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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 $$\|\phi\|=\lim \phi(u_i)=\langle \pi(u_i)h,h\rangle = ... 2 1) In that page the author is not claiming that \eta_{\mathscr{I}}(A) exists (yet), but is rather discussing what properties it should have before constructing it. 2) Note that \eta_{\mathscr{I}}(A) is an operator, not a function. The idea of functional calculus is that the map f\longmapsto f(A) should be a *-homomorphism, i.e. it should preserve ... 2 You have many questions in one1), but let me only calculate the formal adjoint. In the Sobolev space H^k=W^{k,2}(\mathbb R^n), k\geq1, one can use the inner product$$ \langle u,v\rangle_{H^k} = \sum_{|\alpha|\leq k}\langle\partial^\alpha u,\partial^\alpha v\rangle_{L^2}. $$Compactly supported smooth functions are dense, so let us work with u,v\in ... 2 If M is an orthonormal set (the normalisation is not necessary, by the way, that the elements are mutually orthogonal suffices), the set$$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 ... 2 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, $$where \{ E_{j} \} is a finite or countably infinite set of disjoint orthogonal projections onto finite-dimensional subspaces, and the sequence \{ \lambda_{j} \}, if infinite, converges to ... 1 Let F be a closed subspace of H, such that: E \subset F. As F is a closed subspace, we get F^{\perp \perp} = F. But: E \subset F \implies F^{\perp} \subset E^{\perp} \implies E^{\perp \perp} \subset F^{\perp \perp} = F Edit: Proving that F = F^{\perp \perp} Note that since F is a closed subspace, and H is a Hilbert, then H = F \oplus ... 1 The point is that we want to show that we can get \|x_n\| as close to \|x\| as we wish based on the assumption that we can get \|x_n - x\| as close to 0 as we wish. The bound |\|x_n\| - \|x\|| \leq \|x_n -x\| guarantees this, because we can just take the limit on both sides:$$\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: $$\langle T(a_n), (b_n) \rangle = \langle (a_n), ... 1 Assuming you're working on a complex Hilbert space, the range condition and the following condition are equivalent to dissipative:$$ \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 $\eta_\mathscr I^2=\eta_\mathscr ... 1 What does it mean that an eigenvalue is "isolated"? An eigenvalue is isolated if it is an isolated point of the spectrum, i.e., it has a neighborhood in which there are no other points of the spectrum. This is a stronger property than not having other eigenvalues around. The claim is that the equation$ (H_0-\omega_0)X=\Psi $has a solution for all ... 1 The metric you described is the standard metric on the projective space: in the real case it can be visualized as the angle between lines (thinking of the elements as lines). It arises as the quotient of the spherical metric on$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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