New answers tagged

0

After some fiddling, I feel I have a satisfactory answer. Define $\omega := [2,\infty) \cap \sigma(T)$, and $P := E(\omega)$. This is a self-adjoint projector by definition of $E$. By symbolic calculus, we have $$((TP - 2P)x,x) = ((T-2I)Px,x) = \int_{\sigma(T)} (\lambda - 2)\chi_{\omega} d(E_\lambda x,x).$$ We know $(E_\lambda x,x)$ is a positive measure ...


0

\begin{align} TP-2P&=(T-2I)P,\\ 2(I-P)-T(I-P)&=-(T-2I)(I-P) \end{align} $P=E[2,\infty)$ does the job because $I-P=E(-\infty,2]$.


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 ...


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 - ...


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] ...


0

The sequence $x=\{ \cdots,0,0,-1,0,1,0,0,\cdots\}$ where the middle $0$ is $x_0$ is mapped to $\{ \cdots,0,0,\frac{-2}{1^2+1^2},0,\frac{2}{1^2+1^2},0,0,\cdots\}$ under the transformation $T$. That is $Tx=\frac{2}{1^2+1^2}x$. In this way you can see that $\frac{2}{1+n^2}$ is an eigenvalue for $n=1,2,3,\cdots$. And $0$ is an eigenvalue also because $Ty=0$ ...


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

Fourier transform does not exist for every signal application.So by introducing the region of convergence in Fourier transform which is known as Laplace Transform one may have indirectly the Fourier transform of signal.


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 $$ ...


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 ...


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 ...


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 ...


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 ...


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 ...



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