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Assume $\lambda$ is a non-zero eigenvalue. Since $T$ is seladjoint we may assume that $\lambda$ is real. We have some non-zero eigenvector $u\in L_2[0,1]$, such that $$\int_0^x \left(\int_t^1 u(s)ds\right)dt=\lambda u(x)\tag{1}$$ Note that $u(0)=0$. Since left hand side of the equation is absolutely continuous as any Lebesgue integral, then $u\in AC[0,1]$ ...

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That has to be some of the most awkward notation for the Spectral Theorem that I've seen. The author could state that $E$ is the spectral measure for $a$, thereby eliminating $a$ from the decorations added to $E$. Who decorates the eigenfunctions or eigenvalues for an operator with the name of that operator, unless there is some special reason to do so? The ...

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Put $B=A^*A$ which is a Hermite matrix. As a linear transformation of Euclidean vector space $E$ is Hermite iff there exists an orthonormal basis of E consisting of all the eigenvectors of $B$ Let $\lambda_1,...,\lambda_n$ be the eigenvalues of $B$ and $\left \{ e_1,...e_n \right \}$ be an orthonormal basis of $E$ Let $x=a_1e_1+...+a_ne_n$ we have $\left ... 2 Question: does there exist such a$b$with a spectrum$\sigma(b)$also contained in the positive reals$\mathbb{R}_{>0}$? Yes, that's an important point that you have a positive square root of a positive element. You take the principal branch of the square root, defined in$\mathbb{C}\setminus (-\infty,0]$, with positive real part to construct it. 1 Too long to be a comment. Spectrum of a non-closed operator is not defined. If A is not closed, then$A−z, z\in\mathbb C$is never closed and$(A−z)D(A)$cannot be equal to$H$. So far you have shown that$z\notin \overline{N(A)}$implies that$(A-z)^{-1}$is bounded on the range of$(z-A)$which is closed. But the$Ran(z-A)$need not to be the whole$H.$... 1 Spectral theorem for a self-adjoint operator is often formulated as$a=\int\lambda dE_\lambda,$where$E_\lambda:\mathbb R\to B(H)$is the spectral resolution of$a=a^*\in B(H).$In your book$E(\cdot\ ;a)$is the spectral measure (function of Borel subsets$\mathbb R$) associated with$a.$The mapping$\lambda\mapsto E_\lambda=E((-\infty,\lambda];a)$is now ... 1 Either there should be an additional hypothesis, or you should be allowed to choose the eigenvectors cleverly in the case where several eigenvalues are equal. As it stands, there are easy counterexamples: Let$A$be the zero matrix or the identity matrix. then the$x_i$'s and$y_j$'s could be anything, and in particular you could have unwanted ... 1 Solved it. :-) I shall post it here on the off chance somebody else ever has the same problem. Conversely, suppose$(T-\lambda)v=0$. We wish to show that$E_{\{\lambda\}}v=v$. This is achieved by proving that the following representation of the spectral projector$E_{\{\lambda\}}$is valid (it is also known as Riesz's formula): If$\gamma$is a simple, ... 1 The adjoint is defined by $$T^{*}(x_1, x_2, \ldots, x_{2n}, \ldots ) = (0, x_1, \ldots, 0, x_n 0, x_{n+1} \ldots),$$ that is,$x_k \mapsto 0$if$k$is odd,$x_k \mapsto x_{\frac{k}{2}}$if$k$is even. Then you can see that$\left<x, T^*y \right> = \sum_{n=1} x_n z_n$, where$z_n = \left\{ \begin{array}{ll}0 &\mbox{ for } n \mbox{ odd} \\ ...

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