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Given $f$ and $\epsilon$, choose a polynomial $p$ with $\Vert f-p\Vert_{\infty,X}<\epsilon$ (where $\Vert\cdot\Vert_{\infty,X}$ is the supremum norm oin $X$). Now see the corresponding polynomial function in $\mathcal{A}$, $p:\mathcal{A}\to\mathcal{A}$. (Remember: the functional calculus respects this notation, i.e., $p(a)$, in the functional calculus, is ...


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Let $(Mf)(x)=xf(x)$ be the multiplication operator on its natural domain in $L^2(\mathbb{R})$. Then $$ \|(\lambda I-M)^{-1}f\|^2 = \int_{\mathbb{R}}\frac{1}{|\lambda -s|^2}|f(s)|^2ds. $$ As an example, take $f(s)=1/\sqrt{1+s^2}$. For this $f$ and for $\Im\lambda > 0$, \begin{align} \|(\lambda I-M)^{-1}f\|^2 & ...


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I think this is a very good example where the slightly more abstract proof using linear maps (which you use in any case) instead of matrices simplifies the argument and shows more clearly what is going on. Let $(W, \left< \cdot, \cdot \right>)$ be a finite dimensional real inner product space and let $T \colon W \rightarrow W$ be a symmetric operator ...


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$V$ needs to be an invariant subspace under $A$ so that $T$ goes from $V$ to $V$. Otherwise, if $V$ was not invariant, $T$ would go from $V$ to $\Bbb{R}^n$ and then it would not have a matrix representation in $\Bbb{R}^{(n-1)x(n-1)}$. Here's a proof that $w_j$ is an eigenvector of $A$: $$Bu_j=\lambda_ju_j$$ This is because $u_j$ is an eigenvector of ...


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If $k$, $d$ are as specified, then $$ d\|x\|^2 \le |k\|x\|^2-\langle Tx,x\rangle| \\ d\|x\|^2 \le |\langle (kI-T)x,x\rangle| \le\|(kI-T)x\|\|x\| \\ d\|x\| \le \|(kI-T)x\|. $$ That's enough to give you injectivity of $kI-T$, along with a closed range. Then $$ \mathcal{R}(kI-T) = ...


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We are given the system of ODE's $$ \frac{dv_k}{dt} = \sum_{i=0}^{k-1} v_i(t),\qquad k \geq 1, $$ with $v_0(t)$ some integrable function in $[0,\infty)$, with initial condition $v_k(0) = 0 $ for all $k\geq 1$, and $v_0(0)=1$. Let $||v_0||_t = \int_0^t |v_0(\tau)|d\tau$. Claim. The estimate \begin{equation} |v_k(t)| \leq ||v_0||_t + (2^k-k-1)e^t ...


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I have had a glance at the paper. It is not that it is badly written but I consider pages 568-573 constitute a very intricated lecture on orthogonal polynomials, and the attached continued fraction. It is not surprising that you cannot easily find your way in this jungle. The presentation can be vastly simplified by centering all on the positive definite ...


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The scalar product $(f,g)$ is defined in equation (2.2): $$ (f,g)=\int_{0^-}^{\infty} w(\lambda) \, f(\lambda) \, g(\lambda) \,d\lambda . $$ Note that it includes the weight function $w(\lambda)$. The polynomials $p_0(\lambda), p_1(\lambda), p_2(\lambda),p_3(\lambda),\dots$ are what you obtain if you start with the monomials ...


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First notice that $\overline{\mathrm{ran}(A-\lambda)}=\ker(A-\lambda)^\perp$ for $\lambda\in\mathbb{R}$. Thus, it suffices to show that $\ker(A-\lambda)=\{0\}$ and that $\mathrm{ran}(A-\lambda)$ is closed. Let $\lambda<0$. We have $$ \|(A-\lambda)u\|\|u\|\geq\langle (A-\lambda) u,u\rangle\geq -\lambda\|u\|^2 $$ for all $u\in D(A)$. Thus, ...


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I think you make it unncessarily complicated. By the minimality of $E$, you have $P_{\lambda_i}=P_{\lambda_j}$ for all $i,j$. But each $\lambda_i$ is the particular projection corresponding to $\lambda_i$; so $\lambda_i=\lambda_j$ for all $i,j$ (projections corresponding to different eigenvalues are orthogonal to each other). Thus $T=\lambda_1\,E$.


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For a $2 \times 2$ matrix $A$, the characteristic equation is $x^2 - \text{Tr}(A)x + |A| = 0$. If this equation has complex roots, then the discriminant is negative. But if you assume A = [a,b;c,d], then $\text{Tr}(A)^2 - 4|A| = (a-d)^2 + 4bc \geq 0$!!


