Tag Info

Hot answers tagged

2

A square root for an operator is not unique because square roots of complex numbers are not unique. If $A$ is a diagonal matrix on $\mathbb{C}^{n}$, then there are $2^{n}$ possible square roots for $A$ that you can spot right away. It's worse for a general Hilbert space. If $N$ is bounded and normal on a Hilbert space, then the Spectral Theorem for $N$ ...


2

Just as an example, consider the operator $\frac{d^{2}}{dx^{2}}$ on the set $\mathscr{D}$ of all twice absolutely continuous functions $f \in L^{2}[0,2\pi]$ with $f'' \in L^{2}[0,2\pi]$. Then $T_{\alpha,\beta}=\frac{d^{2}}{dx^{2}}$ is selfadjoint on the domain $\mathcal{D}(T_{\alpha,\beta})$, $0 \le \alpha,\beta < \pi$ consisting of all $f \in ...


2

The second claim is indeed true. Take $H = -i\; \log(U)$, where $\log$ is any branch of the natural logarithm that is a bounded Borel function on the unit circle.


2

Yes. Note that $f$ is a function not just of dimension but of the volume of the domain.


2

The span of $\lbrace x_n \rbrace$ isn't obviously a Banach space (it's not clear why it's closed in $E$), and, as one consequence, it's not clear to me why the restriction of $A$ to this space should be "compact" (meaning, takes bounded sets to precompact sets). To give a specific counterexample to show this is a real problem, let $H$ be a separable ...


2

(1) If $T$ is a unitary, then $\|T\| = 1$, so $\sigma(T) \subset B[0,1] \subset \mathbb{C}$. Also, if $\lambda \in \sigma(T)$, then $\overline{\lambda} \in \sigma(T^{\ast}) = \sigma(T^{-1})$. Since $$ (T-\alpha)^{-1} = -\alpha^{-1}(T^{-1} - \alpha^{-1})T^{-1} $$ and so $\overline{\lambda}^{-1} \in \sigma(T)$, whence $\sigma(T) \subset \mathbb{T}$. (2) If ...


1

For any bounded Borel function $f$, and Borel subset $S\subset\mathbb{C}$, $$ E(S)\int f(s)dE(s)x =\int_{S} f(s)dE(s)x,\;\;\; x \in H. $$ And $\int s dE(s)x = Ax$. Therefore, $\mathcal{R}(E(\{\lambda\}))\subseteq\mathcal{N}(A-\lambda I)$ follows from $$ (A-\lambda I)E(\{\lambda\})x=E(\{\lambda\})(A-\lambda ...


1

The holomorphic functional calculus allows holomorphic functions on an open neighborhood of the spectrum. Such functions can be defined independently on disjoint neighborhoods of the closed components. In particular, the holomorphic function $f$ used for defining $f(A)$ may be $0$ on an open neighborhood of one component and non-zero on another. And that's ...


1

This is a copy of my answer to original question at Physics.SE. It seems wrong that although the question has been satisfactorily answered, but is still listed on unanswered list. Then, there is the case that such an operator is defined on the full interval I assume that by "full interval" you mean the whole real line. First question: Do we then ...


1

You left out the factors $1/(2\pi i)$. For the standard unit vector $e_k$ (i.e. $e_k(k) = 1$, $e_k(n) = 0$ otherwise) we have $A e_k = \alpha_k e_k$ so $(z - A)^{-1} e_k = (z - \alpha_k)^{-1} e_k$, and $$\dfrac{1}{2\pi i}\int_\Gamma (z - A)^{-1} e_k \; dz = \left(\dfrac{1}{2\pi i}\int_\Gamma (z - \alpha_k)^{-1} \; dz \right) e_k$$ which is $e_k$ if ...


1

$$(\lambda e-A-P+P)=(\lambda e-P)(e-(\lambda e-P)^{-1}(A-P)).$$ Since $e-a$ is invertible for $\|a\|<1$ with $(e-a)^{-1}=\sum a^k$, it's enough to prove that $$\|(\lambda e-P)^{-1}(A-P)\|\le \delta\|(\lambda e-P)^{-1}\|<1.$$ By Gelfand-Neumark theorem, it's sufficient to consider $\mathcal{A}$ as a subalgebra of the space $L(H)$ of operators on some ...


1

So, I couldn't find Spectral theory and differential operators by D.E. Edmunds and W.D. Evans, which mysteriously vanished from the library. However, a search on related questions on MathOverflow led me to Semi-Fredholm operators, perturbation theory and localized SVEP by P.Aliena. The answer ot my question is on p. 131 (p.141 of the .pdf): the spectrum of ...


1

First show that if $E$ is a spectral measure then so is $UE(\cdot)U^*$. Now note that for a spectral measure $UE(\cdot)U^*$ we have $$ \left<f, \int \limits_{\sigma(N)}z \ UE(\mathrm{d}z)U^* g\right>= \int\limits_{\sigma(N)} z\ F_{f, g}(dz) \ \quad (f \in \mathcal{H}, g \in U \mathcal{D}),$$ where $F_{f, g}(\cdot):=\left<f, UE(\cdot)U^*g\right>$ ...


1

Actually you don't need to assume the condition on pure-point spectrum. The restriction of $T$ to $V$ is a self-adjoint operator from $V$ to $V$, and any self-adjoint operator on a finite-dimensional Hilbert space corresponds to a Hermitian matrix. A Hermitian matrix is diagonalizable. By the way, a consequence of this is that you can determine the ...


1

You already found a non-normal operator $N$ with spectrum $\{0\}$. Now simply note that $1+N$ will be another non-normal operator with spectrum $\{1\}$.


1

Given $Z=X+iY$ with $X$ and $Y$ self-adjoint, note that $X=\frac12(Z+Z^*)$, and $Y=\frac1{2i}(Z-Z^*)$. Take norms, apply the triangle inequality, and recall that $\|Z^*\|=\|Z\|$ to conclude that $\|X\|\leq \|Z\|$ and $\|Y\|\leq \|Z\|$.


1

By Theorem 12.35(b) of Walter Rudin's Functional Analysis, if $T\in B(H)$ is normal, then it has a polar decomposition $T=UP$, where $U$ and $P$ commute with each other and with $T$ and $U$ is unitary and $P≥0$. The spectrum of $P$ is a bounded subset of $[0,\infty)$, hence $\sqrt{z}$ can defined as a nonnegative square root of a nonnegative number. In this ...



Only top voted, non community-wiki answers of a minimum length are eligible