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1

No. Consider the operator $T:\mathbb C^2\to \mathbb C^2,\ T=\left(\matrix{0 & \ast\\ 1 & \ast}\right).$ That is $D(T)=\mathbb C e_1$ and $Te_1=e_2.$ The operator is closed and the unique self-adjoint extension is $\left(\matrix{0 & 1\\ 1 & 0}\right).$

5

Note that $M_n(\mathcal{A})$ is a $C^*$-algebra, so we have $$A\leq B\implies C^*AC\leq C^*BC$$ for all $A,B,C\in M_n(\mathcal{A})$. Since $A\leq \Vert A\Vert 1_{M_n(\mathcal{A})}$ for all $A\in M_n(\mathcal{A})_+$ we get $$C^*AC\leq \Vert A\Vert C^*C$$ Now consider $$A= \begin{pmatrix} a^*a & 0 & \ldots & 0\\ 0 & a^*a & \ldots & ... 1 Sorry to drag up such an old thread, but concerning the first question: For E a Banach space, a theorem of Edward Odell states that the following assertions are equivalent: \ \ \ 1) E contains no copy of \ell_1. \ \ \ 2) Every completely continuous operator on E (to some Banach space) is compact. This result is referenced in Corollary 5 of ... 3 To answer your questions, we use the fact that an operator on a Hilbert space is compact if and only if it maps some orthonormal basis to a sequence converging to 0. Question 1: No. Consider the operator on \ell^2 given by$$T(\{x_k\}) = \{k^{-1}x_k\}$$and denote by e_j the standard basis. T is obviously self-adjoint an Te_j = j^{-1}e_j \to 0 so ... 2 Consider the left shift L on \ell^2(\mathbb N),$$ L((a_1,a_2,a_3,\dots)) = (a_2,a_3,\dots) .$$Suppose for contradiction that L = U|A| where U is unitary. Then L^* L = |A|^2 = \text{diag}(0,1,1,\dots) which implies \sigma(|A|^2) = \{0,1\}. But LL^* = U|A|^2 U^* = I which implies \sigma(|A|^2) = \{1\}. Hence contradiction. 3 Is there a result in C^*-algebras that if b satisfies b = b^* and \sigma(b) \subset [\alpha,\beta] where \alpha,\beta > 0, then \alpha I \le b \le \beta I? Then you could prove it as follows. Consider b = a^* a. Then a short argument shows that \|b\| = \|b^{-1}\| = 1. But b is positive, and \inf\sigma(b) = \sup\sigma(b) = 1. Hence ... 0 If you only need the continuity into L^q, then (I bet) the best thing you can do is to use the Sobolev embedding W^{1,p} \hookrightarrow L^r and the continuity of \eta_a : L^r \times L^p \to L^q. 0 Just a remark on intuition, rather than precise proofs. The operator T_\alpha is a spectral multiplier and you should therefore think of it as an infinite dimensional analogue of a diagonal matrix. Functional calculus with diagonal matrices is trivial: one views them as functions on the set \{1,\ldots, n\} and performs addition and multiplication ... 2 Hint: the following statement solves your problem: Statement: If P_1,\dots,P_n\in\mathcal H are orthogonal projections such that \sum_{i=1}^n P_i is again an orthogonal projection, then P_i are pairwise orthogonal, i.e. P_iP_j=0,\ i\neq j. Edit. Proof. If Q,R are projections such that Q\geq R then Q\supseteq R. Indeed, let x\in Ran R. ... 0 1) If \lambda is in the image of \alpha, you want to show that T_\alpha-\lambda I is not invertible. If \alpha(t_0)=\lambda, consider a continuous f with f(t_0)=1. Then (\alpha(t)-\lambda)g(t)=f(t)  fails at t_0 for all g, so T-\lambda I is not onto. 2) Remember that your f is continuous, so it will be nonzero in a whole interval ... 0 For a bounded operator A with polar decomposition A = UP, U is the (canonical, if you'd like) partial isometry$$ (A^*A)^{\frac{1}{2}}h \stackrel{U}{\mapsto} Ah $$from Ran(A^*A) = Ran(A^*A)^{\frac{1}{2}} to Ran(A). If A is invertible, so is A^*A. This make U unique. Assuming U admits unitary extensions and has more than one extension, ... 1 (a) So, you already have that \|A\|\le 1 (because t\in [0,1] so \sup_t |t|=1 indeed). To prove the equality, it is enough to find any function f which satisfies \|Af\|_\max=\|f\|_\max, simplest might be to consider the constant f(t)=1. (b) For p\in (1,\infty), if f\in L^p(0,1), then, as you stated, we similarly have$$(\|Af\|_p)^p = ...

