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2

It appears you saying that you want to assume that $1=1'$. Also, presumably $A$ and $A'$ sit together within a larger algebra (perhaps you want to think of them as subalgebras of the same $B(H)$?). Let $B$ be the $C^*$-algebra generated by $A$ and $A'$. Then the identity $1$ for $A$ and $A'$ is also the identity for $B$. Thus you know that the spectra ...


2

The equation must be considered in the context of the Hilbert space $L^{2}(-1,1)$ because the filter of requiring eigenfunctions to be in $L^{2}$ is what eliminates the non-regular solutions, and it is what determines the eigenvalues. For $m = 1,2,3,\cdots$, the operators $$ L_{m}f = -\frac{d}{dx}(1-x^{2})\frac{d}{dx}f + \frac{m^{2}}{1-x^{2}}f ...


0

First suppose that $H$ is separable. Fix an orthronormal basis $\{e_n\}$ of $H$. Fix a countable set $B$ of $A$ and enumerate the elements of $B$ by $\lambda_n$. Consider the operator $M\colon H\to H$ defined by $Me_n = \lambda_n e_n$. Show that the spectrum of $M$ is in fact $A$. Then modify this to work for any infinite-dimensional Hilbert space.


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If $H$ is separable, then $H$ is isomorphic to $L^2(A,\mu)$, where $A$ is the given compact set, and $\mu$ a regular probability measure. So we will construct an example here. Let $f$ be a function which is $0$ everywhere outside $A$, and is non zero inside $A$. Then $M_f$ has spectrum equal to $A$.


1

Since $\lVert L\rVert = 1$, it follows that $L-\lambda I$ is invertible for all $\lambda$ with $\lvert\lambda\rvert > 1$, we can see that using the Neumann series. For $\lvert \lambda\rvert > \lVert L\rVert$, the series $$\sum_{n=0}^\infty \lambda^{-n}L^n$$ is absolutely convergent, and since $B(H)$ is a Banach space, it is convergent. One computes ...


1

There is a general method for finding sequences which satisfy linear recurrent relations with constant coefficients. In this case, we have the relation $$ x_{n+1}+x_{n-1} = \mu x_n \text{ for } n\geq 1. \text{ (*)} $$ Here is how we solve it: consider all the sequences (not nesessarily from $l^2$) which satisfy (*). They form a linear space (it's easy to ...


0

Using the formula $\sigma(A)=\lim_{j\to\infty}\lVert A ^j\rVert^{1/j}$ (where $\lVert\cdot\rVert$ is a consistent norm), we can show that $\sigma(\mathcal A) \leqslant 1$. If $\sup_{A\in\mathcal A}\rho(A)\lt 1$, then $\sigma(\mathcal A)=0$. If it is not the case, then we may have $\sigma(\mathcal A)=1$, for example if $\mathcal A=\{(1-1/n)I,n\geqslant ...


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See Corollary 6.9(1), p242 in Invitation To Operator Theory (here).


1

First, in general case the spectrum consists of much more than eigenvalues: the resolvent set is $$R(L)=\{z\in\Bbb C: L-zI \text{ is continuously invertible}\}$$ and $$\sigma(L)=\overline{\Bbb C\setminus R(L)}.$$ Second, in your case, you can explicitly write $$(L-zI)x = (x_2-zx_1,x_3-zx_2,\ldots).$$ Fix $z\in \Bbb C$. If for an arbitrary $\epsilon>0$ ...


1

As you acquire examples, please be sure to include this multiplication operator: $$ X = L^{2}_{\mu}(-1,1),\\ (Af)(x) = xf(x). $$ Here, $\mu$ is a finite Borel or Lesbesgue measure on $[-1,1]$. The reason to include this one in your collection of examples is that bounded selfadjoint operators with ...


