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Here are some thoughts - not sure if there is a clear-cut answer to your question even for matrices (I use $r(T)$ to denote $\text{rad}(T)$) : $T$ has this property iff $r(T) \geq \|T\|$, which happens iff $$\|T(x)\| \leq r(T)\|x\| \quad\forall x$$ $$\Leftrightarrow \langle Tx,Tx\rangle \leq r(T)^2 \langle x,x\rangle$$ $$\Leftrightarrow r(T)^2I - ... 0 This is a bounded form: 0 \le q_{\lambda}(f) \le \|f\|^{2}, regardless of \lambda, and it is defined everywhere on the Hilbert space. So the form norm q_{\lambda}(f)+\|f\|^{2} and the usual norm are equivalent norms. 3 Yes, this does make sense in the context of "rigged Hilbert spaces", e.g., something like what is occasionally called a Gelfand triple H^{+1}\subset L^2\subset H^{-1} of Sobolev spaces on an interval in \mathbb R. Somehow Dirac had a wonderful intuition in this direction already prior to 1930. Also, the possibility of writing "integral kernels" for all ... 1 Here is my understanding. But I know very little about QM, so this explanation may be incorrect. Consider the multiplication operator A defined on L^2[-a, a] by$$A\varphi(x) = x\varphi(x). $$(In quantum mechanics, A corresponds to the position operator.) Since this operator is bounded with the spectrum \sigma(A) = [-a, a], by the spectral ... 0 Spectral methods look for approximations of solutions of an equation as a linear combination of a determined set of fucnctions, ussualy a a complete orthonormal system with respect to some weight. The firs step is to be able to approximate a function in that way . To fix things, fix an interval [a,b] and a complete orthonormal system ... 0 The motivation for Spectral Theory came from Fourier's separation of variables technique that he invented to solve his heat equation. There was one notable example in particular that Historians say triggered the birth of Spectral Theory. In this example, Fourier was looking at the accuracy of a thermometer and, in particular, the "cooling off" problem for a ... 1 Yes, I would say that you have fully described the relationship between A and B. 0 Not sure what you mean by "performing a spectral analysis", but here are some examples that might help you understand the ideas involved: If T is an operator on a finite dimensional Hilbert space, then its spectrum is simply the collection of all eigen-values If T: C[0,1] \to C[0,1] is the operator$$ T(f)(x) = \int_0^x f(t)dt $$Then T is injective, ... 1 \dfrac{\lambda}{c} I- A isn't invertible if and only if \lambda I- cA isn't invertible. Hence, \lambda\in Sp(cA) if and only if \dfrac{\lambda}{c}\in Sp(A) 0 The resolvent operator is$$ (L-\lambda I)^{-1}f = \sum_{k=-\infty}^{\infty}\frac{1}{n-\lambda}(f,e^{inx})e^{inx}. $$where (f,g)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt is the normalized inner product on L^{2}[-\pi,\pi]. If [a,b]\subset\mathbb{R} has endpoints that are not integers, then the spectral measure E is evaluated ... 1 The main reason for posting this was to answer it, thus collecting all this stuff in a single place for future reference -- and present too. The first item on this proof is that a linear operator on a finite-dimensional complex vector space admits an upper triangular representation. This is proved by induction on n:=\dim V, V being the vector space. If ... 1 Historically, integral operators are the prototypical compact operators. Compactness came up in the late 1800's when studying differential operators by recasting them as integral operators. While differential operators are very discontinuous, integral operators are especially continuous on bounded regions because they typically map uniformly bounded ... 1 The first part is a standard result (easy to prove also!) available in most books on operator theory for eg: Proposition 4.6 in Banach Algebra Techniques in Operator Theory by R.G. Douglas, 2nd edition states it as follows If T is an operator on the Hilbert space H, then ker T = (ran T^*)^\perp  and (ran T)^\perp = ker T^* This proposition ... 1 I agree with Jack D'Aurizio. If you think you know enough Graph Theory to understand Spectral Graph Theory, grab a Spectral Graph Theory book and start reading it. (My suggestions: Read either Spectral Graph Theory by Fan R. K. Chung, or Graph Spectra by Miegham) If you dont know a graph theory concept mentioned, search it up and learn it. Here's some ... 1 For 1: note that g^n \in G \subset B for each n. It follows that if \lambda is an eigenvalue of g, then we have$$ |\lambda^n - 1| < 1  for every $n \in \Bbb Z$ (since $\lambda$ can't be zero, negative exponents are fine). If $|\lambda| \neq 1$, then either $|\lambda^n|$ or $|\lambda^{-n}|$ diverges to $\infty$ as $n \to \infty$. You'll need ...

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This is an explicit instance of the No Small Subgroups theorem for real or complex Lie groups, namely, that there is a sufficiently small neighborhood of $I$ so that the only subgroup contained in it is $\{I\}$. The specific estimate is perhaps not so crucial in the bigger scheme of things, but is tangible. Indeed, if there were an eigenvalue $\lambda$ ...

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Write $f_{\epsilon}=cos(\theta_{\epsilon})+i\sin(\theta_{\epsilon})$. Then \begin{align*} \lim_{\epsilon\to0}\theta_{\epsilon}&=0,\\ \Im(a)&=\lim_{\epsilon\to0}\frac{\sin(\theta_{\epsilon})}{\epsilon},\\ ...

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One of the nice things about Functional Analysis is that you can generally reduce to the scalar case by applying a linear functional to everything, rearranging scalar integrals, and then pulling the functional back outside. Then, knowing that you have enough functionals to separate points allows you to remove the functional from both sides of the resulting ...

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