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4

Consider the space $H = L^2([0,1],\mu)$, where $\mu$ is the Lebesgue measure. Define $T$ to be the multiplication with the identity function, i.e. $$(Tf)(x) = x\cdot f(x).$$ Since the identity function is bounded, $T$ is bounded ($\lVert T\rVert \leqslant 1$), and since it is real-valued, $T$ is self-adjoint. Clearly $T$ has no eigenvalues, since $$(T - ...


3

Write $f=Pf+(I-P)f$; the summands are mutually orthogonal. From the assumption by the spectral theorem you get $$1/2\ge\|(T-3I)f\|\ge \|(T-3I)(I-P)f\|\ge\|(I-P)f\|$$ (the last inequality is true because $(I-P)f$ belongs to the spectral subspace of the set $\{x: \; |x-3|>1\}$, which is the complement of the segment $[2, 4]$). Now the claim follows from the ...


3

Assume $X$ is a Banach space and $A$ is a bounded linear operator on $X$. $\lambda$ is in the point spectrum iff $\mathcal{N}(A-\lambda I) \ne \{0\}$. $\lambda$ is in the continuous spectrum iff $\mathcal{N}(A-\lambda I)=\{0\}$ and $\overline{\mathcal{R}(A-\lambda I)}=X$. Everything else is the residual spectrum. You want $\lambda$ to be in the residual ...


3

Let $T=\frac{1}{i}\frac{d}{dt}$ be defined on the domain $\mathcal{D}(T)$ consisting of all absolutely continuous functions $f \in L^2[0,1]$ for which $f(0)=0=f(1)$. More precisely, $f \in \mathcal{D}(T)\subset L^2[0,1]$ is an equivalence class of functions equal a.e. with one element $\tilde{f}$ of the equivalence class that is absolutely continuous on ...


2

In the calculation for the norm of $T$, your evaluation of $(Tx,Tx)$ is wrong. And, in any case, you don't show how the norm would be achieved. Actually, $\|T\|=2$. If you try with $x=(0,1,-1,1,-1,\ldots,1,-1,0,\ldots)$ (with $n$ nonzero entries) you'll get that $$ \|x\|=\sqrt n,\ \ \ \|Tx\|=(1+4(n-1))^{1/2}, $$ so $$ ...


2

Hint: if $[1-\lambda(n^2+1)]x_n=x_{-n}$ for all $n$, then $[1-\lambda(n^2+1)]x_{-n}=x_{n}$ for all $n$ as well.


1

Your conjecture is false. It is always the case that $T^{\star}T$ is densely-defined and selfadjoint if $T$ is a closed densely-defined linear operator on a Hilbert Space $H$. Let $H=L^2[0,1]$ and let $\mathcal{AC}[0,1]$ be the absolutely continuous functions on $[0,1]$. Define $T=\frac{d}{dt}$ on the domain $$ \mathcal{D}(T)=\{ f \in \mathcal{AC}[0,1] ...


1

As discussed in the comments, we are only left to show that $\Delta f_{\lambda}(x) = - \lambda^2 f_{\lambda}(x)$ The following proof will work if $f \in C^2(\mathbb{R}^n)$ and it, together with all of its partial derivatives up to second order are in $L^1$, so that differentiation under the sign is justified. Also, we need $\widehat{f} \in L^1$, so that all ...


1

I'm assuming that the functions $\varphi_j$ are real-valued. The Cauchy-Schwartz inequality on $\ell^2(\mathbb{N})$ gives \begin{align*}\sum_{j=1}^{\infty} |\lambda_j \varphi_{j}(x) \varphi_{j}(y)| &= \sum_{j=1}^{\infty} |\sqrt{\lambda_j} \varphi_{j}(x) \sqrt{\lambda_j}\varphi_{j}(y)| \leq \sum_{j=1}^{\infty} \lambda_j \varphi_j(x)^2 ...


1

I think you can find the spectrum directly. First, $$T((a_j))_i=\sum_{j=2}^{\infty}a_j e_1+\sum_{i=2}^{\infty}a_{i-1} e_i=\sum_{j=2}^{\infty}a_j e_1+a_{i-1}$$ Then $$(T-\lambda I)((a_j))_1=\sum_{j=2}^{\infty}a_j -\lambda a_1,$$ $$(T-\lambda I)((a_j))_i=a_{i-1}-\lambda a_i$$ So if $\lambda$ is an eigenvalue, the second equation implies ...



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