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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 ...


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Since $D = \lambda I = \lambda U U^\top$, we have $$ (D+A)^{-1} = (U (\lambda I + \Sigma) U^\top)^{-1} = U (\lambda I + \Sigma)^{-1} U^\top. $$


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Spectral theory in infinite-dimensional spaces is quite a bit more complicated than in the finite-dimensional case. In particular, we have to distinguish between the spectrum $\sigma(A)$ of an operator and its eigenvalues. Let $A$ be a linear operator on a Banach space $X$ over the scalar field $C$. We have $$ \sigma(A) = \{ \lambda \in C: (\lambda I - A) ...


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The multiplication operator $(Mf)(x)=xf(x)$ on $L^{2}[0,1]$ is a classical example of an operator with no eigenvalues, but its spectrum is $[0,1]$. $M$ has no eigenvalue because $Mf=\lambda f$ gives $(x-\lambda)f=0$, which forces $f(x)=0$ a.e.. To see that $[0,1]\in\sigma(M)$, note that the constant function $1$ is not in the range of $(M-\lambda I)$ for ...


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I'll give here another proof which makes the (admittedly, strong) additional assumption that the operators are compact and self-adjoint. For such operators, we have a very useful tool in the Courant-Fischer min-max principle, which we will use here in the form $$ \lambda_k(A) = \min_{\dim V=k-1} \max_{x \in V^\bot, \|x\|=1} \langle A x, x \rangle. $$ Here ...


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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 ...


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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 ...


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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 + ...


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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, $E\{\lambda\}H \subseteq \mbox{ker}(u-\lambda I)$. ...


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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. ...


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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.


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What about the following generalisation (it works, but is it useful?) ? Let's deal with polynomials of the form $P(A,B,z)=A-Bz$ first and let's give a modified definition to the "generalised resolvent set" $\rho(A,B)$: \begin{equation} \left\{z\in \mathbb{C}\left| \right. \left(A-zB\right)^{-1}\text{,}B\left(A-zB \right)^{-1}\text{exist on a dense (common) ...


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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} ...


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For $\lambda \notin \{0\} \cup \sigma_p(T)$, you can explicitly write down $(\lambda I - T)^{-1}$.



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