1

I'm going to build on another problem of yours, Tobias: Why is this operator self-adoint . In the above problem, it is shown that, if $A : \mathcal{D}(A)\subseteq H\rightarrow H$ is symmetric with $(A\pm iI)$ surjective, then $A$ is densely-defined and selfadjoint. As you noted, your operator $O$ is symmetric on its domain. To see that $(O\pm iI)$ are ...


1

Consider the unilateral shift on $\ell^2(\mathbb N)$, i.e. $$ S(a_1,a_2,\ldots)=(0,a_1,a_2,\ldots). $$ It is easy to verify that $S$ is injective, so $0$ is not an eigenvalue of $S$. But $S$ is not invertible (because it is not surjective). So we have $0$ as an element of the spectrum of $S$, but not an eigenvalue. In finite dimension, two things happen: ...


0

Essentially, you need to check that $\forall f,g\in dom(G)\cap dom(T)$ you have $$(f,T[g]/\phi) = (f,T[g/\phi]),$$ where $(\cdot,\cdot)$ is a scalar product in $L^2$ and $\phi$ is the function $\frac{1}{1-x^2}$. I don't quite see how it could be possible for generic $T$.


0

I do not think it is self-adjoint in general. If you define the operator $\mathcal{O}$ by $\mathcal{O}f(x) = \dfrac{f(x)}{1-x^2}$, then $G = T\mathcal{O}$. $\mathcal{O}$ is a self-adjoint operator which you can see pretty easily. It is a general result that if $A,B$ are self-adjoint, $AB$ is self-adjoint if and only if $A$ and $B$ commute. Hence $G = ...


1

The spectral theorem for an unbounded selfadjoint operator allows you to represent $T$ as $$ Tx = \int_{0}^{\infty}\lambda dE(\lambda)x,\\ \mathcal{D}(T) = \left\{ x \in H : \int_{0}^{\infty}\lambda^{2}d\|E(\lambda)x\|^{2} < \infty \right\}. $$ The unique positive square root of $T$ is $$ \sqrt{T}y = ...


1

$\text{dom}(A) = \text{dom}(T)$ is wrong. By the Spectral Theorem, you can essentially assume $T$ is multiplication by the variable $x$ on $L^2(\mu)$ for some positive measure $\mu$ on $[0,\infty)$, with $\text{dom}(T) = \{f \in L^2(\mu): x f \in L^2(\mu)\}$. Then you want $A$ to be multiplication by $\sqrt{x}$, with $\text{dom}(A) = \{f \in L^2(\mu): ...


0

I prefer the Mathematician's inner product which is linear in the first coordinate. For each $w \in H$, there is a unique regular Borel measure $\mu_{w}$ on $\sigma(A)$ such that $$ (f(A)w,w) = \int_{\sigma} f\,d\mu_{w},\;\;\; f \in C(\sigma). $$ Because of this, $\|w\|^{2}=(Iw,w)=\mu_{w}(\sigma)$. Then $$ \begin{align} ...


1

For a regular Sturm-Liouville eigenvalue problem on a finite interval $[a,b]$, say $$ Lf = \left[-\frac{d}{dx}p\frac{d}{dx}+q\right]f = \lambda f, $$ there are two types of standard endpoint conditions: Separated conditions $$ \cos\alpha f(a)+\sin\alpha f'(a) = 0 \\ \cos\beta f(b) + \sin\beta f'(b) = 0. $$ The linear ...


0

Let me add an example of operators with spectrum consisting of a single eigenvalue. For example $$ \begin{bmatrix}0&0\\1&0\end{bmatrix} $$ has spectrum consisting of just its eigenvalue $0$. Similarly, $$ \begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}, $$ and we can construct examples in $n\times n$ for any $n$.


4

A good example of an operator with one point in the spectrum is the integral operator $$ Lf=\int_{0}^{t}f(u)\,du $$ defined on $X=L^{2}[0,1]$, or defined on $X=C[0,1]$. In both cases $\sigma(L)=\{0\}$. This operator is also compact but it has no eigenvalues, which makes it a good counterexample to remember when studying compact operators.


