# Tag Info

What I am going to say isn't anything new, but sometimes it helps to look more abstractly to see the big picture. Consider a linear map $A : \mathcal{D}(A) \subseteq X \rightarrow Y$ where $X$ and $Y$ are Banach spaces. The graph of $A$ is $\mathscr{G}(A)=\{ (x,Ax) \in X \times Y : x \in \mathcal{D}(A) \}$. The graph is a linear subspace of $X\times Y$. ...
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 $\eta_\mathscr I^2=\eta_\mathscr ... 2 1) In that page the author is not claiming that$\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)$should be a$*$-homomorphism, i.e. it should preserve ... 3 Now that you ask the question, I have to admit that I had to think about it for a little bit. It seems that most courses/books focus more on the magnitude portion of the Fourier Transform, and tend to neglect the phase spectrum despite its overwhelming importance. Most of my examples are drawn from electrical engineering and signal processing, so hopefully ... 1 No. If$\lim_{n} \|A\psi_{n}\|=0$implies$\lim_{n}\|\psi_{n}\|=0$, then the inverse of$A$is continuous. It's easy to construct a counterexample. For example, on$L^{2}[0,\infty)$, $$Af = xf,\;\;\; f \in \mathcal{D}(A)=\{ f \in L^{2} : xf \in L^{2} \}.$$ In this case,$\psi_{n}=\sqrt{n}\chi_{[0,1/n]}$is a unit vector in$L^{2}[0,\infty)$, and $$... 0 By definition, the graph of \overline{A} is the closure of the graph of A. That means (x,y) is in the graph of \overline{A} iff there is a sequence x_n \in D(A) such that ... 0 The easiest proof I have seen relies on knowing the dual space C[a,b]^{\star} of C[a,b]. The characterization of this dual was one of the oldest results in functional Analysis, due to one of the Riesz brothers and probably proved in a fairly horrible way. Here's a simple way using Hahn-Banach: Theorem: Let \Phi be a continuous linear functional on ... 0 Let \varphi = d\mu/d\gamma be the Radon-Nikodym derivative, then for any f \geq 0 we have$$ \int fd\gamma = \int f\varphi d\mu \qquad (\ast) $$In addition to absolute continuity, you also need the fact that$$ \varphi \in L^{\infty}(\mu) $$The proof is in the following steps : - Since \mu and \gamma are both positive measures, \varphi \geq 0 ... 0 Hint: Consider the polynomial f(X)=X^2. 0 Hilbert-Schmidt operators are compact. The resolvent (\Delta-\lambda I)^{-1} is not compact and \Delta is not compact. Otherwise you would end up discrete spectrum, which you do not have. The spectral resolution of the identity for -\Delta is not as simple as f=\int_{0}^{\infty}(f,\phi_{\lambda})\phi_{\lambda}d\lambda because there are so many ... 1 You should use Fourier transform. Taking the Fourier transform we see that is equivalent to$$(-\lvert\xi\rvert^2 +z)\hat{u}=\hat{g}, $$which has the unique formal solution$$\tag{2}\hat{u}=-\frac{\hat{g}}{-z+ \lvert\xi\rvert^2}.$$Now you have to prove that this formal solution make sense in H^2 if and only if z\leq 0. 1 Any strongly regular graph that is not vertex transitive provides examples. If G is strongly regular, all subgraphs obtained by deleting one vertex have the same spectrum. The smallest such strongly regular graphs are on 25 vertices, and can be constructed from Latin squares of order five. Why are these subgraphs cospectral. If a\in V(G), then the ... 1 I hope its ok to answer my own question (the reduced one which i labelled with 'Edit'). I would like to know if the following elementary argument works. The spectral theorem tells us that there is a projection-valued measure P such that A = \int \lambda dP(\lambda). For a fixed vector x this gives a finite Borel measure, defined by \mu_x(\Omega) = ... 