Tag Info

Hot answers tagged

7

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 map $L$ ...


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 - ...



Only top voted, non community-wiki answers of a minimum length are eligible