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If something is non-invertible, there's two (non-disjoint) possibilities: it fails to be injective, or it fails to be surjective. In finite dimension, these are the same, but in infinite-dimensional spaces, weird things can happen. If it fails to be injective, there's $x \ne y$ such that $(T - \lambda I)(x) = (T - \lambda I)(y)$. So $(T - \lambda I)(x - y) ... 5 For finite-dimensional vector spaces, injectivity and surjectivity are equivalent. That's not the case for an arbitrary Hilbert space. The classic examples are the left- and right-shift operators$L, R:\ell^2 \to \ell^2, given by \begin{align*} L(x_1, x_2, \dots) &= (x_2, \dots) \\ R(x_1, x_2, \dots) &= (0, x_1, x_2, \dots). \end{align*} The mapL$... 4 Let$J $be an index for the cardinality of an orthonormal basis of$H $. Then$H $is isometrically isomorphic to$\ell^2 (J) $, so it is enough to discuss the problem on this latter space. Define the product$fg $pointwise, i.e.$fg (j):=f (j)g (j) $. The question is whether this product stays in$\ell^2$, and whether the norm is submultiplicative. We ... 3 Another Banach-algebra structure on the Hilbert space$\mathsf{hs}(H)$of Hilbert-Schmidt operators on a Hilbert space$H$is just operator multiplication (composition). There is a natural involution on this algebra but it does not make it a C*-algebra. 3$T-\lambda I$being non-invertible does not imply there is a non-zero$x$with$(T-\lambda I)x=0$. That is true when$H$is finite-dimensional, but not necessarily when$H$is infinite-dimensional. The classic counterexample is the right-shift operator$R:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N})$. Take a look at the Wikipedia article on the notion of ... 2 Left/right shift operators are the standard examples, but I personally think that the multiplication operator is the easiest way to see that there may be something else in the spectrum besides eigenvalues. Consider a multiplication operator$A_c$on$\ell^\infty$($c\in\ell^\infty$) $$(A_c x)_n=c_n x_n.$$ The inverse if exists is clearly a multiplication ... 2 As noted in my comment,$E$has to be a Hilbert space. To see this, fix a linear functional$\varphi \in E'$with$\Vert \varphi \Vert = 1$. For$z \in E$, define $$A_z : E \to E, x\mapsto \varphi(x) \cdot z.$$ It is not hard to see that$E \to B(E), z \mapsto A_z$is linear and isometric. Thus, if$B(E) = H$is a Hilbert space, we see that$E$is ... 2 I know a very special case, and I guess it is extensible. Suppose that$H$is Hilbert space such that$H\cong (H_1\hat{\otimes}(H_2)^*)^*$, where$\hat{\otimes}$is projective tensor product, and$H_1,H_2$is some Hilbert spaces. For any Hilbert space$\mathcal{H}$, always$\mathcal{H}^{**}=\mathcal{H}$. Also for two Banach space$E,F$always ... 1 For your second question, the answer is always yes. More generally, given any Banach algebra$A$, let$\tilde{A}=A\oplus \mathbb{C}$be its unitization (with norm$\|(a,z)\|=\|a\|+|z|$). Given$a\in A$, let$L_a\in B(\tilde{A})$be left multiplication by$a$. Then$a\mapsto L_a$is an isometric isomorphism from$A$to a subalgebra of$B(\tilde{A})$. ... 1 With a clear domain definition, and$\|\frac{1}{\Delta t}\{U_{\alpha}(\Delta t)-1_\alpha\}\varphi_{\alpha}-H_{\alpha}\varphi_{\alpha}\|=\|\frac{1}{\Delta t}\int_{0}^{\Delta t}(U_{\alpha}(t)-1_\alpha)H_{\alpha}\varphi_\alpha dt\|$, can you now better establish convergence? 1 If$\alpha\neq 1$, then it fails for$x=y$, since the right hand side is$0$, and the left hand side is positive. 1 Yes. The functional calculus preserves approximation by polynomials. So, from $$T^n=\begin{bmatrix}A^n&0\\0&B^n\end{bmatrix},$$ you get that $$p(T)=\begin{bmatrix}p(A)&0\\0&p(B)\end{bmatrix}$$ for any polynomial$p$. Now using a sequence of polynomials that converges uniformly to$f$, you get that $$... 1 The implication 0\leq A\leq B \implies \mathcal RA\subset\mathcal RB can be proven as follows. From 0\leq A\leq B, we easily see that if Bx=0, then$$ 0\leq\langle Ax,x\rangle\leq\langle Bx,x\rangle=0, $$so A^{1/2}x=0, and thus Ax=0. In other words, \ker B\subset\ker A. Then$$ \overline{\mathcal RA}=(\ker A)^\perp\subset(\ker ... 1 Although$0 \le A \le B$implies$\overline{{\cal R}(A)} \subseteq \overline{{\cal R}(B)}$, it's not true without the closures. For a counterexample, take$L^2[0,1]$. Let$A$be multiplication by$x$(i.e.$A f(x) = x f(x)$), and$B = A + u u^*$where$u(t) = t^{1/4}$and$u^*\$ is the corresponding linear functional, i.e. $$B f(x) = A f(x) + u^*(f) u(x) = ... 1 This is always true, as \Omega \setminus (\Omega_1\cup \Omega_2) can only contain the boundaries of \Omega_1 and \Omega_2, which have zero measure due to the regularity assumptions on the domains. 1 Let \mathcal S be an orthonormal basis of eigenvectors for |A| (this exists, since |A| is trace-class and thus compact). Let \mathcal T be any other orthonormal basis. For any \sigma\in\mathcal S, we have$$ |A|\sigma=s_\sigma(A)\,\sigma $$and \sum_\sigma s_\sigma(A)=\sum_\sigma\langle |A|\sigma,\sigma\rangle<\infty. Then, for any B\in ... 1 Presumably, you mean for H to be a vector space over \Bbb C. Note that \operatorname{Herm}(H^k) is a vector space over \Bbb R (as opposed to \Bbb C), and that \dim \operatorname{Herm}(H^k) = k^2. From there, we can simply apply your earlier reasoning to find that$$ \mathrm{dim}(\mathrm{Sym}(\mathrm{Herm}(H^k)^{\otimes N})) = \binom{N + k^2 - ...

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