1 To see that S - \lambda I has trivial kernel for \lambda \ne 0, let x \in \ell^2 with Sx = \lambda x, we have$$ \lambda x_j = (Sx)_j = \frac{1}{j}\cdot x_{j+1} $$That is, for any j, we have$$ x_j = (j-1)!\lambda^{j-1} x_1 $$As x \in \ell^2, we have$$ \|x\|^2_2 = \sum_{j=1}^\infty \bigl((j-1)!\lambda^{j-1}\bigr)^2 x_1^2 < \infty $$But ... 1 As already mentioned,$$ (\spadesuit) \qquad C(\Bbb{T}) \rtimes_{\alpha} \Bbb{Z}_{2} \cong \{ f \in C([0,1] \to {\text{M}_{2}}(\Bbb{C})) \mid \text{$ f(0) $and$ f(1) $are diagonal} \}. $$Hence, by the definition of a continuous field of C^{*} -algebras, C(\Bbb{T}) \rtimes_{\alpha} \Bbb{Z}_{2} is a continuous field of C^{*} -algebras over the ... 1 Usually the point spectrum consists only of the eigenvalues of T, that is,$$\sigma_p(T) = \{\lambda\in\mathbb{C}\ |\ T-\lambda I\text{ is not injective }\}.$$Notice how this corresponds to the eigenvalues of T, since when T-\lambda I is not injective, there exists some x\neq 0, such that (T-\lambda I)x=0, or Tx=\lambda x. My definition of the ... 3 The difference is due to the fact that Kreyszig consider the spectrum of unbounded operators. So for \lambda in the continuous or residual spectrum, R_\lambda exists (as an unbounded operator). The usual approach in functional analysis is for bounded operators, where you wouldn't allow for R_\lambda to be unbounded, and so you would say it doesn't ... 2 A regular graph has its valency as its largest eigenvalue, if the graph is not regular, the largest eigenvalue is less than the valency. But this is practically the only thing that distinguishes the spectral theory of regular graphs from that of general graphs. In particular there is no formula of any sorts for the eigenvalues of regular graphs in general. ... 1 You prove it for monomials, then polynomials, then continuous functions. Then you can use Luzin to show that if f is bounded Borel, then f (A) is a wot limit of a net f_j (A) with f_j continuous. 2 Let's first see if it has any eigenvalues. Let x = (x_1,x_2,\ldots)\in H (in terms of the basis \{u_n\} given above) and suppose that it is an eigenvector, then$$Ax = \mu x \Longrightarrow (\lambda x_1 - x_2,\lambda x_2-x_3,\ldots) = \mu (x_1,x_2,\ldots).$$Then \lambda x_i - x_{i+1} = \mu x_i, i.e. x_{i+1} = (\lambda - \mu)x_i. Inductively, this ... 1 The spectral measure is a discrete measure concentrated on \sigma(A)=\{ \lambda_1,\cdots,\lambda_{m}\}, and given by$$ P(S) = \sum_{\{ j : \lambda_{j} \in S\}}P_{j}. $$The projections are a mutually orthogonal partition of unity:$$ I = P_{1}+\cdots+P_{m}, \\ P_{j}=P_{j}^{\star}=P_{j}^{2}, \\ ... 1 1) You have, when$A=\sum_j\lambda_jP_j$, that $$P_A(\Omega)=\int1_\Omega(\lambda)\,dP_A(\lambda)=1_\Omega(A)=\sum_j1_\Omega(\lambda_j)P_j.$$ In particular $$\mu_\psi(\{\lambda\})=\langle\psi,P_A(\{\lambda\})\psi\rangle=\sum_j1_{\{\lambda\}}(\lambda_j)\langle\psi,P_j\psi\rangle=\sum_j\delta(\lambda-\lambda_j)\langle P_j\psi,P_j\psi\rangle\\ ... 0 Assuming:$$ A(f)(t)=\int_{0}^{t^2}f(s)\,ds = \int_{0}^{t}2u\, f(u^2)\,du=\lambda\cdot f(t) \tag{1}$$we may look for first for analytic solutions in a neighbourhood of zero. If$$ f(z) = \sum_{n\geq 1} a_n\, z^{2n}\tag{2}$$then:$$ A(f)(z) = \sum_{n\geq 1} \frac{a_n}{n+1} z^{2n+2} = \sum_{n\geq 2}\frac{a_{n-1}}{n} z^{2n} \tag{3}$$so the only analytic ... 0 Since D = \lambda I = \lambda U U^\top, we have$$ (D+A)^{-1} = (U (\lambda I + \Sigma) U^\top)^{-1} = U (\lambda I + \Sigma)^{-1} U^\top.$\$