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4

There are two scales at play here. Let $M$ be a closed Riemannian manifold and let $\Delta$ be the Laplace-Beltrami operator on it. There is an orthonormal eigenbasis $\{\phi_k\}_{k=1}^\infty$ of $L^2(M)$ and an increasing sequence of eigenvalues $\lambda_k\geq0$ so that $\Delta\phi_k=-\lambda_k\phi_k$. The piece of an article you linked to discusses the ...


2

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


2

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 ...


2

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.


2

Your reasoning is correct, for a continuous (bounded) everywhere defined operator $A$ on a Banach (in particular on a Hilbert) space, the denseness of the range of $\lambda\mathbb{I} - A$ together with the boundedness of the inverse already implies the surjectivity of $\lambda\mathbb{I} - A$, and an equivalent definition of the resolvent set in this setting ...


1

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, ...


1

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 ...


1

Note that all of the work is in Exercise 8, which says that if $0\leq x\leq y$ in a C*-algebra and $x$ is invertible, then $y^{-1}\leq x^{-1}$. You will apply this with $x=1+\alpha a$ and $y=1+\alpha b$, and the rest follows easily. For a solution to Exercise 8, see Inversion in a unital C* algebra or positive invertible operators or Inverse of ...


1

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 $$ ...


1

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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