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3

You assumed that $T_\lambda$ is surjective (when applying bounded inverse theorem). It is not true. In fact a small modification of your second argument shows precisely that $T_\lambda$ cannot be surjective.


2

One can show that $T$ is a compact operator: define $S_n:\ell^2(\mathbb{C}) \to \ell^2(\mathbb{C})$ by $$(S_nx)_m = \left\{ \begin{matrix} \frac{x_m}{m} & m \leq n \\ 0 & m > n \end{matrix} \right. $$ Note that the $S_n$ are finite-rank and that $$(T-S_nx)_m = \left\{ \begin{matrix} 0 & m \leq n \\ \frac{x_m}{m} & m > n \end{matrix} ...


2

This is a little too long for a comment. I wonder if the standard Picard iteration would work. If you want a solution $u$ with $u(0)=A$ and $u'(0)=B$, then $$ \begin{align} u'(x) & = B+\int_{0}^{x}(q(x_1)-\lambda)u(x_1)dx_1, \\ u(x) & = A+Bx+\int_{0}^{x}\int_{0}^{x_2}(q(x_1)-\lambda)u(x_1)dx_1 dx_2 \\ & = A+Bx + ...


2

Yes, it follows from the following general theorem (see Dixmier "$C^*$-algebras and representations", 2.10.2): Every irreducible representation $\rho$ of a $C^*$-subalgebra $C$ of a $C^*$-algebra $A$ can be continued to an irreducible representation $\pi$ of $A$ in a possibly larger Hilbert space. You can also apply the Lemma 2.10.1 from Dixmier directly. ...


1

Here is another partial answer. From the book by Kato, "Perturbation theory for linear operators", this is Theorem 4.10 in Chapter 5 (p. 291); I'm paraphrasing a bit: Let $T$ be selfadjoint and $A$ be selfadjoint and bounded operators in a Hilbert space. Then $$ \operatorname{dist}(\Sigma(T + A), \Sigma(T)) \le \| A \|. $$ Here $\Sigma(T)$ denotes the ...


1

After integrating by parts, you get $$ \langle f,Bg\rangle =-\int_0^1\int_0^x\overline{f(t)}g(x)\,dt\,dx +\big[\int_0^xg(t)\,dt \,\int_0^x\overline{f(t)}\,dt\big]_0^1 $$ Continuing this calculation (inserting the limits, noting that $x=0$ gives nothing, and renaming the integrating variable in the $g$ integral), we get $$ \begin{aligned} \langle f,Bg\rangle ...


1

It's not quite stated precisely. This should be more-or-less in Kato. The spectral projection for the isolated eigenvalue $\lambda$ is $$ P = \dfrac{1}{2\pi i} \oint_\Gamma (z I-A)^{-1}\; dz $$ where $\Gamma$ is a small circle centred at $\lambda$. By assumption, this is a projection of finite rank. It is the limit (in operator norm) of the ...


1

Starting with the comment, $$ E(S)\left(\int f(\mu)dE(\mu)\right)=\left(\int f(\mu)dE(\mu)\right) E(S)=\int_{S}f(\mu)dE(\mu). $$ Let $f(\mu)=\mu-\lambda$ and let $S$ be the singleton set $\{\lambda\}$. Then the second equality gives $$ (u-\lambda I)E\{\lambda\} = 0. $$ So, as you deduced, $E\{\lambda\}H \subseteq \mbox{ker}(u-\lambda I)$. ...


1

Lanczos and randomized SVD are both able to compute a rank-$r$ truncated SVD based on application of the underlying matrix to $O(r)$ vectors. Neither require the actual matrix to be formed.



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