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I'll assume your matrix is $$ M=\left( \begin{array}{ccc} a & b\\ c & d\\\end{array} \right). $$ The the characteristic polynomial is $$ |M-\lambda I|=(a-x)(d-x)-bc $$ and setting the characteristic polynomial to zero yields the solution $$ \lambda = \frac 12 \left(a + d \pm\sqrt{4bc + (a-d)^2}\right). $$ The eigenvalues are only imaginary if ...


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Let $X = L^2[-\pi,\pi]$, and let $A=\frac{d^2}{dx^2}$ on the domain consisting of all polynomials. Let $B=\frac{d^2}{dx^2}$ on the domain $\mathcal{D}(B)$ of all linear combinations of $\{ \sin(t),\sin(2t),\ldots \}$. Both $A$ and $B$ are densely-defined, and they're both closable. However $\mathcal{D}(A)\cap\mathcal{D}(B)=\{0\}$, which forces ...


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Disclaimer : I am not an expert in this area. From what I have understood by some self-reading is that given a Riemannian manifold $M$ we have the Laplace-Beltrami operator $\Delta$ acting on $C^{\infty}(M)$. We can look at the spectrum of this operator. The main idea is that how much of geometry of $M$ can be recovered from analysisng the spectrum of ...


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Here's a couple key books. I took a course on spectral geometry a few years ago, but I can't find my notes, which had a list of classical papers like Kac 1966 "Can one hear the shape of a drum?" (you should read this one regardless - it's well written and considered a landmark). "Eigenvalues in Riemannian Geometry" by Chavel "Old and New Aspects in ...


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If $\|\varphi(A)\|\leq c\,\|A\|$ when $A$ is selfadjoint, then for arbitrary $A$ you have $$ \|\varphi(A)\|=\|\varphi(\text{Re}\,A)+i\varphi(\text{Im}\,A)\| \leq\|\varphi(\text{Re}\,A)\|+\|\varphi(\text{Im}\,A)\|\\ \leq c\,(\|\text{Re}\,A\|+\|\text{Im}\,A)\|\leq 2c\|A\|, $$ since $$ \|\text{Re}A\|=\frac12\,\|A+A^*\|\leq\frac12\,(\|A\|+\|A^*\|)=\|A\|. $$


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If $\mu$ is sigma-finite, then there are disjoint Borel subsets $\{ A_j \}_{j=1}^{\infty}$ of finite $\mu$-measure such that $\bigcup_j A_j=\Omega$. Let $f_j = T1_{A_j}$. Then $$ 1_{A_k}f_j = 1_{A_k}T1_{A_j}=T1_{A_k\cap A_j} =0 ,\;\;\; k \ne j,\\ 1_{A_j}f_j = f_j. $$ So each function $f_j$ is supported in $A_j$. Let $f$ be the a.e. unique ...


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ou have shown that the sequence of positive operators $$(b+1/n)^{1/2}-(a+1/n)^{1/2} $$ converges to $b^{1/2}-a^{1/2} $. So now all you have to show is that a limit of positives is positive. The two ways that come to mind are: by representing $A\subset B (H) $ (and then using wot convergence); or, if you want to stay abstract, by looking at states. Edit: ...


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Assume $V_n \subseteq V_{n+1}$ for all $n$. Then $P_{n+1}P_n=P_n$. Using adjoint, $P_nP_{n+1}=P_n$ must also hold because $P_k^*=P_k$ for all $k$. Therefore, for all $x$, $$ (P_nx,x) = (P_nx,P_nx)=\|P_nx\|^2=\|P_nP_{n+1}x\|^2 \le \|P_{n+1}x\|^2=(P_{n+1}x,x). $$


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Assuming you mean the closure in the natural topology on $AC$, that being the one given by the norm $|f(0)|+||f'||_1$, then yes. This is clear because any $L^1$ function (of mean zero) can be approximated in $L^1$ by continuous functions (of mean zero). In detail: Choose a sequence $g_n$ of continuous functions such that $\int_0^{2\pi}g_n=0$ and ...


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For simplicity, suppose that $0\in D$ and $T = -J(0)^{-1}$. We prove that for any $z\in D$, $J(z) = (zI - T)^{-1}$. Note that by definition of $J$, $J(z)J(w) = J(w)J(z)$ for $w\in D$, $z\in D$. Also, $J(z) = -zJ(z)J(0) + J(0)$. Hence \begin{align*} (zI - T)J(z) = zJ(z) + zTJ(z)J(0) - TJ(0)= I. \end{align*} Similarly, $J(z)(zI - T) = I$. That is, $J(z) = (zI ...



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