0

You want to show that $\int_\Omega|f|^4\,dE_{x,x}<\infty$ implies $\int_\Omega|f|^2\,dE_{x,x}<\infty$. and $$\int_\Omega|f|^2dE_{x,x}=\int_{\Omega\cap\{|f|<1\}}|f|^2dE_{x,x} +\int_{\Omega\cap\{|f|\geq1\}}|f|^2dE_{x,x}\leq E_{x,x}(\Omega)+\int_{\Omega\cap\{|f|\geq1\}}|f|^4dE_{x,x}\\<\|x\|^2+\int_{\Omega}|f|^2dE_{x,x}<\infty,$$ since the ...

2

In general you would use the holomorphic functional calculus $$f(T)=\frac1{2\pi i}\,\int_\Gamma \,f(z)\,(z-T)^{-1}\,dz,$$ where $\Gamma$ is a curve in the complex plane with the spectrum of $T$ inside the region it delimits, and $f$ is an analytic function on that region. Now, if $H$ is finite-dimensional, things are way easier. Then you have ...

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The existence of $J$ would imply that $P_1D$ and $P_2D$ are isometric. This is not the case in general: let $H_1=\mathbb C^2$, $H_2=\mathbb C^3$, and $$D=\{(0,a)\oplus(b,c,0):\ a,b,c\in\mathbb C\}.$$ Then $P_1D$ is one-dimensional, while $P_2D$ is two-dimensional.

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$\alpha$ implies that $I - K$ is bijective, hence injective, giving in turn "not $\beta$". Not $\beta$ says that $I - K$ is injective, hence you get the bijectivity from 4. This gives $\alpha$.

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The theorem means that the inequality holds for all $x\in X$: $$\bigl(\forall x\in X\bigr)\left(p_U(x) \leqslant \frac{1}{\varepsilon} \lVert x\rVert\right).$$ That inequality is easy to see since $r^{-1}\cdot \lVert\,\cdot\,\rVert$ is the Minkowski functional of $B_r(0)$ (very easy to see), and one has the implication $$A \subset B \Rightarrow p_B ... 3 Ad (1): Gram–Schmidt can readily be applied to any countable collection of non-zero vectors. Just remember how Gram–Schmidt orthonormalisation works: you simply replace v_{k+1} with its orthogonal projection onto \{v_1,\dotsc,v_k\}^\perp, and then normalise (if non-zero). Now, suppose that N(1-K) is infinite-dimensional, and that \{v_k\} is a ... 0 if H_1=H_2 then B(H)_*=L^1(H)(trace class), so, in my opinion, B(H_1,H_2)_*=L^1(H_1,H_2). takesaki shows that K(H)^*=L^1(H) , K(H)^{**}=L^1(H)^*=B(H) 3 But the key point is exactly that your matrix is diagonalizable more specially that you can find an orthonormal basis of eigenvector e_i for which you have A e_i=\lambda_i e_i. Then your write for x=\sum x_ie_i and you have Ax=\sum x_i\lambda_i e_i so that \|Ax\|^2=\sum\lambda_i^2x_i^2 now by definition of the norm of matrix it gives you ... 3 The norm of a matrix is defined as $$\|A\| = \sup_{\|u\| = 1} \|Au\|$$ Taking the singular value decomposition of the matrix A, we have $$A = VD W^T$$ where V and W are orthonormal and D is a diagonal matrix. Since V and W are orthonormal, we have \|V\| = 1 and \|W\| = 1. Then \|Av\| = ... 2 It depends on how you define Y valued integrals. Such integrals usually satisfy (or are defined by) the condition that \phi(\int g(x) d \mu (x)) = \int \phi(g(x)) d \mu (x) for all \phi \in Y^*. If this is the case, then we have \begin{eqnarray} \phi(\int_{\mathbb{R}}{L_{x}L_{y}f(y)dy} ) &=& \int_{\mathbb{R}}{\phi(L_{x}L_{y}f(y))dy} \\ ... 2 The positive square roots are defined by the continuous functional calculus. For a normal operator T, a continuous function f(T) is the norm-limit of polynomials in T and T^*. (You need the identity operator I for f's that don't vanish at 0 but that doesn't apply here.) But by your assumption, any polynomials in T and S commute. So ... 1 Fix a linear functional f \in V^* with norm 1 (whose existence is guaranteed by Hahn-Banach), and v_0 \in V with f(v_0) = 1. Define \Phi : W \to L(V,W) by (\Phi w)(v) = f(v) w. Verify that \Phi is a linear isometry. Now if \{w_n\} is a Cauchy sequence in W, then \{\Phi w_n\} is Cauchy in L(V,W). By assumption \{\Phi w_n\} converges ... 2 You are right, the norm of T_K is indeed$$\lVert T_K\rVert = \sup_{x\in U_1} \int_{U_2} \lvert K(x,y)\rvert\,dy.$$However, in general, there is no f \neq 0 with \lVert T_K(f)\rVert = \lVert T_K\rVert\cdot \lVert f\rVert, so you need to see that you can come arbitrarily close. Given \varepsilon > 0, pick an x\in U_1 with$$\int_{U_2} \lvert ...

