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The answer to your question is no, the spectrum can be infinite, and pairs of self-adjoint involutions are actually a good class of counterexamples because they can be fully described using the spectral theorem. The irreducible pairs of involutions occur in dimensions $1$ and $2$, and the rest are direct integrals of irreducibles. So every such pair is ...


2

Any normal operator $T$ gives rise to some spectral measure $E:Bor(\sigma(T))\to\mathcal{P}(H)$ which maps Borel subsets of the spectrum of $T$ into orthogonal projections in $H$. If you take Borel subset $A\subset\sigma(T)$, then $E(A)$ is called a spectral projection. Search spectral theorem on this site.


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It is often difficult to prove that an operator is self-adjoint, which seems reasonable since you know a lot about an operator when it is self-adjoint. For instance it is unitarily equivalent to a multiplication operator by a real function. There are many possible approaches, one which is sometimes useful is the following Show that $D$ is symmetric, i.e. ...


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Unfortunately, the multipliers of the Dirichlet space are not as easy to describe as of Hardy and Bergman spaces. The characterization was obtained by Stegenga in Multipliers of the Dirichlet space (free access). In addition to boundedness, it requires a Carleson-type condition in terms of capacity: $$ \iint_{\bigcup S(I_j)} |f'|^2 \le A ...


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What involution do you consider? If just complex conjugation, then $x\mapsto \overline{x}$ is not even differentiable. If you consider $f\mapsto f^*$ where $f^*(z) = \overline{f(\overline{z})}$ then it does not satisfy the C*-identity.


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The Calculus distance scale factors in spherical coordinates are $s_{r}=1$, $s_{\theta}=r$, $s_{\phi}=r\sin\theta$. So the Laplacian is $$ \Delta=\frac{1}{s_{r}s_{\theta}s_{\phi}}\left[ \frac{\partial}{\partial r}\frac{s_{\theta}s_{\phi}}{s_{r}}\frac{\partial}{\partial r} ...


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I now looked up the definition of a core of an operator and it could be that the answer actually depends on the exact definition. In Wikipedia, a core (of a closed operator) is defined as a subset $D$ of the domain of $A$ such that $A$ is the closure of $A|D$. If we replace this by requiring that $D$ be a subspace, I can prove your claim. By symmetry, it ...


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Assume that the property does not hold. This means that for each orthonormal sequence $(e_n)$, there exists a subsequence $(e_{n_k})$ such that $\lVert Te_{n_k}\rVert\to 0$. Claim. If $(v_n)$ is an orthonormal subsequence, then $\lVert Tv_n\rVert\to 0$. Indeed, suppose it is not true. There for some sequence $n_k\uparrow\infty$ and some positive ...


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It is convinient to use the following notation $\mu_{x, y}(S) := \left<P(S)x, y\right>$ and shortly $\mu_{x}$ for $\mu_{x,x}$. Let $B(\mathbb{R})$ denote the abelian C*-algebra of bounded Borel functions on $\mathbb{R}$. Define a map $\Phi_P \colon B(\mathbb{R}) \rightarrow \mathcal{L}(\mathcal{H})$ by $$\left<\Phi_P(f)x, y\right> = ...


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HINT: Imagine for a moment that an extension $T$ of $T_0$ exists and take $x\in E\setminus E_0$. Since $E_0$ is dense, you can approximate $x$ with a sequence $x_n\in E_0$. Since our imaginary operator $T$ is continuous, it must hold that $$\tag{1}Tx=\lim_{n\to \infty} T_0 x_n.$$ Now go back to reality, where $T$ does not exist yet. You need to construct it. ...


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If you start with any topological space $\Omega $ then $ C_b (\Omega) $, the set of bounded continuous functions on $\Omega $, is a C $^*$-algebra. But the Gelfand transform allows you to show that there exists a locally compact $\Omega'$ with $ C_b (\Omega)\simeq C_b (\Omega') $. So when talking in abstract, you gain nothing by considering ...


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Since \begin{align} \left\|Af \right\|_{\infty} := \sup_{x\in [0,1]} \left| \int_{0}^{x} f(t) \, \mathrm{d}t \right| \le \sup_{x\in [0,1]} \int_{0}^{x} |f(t)| \, \mathrm{d}t \le \int_{0}^{1}|f(t)|\, \mathrm{d}t =: \left\|f\right\|_{L_1} \le \left\|f\right\|_{\infty}, \end{align} then \begin{align} \left\| A \right\| &:= \sup_{f \in C^{1}[0,1]} ...



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