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For $a\in A$, $$\phi(a^*a) = \langle \pi(a^*a)h,h\rangle = ||\pi(a)h||^2\geq 0$$ Which shows that $\phi$ is a positive linear functional. By theorem 3.3.3 of Murphy's C*-algebras and operator theory, if $\{u_i\}$ is approximate unit of C*-algebra $A$, then $\|\phi\|=\lim \phi(u_i)$. Using this we have $$\|\phi\|=\lim \phi(u_i)=\langle \pi(u_i)h,h\rangle = ...


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1) In that page the author is not claiming that $\eta_{\mathscr{I}}(A)$ exists (yet), but is rather discussing what properties it should have before constructing it. 2) Note that $\eta_{\mathscr{I}}(A)$ is an operator, not a function. The idea of functional calculus is that the map $f\longmapsto f(A)$ should be a $*$-homomorphism, i.e. it should preserve ...


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You have many questions in one1), but let me only calculate the formal adjoint. In the Sobolev space $H^k=W^{k,2}(\mathbb R^n)$, $k\geq1$, one can use the inner product $$ \langle u,v\rangle_{H^k} = \sum_{|\alpha|\leq k}\langle\partial^\alpha u,\partial^\alpha v\rangle_{L^2}. $$ Compactly supported smooth functions are dense, so let us work with $u,v\in ...


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If $M$ is an orthonormal set (the normalisation is not necessary, by the way, that the elements are mutually orthogonal suffices), the set $$N(x) = \{ v\in M : \langle x,v\rangle \neq 0\}$$ is at most countable. For a finite subset $F\subset M$, we define $$x_F := \sum_{v\in F} \langle x,v\rangle\cdot v.$$ Then \begin{align} 0 &\leqslant \lVert ...


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Using the spectral theorem: A selfadjoint operator $T \ne 0$ on a Hilbert space is compact iff $$ T = \lambda_1 E_{1} + \lambda_2 E_{2} + \cdots, $$ where $\{ E_{j} \}$ is a finite or countably infinite set of disjoint orthogonal projections onto finite-dimensional subspaces, and the sequence $\{ \lambda_{j} \}$, if infinite, converges to ...


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Let $F$ be a closed subspace of $H$, such that: $E \subset F$. As $F$ is a closed subspace, we get $F^{\perp \perp} = F$. But: $E \subset F \implies F^{\perp} \subset E^{\perp} \implies E^{\perp \perp} \subset F^{\perp \perp} = F$ Edit: Proving that $F = F^{\perp \perp}$ Note that since $F$ is a closed subspace, and $H$ is a Hilbert, then $H = F \oplus ...


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The point is that we want to show that we can get $\|x_n\|$ as close to $\|x\|$ as we wish based on the assumption that we can get $\|x_n - x\|$ as close to $0$ as we wish. The bound $|\|x_n\| - \|x\|| \leq \|x_n -x\|$ guarantees this, because we can just take the limit on both sides: $$\lim_{n \to \infty} |\|x_n\| - \|x\|| \leq \lim_{n \to \infty} \|x_n - ...


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Here are some thoughts: Let $(a_n), (b_n)\in l^2$. Their inner product is given by: $$ \langle (a_n), (b_n) \rangle = a_1\bar{b_1} + a_2\bar{b_2} + \cdots $$ Hence, $$ \langle T(a_n), (b_n) \rangle = \alpha_1a_1\bar{b_1} + \alpha_2a_2\bar{b_2} + \cdots $$ The defining property of $T^*$ is that it satisfies: $$ \langle T(a_n), (b_n) \rangle = \langle (a_n), ...


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Assuming you're working on a complex Hilbert space, the range condition and the following condition are equivalent to dissipative: $$ \Re(Ax,x) \le 0,\;\;\; x \in \mathcal{D}(A). $$ This last condition persists if you add another bounded operator $C$ satisfying the same condition. In your case, because $D$ is bounded, $$ ...


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You only need to use that The sum and product of two such operators correspond to the sum and product of the corresponding functions. That said, for any operator $A$, the square of the operator $\eta_\mathscr I(A)$ equals to $\eta_\mathscr I^2(A)$ -- whatever it will mean --, but as a real (or complex) function, we have $\eta_\mathscr I^2=\eta_\mathscr ...


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What does it mean that an eigenvalue is "isolated"? An eigenvalue is isolated if it is an isolated point of the spectrum, i.e., it has a neighborhood in which there are no other points of the spectrum. This is a stronger property than not having other eigenvalues around. The claim is that the equation $ (H_0-\omega_0)X=\Psi $ has a solution for all ...


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The metric you described is the standard metric on the projective space: in the real case it can be visualized as the angle between lines (thinking of the elements as lines). It arises as the quotient of the spherical metric on $S^n$ by the group of isometries $\{x\mapsto \alpha x, \ |\alpha|=1\}$ where $\alpha$ belongs to the ground field, $\mathbb{R}$ or ...



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