0

Here's an alternate approach to the problem: once you've shown that $ST$ is self-adjoint (using $ST = TS$), it is enough to show that $(STx,x)>0$ for all $x$ for which $(x,x)=1$. Note that for any such $x$, we have \begin{align} 0 &< (\inf_{(x,x)=1} \left\{(Sx,x)\right\})( \inf_{(x,x)=1} \left\{(Tx,x)\right\})\\ & \leq (Sx,x)(Tx,x)\\ &= ... 0 Proposition 3.2c in Conway's book indeed says that for every idempotent E acting on a Hilbert space \mathscr{H} there holds\mathscr{H} = \mathscr{M}+\mathscr{N} \tag{1}$$where \mathscr{M}=\operatorname{ran}E and \mathscr{N}=\operatorname{ker}E . If in (1), "+" meant orthogonal sum, your reasoning would be correct. But this is not what Conway ... 1 The notion of idempotent does not require a Hilbert space structure, you simply require an operator, say acting on a Banach space, to satisfy T^2 = T. So T restricted to Ran(T) is the identity. On the other hand, when you write ^{\perp}, implicitly there is a Hilbert space structure. Then your condition Ker (P) = Ran (P)^{\perp} means P is a ... 2 For the first part, notice that if x\in\ell^1, then \sum_{k\geqslant n}|x_k|\to 0 as n goes to infinity. For the second part, take x the sequence whose (n+1)-th term is 1 and all the other 0: we find that the norm of A_n is 1. 2 Hint. We can be write$$ T = A + iB $$where A and B are self-adjoint operators defined by$$ A := \frac 1 2 (T + T^+)\\ B := -\frac i 2 (T - T^+) $$2 You can use the sesquilinear forms.Especially a square form. So let's say that f:H \times H-> \Bbb C is a square form that f(x,x)=<Tx,x> then you can so that \lVert f \rVert \leq \lVert T \rVert \leq 2\lVert f \rVert. By sesquilinear map we mean a map φ:H\times H->\Bbb C with properties: 1)The map is linear to the first variable,the ... 0 For A as you state, if A^{\star}Af=0 for some non-zero f in the space, then$$ 0 = (A^{\star}Af,f) = (Af,Af)=\|Af\|^{2}, $$implies that Af=0. So the null space of A^{\star}A is the same as the null space of A. This has a definite effect on the ground state solutions. For example, consider H=-\frac{d^{2}}{dx^{2}} on the domain consisting ... 1 The question seems to be, "Show that zero can be a nontrivial eigenvalue of a positive unbounded operator." To do this, modify Martin Argerami's answer: Choose an orthonormal basis \{e_n\}_{n=1}^\infty of your Hilbert space and define Ne_n = (n-1)e_n. This in particular has Ne_1 = 0 = 0e_1, so 0 is a nontrivial eigenvalue. (This operator N is ... 1 Having zero as an eigenvalue is a synonym of having non-trivial kernel. But you can easily have a positive operator that is injective, i.e. such that zero is not an eigenvalue. For example, fix an orthonormal basis \{e_n\} and let Ne_n=ne_n. 0 This is not true. The functions \{e^{inx}\} do not span C^1_{\text{per}}; indeed, no countable set can. So using the axiom of choice, one can show the existence of a linear functional that vanishes on all the e^{inx} but is not continuous. You can't tell whether a linear functional is continuous by looking at a proper subspace, even if that subspace ... 2 The proof uses the easier half of the Eberlein-Šmulian Theorem, the closed unit ball of a reflexive Banach space is weakly sequentially compact. Accepting that as given, from any bounded sequence (x_n) we can extract a weakly convergent subsequence (x_{n_k}), and by the complete continuity, T(x_{n_k}) is convergent. Thus the image of the closed unit ... 1 I completely agree with the previous answer but I would like to add that in general there are even infinitely many extensions of the Laplacian - not only Dirichlet and Neumann. E.g., the Laplacian with all Robin-type boundary conditions$$ \frac{\partial u}{\partial n}=pu_{|\partial X} $$will do the job. 1 It is not hard to see that it is as you say when \mathcal A is a von Neumann algebra. But there are C^*-algebras with few to no projections, and so none of those masas can live there. Consider for instance C^*_r(\mathbb F_2), the reduced C^*-algebra of the free group on two generators \mathbb F=\langle a,b\rangle. This algebra is known to be ... 1 There certainly exist references for this topic. Here is the one I found the most helpful. Stone's Theorem in C*-algebras, J. HOLLEVOET, J. QUAEGEBEUR and S. VANKEER I found it in the references of the paper On generalized Stone's Theorem, Massoud Amini which is suggested by user "Argument" in the comments. There is also apparently a book, though I ... 0 See Trace Ideals and Their Applications by Barry Simon. He covers the general theory of Schatten ideals and their applications, with particular attention to applications in quantum physics. 0 I will use complex Fourier series for ease of writing; everything works the same for real series. Suppose f\in C^1_{\rm per}, f(x)=\sum c_n e^{i n x}. Since f' is continuous, it belongs to L^2. The Fourier series of f' is f'(x)=\sum inc_n e^{i n x}. By Parseval's identity,$$\sum n^2 |c_n|^2 = (2\pi)^{-1} \int_0^{2\pi} |f' |^2 \le ...

