Springer, 2024.

[БСШ17] Бородин П.А., Савчук А.М., Шейпак И.А. Задачи по функциональному анализу. МЦНМО, 2017.

[Hal82] Halmos P.R. A Hilbert Space Problem Book. 2nd ed., Springer-Verlag, 1982.

[KiGv88] Kirillov A.A., Gvishiani A.D. Теоремы и задачи функционального анализа. 2-е изд., Наука, 1988.

[Le13] Le A. Functional Analysis Problems with Solutions. 2013.

[Rud87] Rudin W. Real and Complex Analysis. 3rd ed., McGraw-Hill, 1987.

[StHa17] Steeb W., Hardy Y. Problems and Solutions in Introductory and Advanced Matrix Calculus. 2nd ed., World Scientific Publishing, 2017.