2

It turns out that for any bounded linear operator on a Hilbert space, the spectrum of that operator is a compact subset of $\mathbb{C}$. In turn, any compact subset, $K$, of $\mathbb{C}$ is the spectrum of some linear operator. In particular, if $\chi(z)$ is the indicator function of that set, then $M_{\chi}$ has spectrum $K$. Here $M_{\chi}$ is the ...


1

Definitely not. A self-adjoint operator just need $A^T=A$. Note the matrix of $T^*$ is the transposed of the matrix of $T$.


1

Ok, so I think I've got it (I use $\Bbb N = \{1,2,...\}$): Let $S' \in \mathcal{B}$ be given by $S'y = Sy - \langle e_1,Sy \rangle e_1$. $T + \epsilon S'$ is an isomorphism for small enough $\epsilon \ (>0)$ from $H_0 = \ell_2$ to $H_1 = \{ x \in \ell_2 | \langle e_1,x \rangle = 0 \}$. Now suppose that $e_1 \in \text{range}(T + \epsilon S)$: there exists ...


1

I suppose that $0\in\mathbb{N}$. If you follow the other convention, replace $e_0$ with $e_1$ in the following. If $T+ \epsilon S$ is surjective, then there is an $x\in \ell_2$ with $\lVert x\rVert = 1$ and $$(T+\epsilon S)(x) = c\cdot e_0$$ for some $c\neq 0$. Then $\lVert\epsilon Sx\rVert^2 = \lVert c\cdot e_0 - Tx\rVert^2 = \lvert c\rvert^2 + 1$ and ...


2

If you are talking about $T^{-1}$, it means you are assuming that $T$ is invertible. So $0\not\in\sigma(T)\cup\sigma(T^{-1})$ You have $$ T^{-1}-\lambda I=-\lambda T^{-1}(T-\lambda^{-1} I). $$ So $\lambda\in\sigma(T^{-1})$ if and only if $1/\lambda\in\sigma(T)$. That is $$ \sigma(T^{-1})=\{1/\lambda:\ \lambda\in\sigma(T)\}. $$


2

Your definition is not the unilateral shift. If your book gave you $Se_{n}=e_{n+1}$, then that is correct. The coefficient of $e_{1}$ is mapped to become the coefficient of $e_{2}$. So, $$ S(\alpha_{1},\alpha_{2},\alpha_{3},\cdots)=(0,\alpha_1,\alpha_2,\alpha_3,\cdots). $$ The operator $S$ never sends anything to $0$ except $0$ because $\|Sx\|=\|x\|$. ...


2

I think you are mixing up notation from two different sources. Your work shows that there are indeed eigenvalues (the open unit disc) for this definition of the unilateral shift. However there is another definition for the unilateral shift: $S(\alpha_1,\alpha_2,\ldots) = (0,\alpha_1,\alpha_2,\ldots)$. This incarnation has no eigenvalues. This incarnation is ...


1

For $\lvert\lambda\rvert > \lVert T\rVert$, we have a non-negative lower bound (strictly positive for $x\neq 0$) for $\lVert (T-\lambda I)x\rVert$, namely $$\lVert (T-\lambda I)x\rVert \geqslant (\lvert\lambda\rvert - \lVert T\rVert)\lVert x\rVert.$$ Now set $x = (T-\lambda I)^{-1} y$ to obtain the inequality $$\lVert y\rVert = \lVert(T-\lambda ...


0

You shouldn't open parantheses for Leibnitz rule, use the boundary conditions with integration by parts instead. Take $y$ that satisfies the same boundary data: $$y^{(j)}(a)=y^{(j)}(b)=0,\, j=0,1,\dots,n-1,$$then $$\int_a^b (px^{(j)})^{(j)}y\ dt =(px^{(j)})^{(j-1)}y\big|_{t=a}^{t=b} - \int_a^b (px^{(j)})^{(j-1)}y^{(1)}\ dt $$ $$=\dots= (-1)^j\int_a^b ...


