# Tag Info

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A closed densely-defined linear operator on a Hilbert space can have empty spectrum. For example, let $H=L^{2}[0,1]$ and let $A=\frac{d}{dx}$ on the domain consisting of absolutely continuous $f \in L^{2}$ for which $f(0)=0$ and $f' \in L^{2}$. To show that $A$ has no spectrum, it is enough to prove that the resolvent $R(\lambda)=(A-\lambda I)^{-1}$ exists ...

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This is only a check! The spectrum is neglible: $$0\leq\lambda_\mathbb{C}(\sigma(H))\leq\lambda_\mathbb{C}(\mathbb{R})=0$$ By functional calculus: ...

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The operator $K$ is called the Volterra operator It is a bounded linear operator on $L^2$ space and is an indefinite integral. The easiest way to prove the Volterra operator has spectrum $\{0\}$ is showing that the spectral radius is zero. To get you started, we must determine that the limit of $K^n$ under the operator norm to the power $1/n$ is $0$, ie, ...

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The function $u_\delta$ is defined so that it is continuous and has compact support. You can make a $C^\infty$ function with compact support out of it by mollification, since the support of $u_\delta$ is at positive distance from $\partial \Omega$. (This is helpful if $H_0^1$ is defined as the closure of $C^\infty$ functions with compact support.) It ...

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Yes, you are correct. For reference: The Spectral Theorem For a Pair of Commuting Operators http://www.mi.ras.ru/~snovikov/78.pdf

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the matters are not o straightforward. Please consult R. K. GOODRICH, The spectral theorem for real Hilbert space, Acta Sci. Math. (Szeged), 33 (1972), 123–127. (open access) Miroslav

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I can answer question 5a*. I'm not sure about 5a** because I don't know any functional analysis. Consider a polynomial $p(t) = a_nt^n + a_{n-1} t^{n-1} + \ldots + a_0 \in P(\mathbb R)$ with $a_n \neq 0$. Then $Tp(t) = p(t+1) = a_n(t+1)^n + \ldots + a_0$. Using the binomial theorem, we can expand this and find the first two coefficients. We then have $$Tp(t) ... 1 Convolution of the Scaled Probability Measure Suppose that f is a probability density with mean 0 and variance 1. Consider the Fourier Transform of f$$ \widehat{f}(\xi)=\int_{\mathbb{R}}f(x)\,e^{-2\pi ix\xi}\,\mathrm{d}x\tag{1} $$The assumptions we've made allow us to say that$$ ...

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Can someone please proof-read?? Denote for readability: $$N:=M^*\quad N':={M'}^*$$ Regard the dense elements:* $$\chi\in\mathcal{D}_0\cap\mathcal{D}'_0:=\bigcup_{R>0}\mathcal{R}E(B_R)\cap\bigcup_{R'>0}\mathcal{R}E'(B_R')$$ ...

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You are sort of on the right track. Let's write things out a little bit more explicitly. Consider $$(\lambda I-T)f = \lambda f - Tf = \lambda f - zf = \sum_{n=-\infty}^{\infty} \lambda \alpha_n z^n - \sum_{n=-\infty}^{\infty} \alpha_n z^{n+1}.$$ Reindexing, the second sum becomes $\sum\limits_{n=-\infty}^{\infty} \alpha_{n-1} z^n$ so the above becomes ...

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Here is a helpful bit of knowledge: an upper-triangular matrix is normal if and only if it is diagonal. By the Schur triangularization theorem, every matrix is unitarily similar to an upper-triangular matrix. So, up to unitary similarity, every normal matrix diagonal, and every non-normal matrix is upper-triangular, but not diagonal. So, in particular, ...

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In the finite-dimensional case, spectral theory says that $A$ is normal iff it is diagonalizable and its different eigenspaces are orthogonal to each other. So you can find non-normal matrices by violating either of these conditions. For instance, you can just choose any non-orthogonal basis for your inner product space and take a matrix that is diagonal ...

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Suppose for some vector $x$ we have $A^2 x = 0$ but $Ax \ne 0$. Then $A$ is not normal. In particular, any nonzero nilpotent operator is non-normal.

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The spectral theorem gives $$(\lambda I-A)^{-1} =\int_{\sigma(A)}\frac{1}{\lambda -t}dE(t), \;\;\; \lambda\in\rho(A).$$ If $C$ is a simple closed piecewise smooth positively oriented contour in $\rho(A)$ that encloses part of the spectrum, then $C$ crosses the real axis where the spectral measure is $0$ in a neighborhood of that crossing. So ...

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You are almost there. The ideals of $C (Y)$ are given by its closed subsets, and the essential ideals are those that correspond to closed nowhere dense subsets. In other words, the essential ideals of $C (Y)$ are precisely $C _0 (T)$, where $T\subset Y$ is open and dense.

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It looks like your first fit is trying to fit all the data to a single spectral line. The problem is the bumps in your data around $6.5$ and $15$. Depending on how you assess the probable error on each point, those bumps are going to pull your fit very hard. This is especially true if you take the error to be something like the square root of the expected ...

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An operator $T$ is normal if $TT^*=T^*T$, where $T^*$ is the adjoint operator. A matrix $M$ is normal if $MM^*=M^*M$, where $M^*$ is the conjugate-transpose matrix. These are equivalent, because if in some orthonormal basis $T$ is represented by $M$, then $T^*$ is represented by $M^*$. Indeed the $(i,j)$ entry of $M^*$ is $\langle T^*e_j, e_i\rangle$, ...

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No. As a simple example, consider $$M=\pmatrix{1 & 1/2 \\ -1/2 & 0}.$$ with $\rho(M)=1/2<1$ and the simple choice $$D_s=D=\pmatrix{1&0\\0&-1}$$ for all $s$. Then $\rho(DM)=(2^{1/2}+1)/2>1$ so $(DM)^t$ diverges with $t\rightarrow\infty$. It is not generally true even if the diagonal entries of $D$'s are in $(0,1)$. E.g., with  ...

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Take $\mathbb C(x)$ - the field of rational functions in one variable over $\mathbb C$. The elements of the algebra are of the form $\frac{p(x)}{q(x)},$ where $p,q\in\mathbb C[x]$ are polynomials with $GCD(p,q)=1$. Then the $(x-\lambda)$ is invertible for every $\lambda \in \mathbb C$. Thus $\sigma(x)$ is empty.

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