1

For your question The statement $$\| P(A)\|^2 = \|P(A)^* P(A)\|$$ holds since, for any bounded operator $X$ on a Hilbert space $H$ the following holds $\|X\|^2 = \|X^*X\|$ (C*-identity). Proof Let $h \in H$ have norm less or equal $1$, $$\|Xh\|^2 = \left<Xh, Xh\right> = \left<X^*Xh, h\right> \leq \|X^*Xh\|\|h\| \leq \|X^*X\|\leq \|X^* ... 1 I believe you need that X is reflexive. Let's replace the subsequence x_{n_k} by x_n for ease of notation. If X is reflexive, then the sequence x_n converges weakly to some x \in X. (every bounded sequence in a reflexive Banach space has a weakly convergent subsequence) Then, because x_n \to x weakly, Kx_n \to Kx strongly because K is ... 2 Sufficient criteria for the existence of a w\in X with Tx_{n_k}\to Tw are \mathcal{R}(T) is finite-dimensional (well, closed, but if a compact operator has closed range, the range is finite-dimensional). X is a reflexive Banach space. Then x_{n_k} has a weakly convergent subsequence, say to w, and since a continuous operator is also weakly ... 0 Consider x=(x_1,x_2,\dots ,x_n)^T. Then Mx=(\sum_{i=1}^{n}1x_i, \sum_{i=1}^{n}1x_i,\dots ,\sum_{i=1}^{n}1x_i)^T and ||M||=\sup_{||x||=1}{||Mx||}=\sup_{||x||=1}\{n|\sum_{i=1}^{n}x_i|\}\leq \sup_{||x||=1}\{n\sum_{i=1}^{n}|x_i|\}=n ||x||=n, if using p-norm with p=1 for \mathbb{C}^n (inequality follows from the triangle inequality). Equality is ... 3 You have Mx = (e^T x) e, where e=(1,...,1)^T. Hence \|Mx\| = |e^Tx | \|e\| = \sqrt{n} |e^T x| \le n \|x\|. Setting x={1 \over \sqrt{n}}e gives Mx = \sqrt{n}e = n x, hence \|M\| = n. 0 Hint: your matrix is proportional to a one-dimensional projector, i.e.$$ M = n |\chi\rangle \langle\chi|$$with$$ |\chi\rangle = \frac{1}{\sqrt{n}} \sum_{j=1}^n |j\rangle $$1 For the equality: as you noted, \|ST\|\le\|S\|\,\|T\| for all S,T, so we only need to show \|AA^\ast\| \ge \|A\|\,\|A^\ast\|. Here we go:$$ \|A^\ast x\|^2 = \langle A^\ast x,A^\ast x\rangle = \langle x,AA^\ast x\rangle \le \|x\|\,\|AA^\ast x\| \le \|AA^\ast\|\,\|x\|^2 $$Dividing by \|x\|^2 and taking the supremum over nonzero x yields ... 1 Choose any subsequence \{n_k e_{n_k}\}, and define$$ y := \sum_k \frac{1}{n_k} e_{n_k} $$Check that y is a well-defined element of H, and that$$ \lim_{k\to \infty} \langle n_k e_{n_k}, y\rangle = 1 \neq 0  Hence, $n_k e_{n_k}$ does not converge to zero weakly.

2

a) You use that the compacts are an ideal. b) Let $T$ be the operator defined on the canonical basis by $Te_{2j-1}=0$, $Te_{2j}=e_{2j+1}$, $j\in\mathbb N$. Then $T$ is not compact, as it maps an infinite orthonormal set into another. But $T^2=0$, which of course is compact. c) For any $T$, if $T^*T$ is compact, then so is $T$. Because then ...

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