1

Hint: $T-\lambda I$ is invertible with inverse $A \in \mathcal{B}(H)$ iff $$ I = (T-\lambda I)A = A(T-\lambda I) $$ The above holds iff $$ I = A^{\star}(T^{\star}-\overline{\lambda}I) = (T^{\star}-\overline{\lambda}I)A^{\star}. $$


1

Is there any relationship between $V_{\lambda_i}$ and $ E_{\lambda_i}$? Yes. $V_{\lambda_i} \subseteq E_{\lambda_i}$, with equality iff the algebraic and geometric multiplicities of $\lambda_i$ are equal Is there any relationship among $V, V_{\lambda_i},E_{\lambda_i}$ (maybe something with direct sum) ? If $\mathcal F$ is algebraically closed, ...


0

'Variation' to T.A.E.'s answer: $\mathbb{C}\setminus\mathcal{R}(v)\subseteq\rho(T)$ Suppose $\lambda\notin\mathcal{R}(v)$. Regard $w:=\frac{1}{v-\lambda}$. Then it exists and it is continuous. Moreover, it is bounded as it lives on a compact set. Thus, its multiplication is an everywhere welldefined and bounded operator: $$\|M_w ...


0

Goal The pure-point-space agrees with the eigenspace: $$\mathcal{H}_\text{pp}(E)=\overline{\langle\mathcal{E}(T)\rangle}$$ with pure-point-space and eigenspace being: $$\mathcal{H}_\text{pp}(E)=\{\varphi:\exists\#\Lambda_0\leq\aleph_0:\nu_\varphi(\Lambda_0)=\nu_\varphi(\Omega)\}$$ ...


0

Goal The spectral subspaces decompose the Hilbert space: $$\mathcal{H}=\mathcal{H}_\text{ac}\oplus\mathcal{H}_\text{sc}\oplus\mathcal{H}_\text{pp}$$ with the spectral subspaces being: $$\mathcal{H}_\alpha=\{\varphi:\nu_{\varphi,\alpha}=\nu_\varphi\}$$ Preparation Check the equivalences: $$\nu_\varphi(A)=0\iff E(A)\varphi=0$$ ...


1

I'll be verbose. For a bounded operator $T$, the resolvent set $\rho(T)$ is the set of $\lambda$ for which $(T-\lambda I)$ is injective and surjective. That is, $\lambda\in\rho(T)$ iff $(T-\lambda I)x=y$ has a unique solution $x\in C[0,1]$, regardless of the choice of $y\in C[0,1]$. Claim: $\sigma(T) = \mathcal{R}(v)$ where $\mathcal{R}(v)=\{ v(t) : 0 ...


1

The map $(T-\lambda I)^{-1}$ is defined as the solution mapping of the equation $$ (T-\lambda I) y = z, $$ i.e. $ y= (T-\lambda I)^{-1}z$. In case of the multiplication with $v$, this is equivalent to $$ (v(t)-\lambda) y(t) = z(t). $$ First consider the case that $\lambda\ne v(t)$ for all $t\in [0,1]$. Then the above equation has a unique solution, ...


1

In the indicated line, we can just expand the right hand side \begin{align} (x-\lambda)^{-1}[(x-\lambda_0) - (x-\lambda)](x-\lambda_0)^{-1} &= [(x-\lambda)^{-1}(x-\lambda_0) - (x-\lambda)^{-1}(x-\lambda)](x-\lambda_0)^{-1}\\ &=[(x-\lambda)^{-1}(x-\lambda_0) - I](x-\lambda_0)^{-1}\\ &= (x-\lambda)^{-1}(x-\lambda_0)(x-\lambda_0)^{-1} ...



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