New answers tagged

0

What you are missing from the statement in the book is that $H$ is separable. Consider first $B(H)$. Then $K(H)$ is weakly dense in $B(H)$, and it is not hard to show that $K(H)$ is separable. Now consider a von Neumann algebra $M\subset B(H)$. You have an inclusion of unit balls $M_1\subset B(H)_1$. The previous exercise in the book proves that, since ...


2

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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Note: There is nothing about completeness of $\mathcal{H}$ needed to carry out of the following steps. Because $P_n$ is monotone, then $(P_nx,x)$ is monotone in $n$ for each fixed $x$, and is bounded above by $(x,x)$, which forces convergence of $\lim_n(P_n x,x)$ for all $x$. Then, using polarization, the following expression must also have a limit in $n$ ...


0

Consider the two cases of $V_n$ being monotone increasing and monotone decreasing in $n$. In the first case you have $P_n P_{n+1}=P_n$ and in the second you have $P_n P_{n+1}=P_{n+1}$. At any rate you have either $P_n(P_{n+1}- P_{n})=0$ or $P_{n+1}(P_{n+1}-P_{n})=0$. First consider $V_n$ is increasing. So $(P_{n+1}-P_n)(z)$ is in $V_n^\perp$. This means ...


1

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 $x$, $$ (P_nx,x) = (P_nx,P_nx)=\|P_nx\|^2=\|P_nP_{n+1}x\|^2 \le \|P_{n+1}x\|^2=(P_{n+1}x,x). $$


1

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 ...


1

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 $\text{Tr}(A)^2 - 4|A| = (a-d)^2 + 4bc \geq 0$!!


-1

Hint: Calculate the discriminant of the characteristic polynomial of $$\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ after some easy calculations to be $$\frac{1}{4}\left((a-d)^2+4bc\right).$$ Remark: Complex-conjugated eigenvalues may only exist if $bc<0$ regardless of the signs of $a$ and $b$.


0

The functions $$ \varphi_{\lambda}(x)= \cos(\sqrt{\lambda}x),\;\;\;\psi_{\lambda}(x)=\cos(\sqrt{\lambda}(x-l)) $$ satisfy the differential equation $-f''+\lambda f = 0$ subject to the conditions $$ \varphi_{\lambda}(0)=1,\;\varphi_{\lambda}'(0)=0,\;\;\;\;\;\;\psi_{\lambda}(l)=0,\;\psi_{\lambda}'(l)=0. $$ These solutions are linearly independent ...


1

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 $T=\lambda_1\,E$.


2

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 ...


1

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 ...


1

$V$ 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 ...


1

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 ...


1

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 ...


1

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 ...


1

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 ...


1

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, ...


0

As others have pointed out, your statement is false ($\rho(A) \le \|A\|$). However, if $\| A\|<1$, one can show that $$ \| (I - A)^{-1}\| \le \frac{1}{1-\|A\|}. $$


0

No, consider $$A=\left[\begin{array}{cc}4/5& 4/5\\0&4/5 \end{array}\right].$$ The spectral radius of $A$ is $4/5$ and its norm is $2/5\sqrt{2(3+\sqrt{5})}>1$ so the quantity on RHS of the proposed inequality is negative, an impossibility.


0

We can actually solve the system explicitly to order $t^2$, obtaining $$ v_1(t) = 1, \; v_j(t) \ge t + \dfrac{j-2}{2} t^2 \ \text{for}\ j \ge 2, \; t \ge 0$$ (i.e. $v_j(t)$ is a polynomial in $t$ with nonnegative coefficients and lowest-order terms $t + ((j-2)/2) t^2$). Note also that $v_j(t)$ is the same for all $n \ge j$. Thus it is impossible to get a ...


1

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 ...


0

Yes. Hint. Let $B=A^{-1}$. Clearly, $B$ is positive definite. By using Sylvester's criterion, show that for every off-diagonal entry $b_{ij}$, we have $|b_{ij}|\le\max\{b_{ii},b_{jj}\}\le\max_kb_{kk}=\max_ke_k^TBe_k$, where $\{e_1,\ldots,e_n\}$ is the standard basis of $\mathbb R^n$. By using the fact that every real symmetric matrix is orthogonally ...


0

Thank you. If I understood it right then a possible solution is: There is a $*$-isomorphism $\phi : A \to L$ , where $L$ is a C*-subalgebra of B(H) for some Hilbert space $H$. *-isomorphism preserves positive elements, so we just need to verify that for any sequence of positive operators $T_n : H \to H$ that converges to $T$ in $B(H)$, $T$ is positive. ...


1

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: ...


1

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 ...


1

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 ...


2

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 & ...


1

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 ...


0

$V^{*}=$ completion of $D\left(H\right)$ w.r.t. \begin{eqnarray*} \left\Vert x\right\Vert _{V^{*}} & := & \left\Vert H^{-1/2}x\right\Vert ,\;x\in D\left(H\right). \end{eqnarray*} For $x\in D\left(H\right)$, \begin{eqnarray*} \left\Vert x\right\Vert _{V} & = & \left\Vert H^{1/2}x\right\Vert \\ \left\Vert Hx\right\Vert _{V^{*}} & = & ...


1

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 $$ \|\text{Re}A\|=\frac12\,\|A+A^*\|\leq\frac12\,(\|A\|+\|A^*\|)=\|A\|. $$


1

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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