New answers tagged

0

A good example to think about this is to take $X=Y=\ell^2(\mathbb N)$, and $$ Ax=(x_1,\frac{x_2}2,\frac{x_3}3,\ldots) $$ Then $A$ is compact (it is a norm limit of finite-rank. The image of $A$ is dense, because if contains all the elements of the canonical orthonormal basis. And $$ \overline{A(B(0,1))}=\overline{\{x\in\ell^2(\mathbb N):\ \sum_n n^2|x_n|^2&...


2

There are a few subtle points. Typically, the domain for $L$ would consist of all absolutely continuous $f \in L^2$ such that $f' \in L^2$. Absolute continuity is required in order to integrate by parts, for example. This operator $L$ is closed, and that's useful to know in cases like this. And $\mathcal{D}(L)$ is dense in $L^2$. The adjoint $L^*$ consists ...


1

No, because not every set is/contains a ball, e.g. the Hilbert cube.


1

To your first question, let $$ p=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ \ q=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}. $$ Then $$ p\vee q=\begin{bmatrix}1&0\\0&1\end{bmatrix},\ \ \ p+q=\begin{bmatrix}3/2&1/2\\1/2&1/2\end{bmatrix}, $$ so $$ p+q-p\vee q=\begin{bmatrix}1/2&1/2\\1/2&-1/2\end{bmatrix}, $$ which is ...


1

(You don't say how you got the second equality; since it is not trivial, I'm not sure how you did it and so it is done below) Since $A^*A$ is positive and compact, it is orthogonally diagonalizable (spectral theorem): $A^*A=U^*D^2U$ for some unitary $U$ and $D$ diagonal with diagonal $s_1(A),s_2(A),\ldots$ Assume $s_1(A)\geq s_2(A)\geq \cdots$ Since $U$ ...


1

Note that the two equalities $A=AA^*A$ and $A^*=A^*AA^*$ are the same, since you can obtain one from the other by taking adjoints. Assume first that $A=AA^*A$. By multiplying by $A^*$ on the left, we get $$ A^*A=(A^*A)^2.$$ It follows that the eigenvalues of $A^*A$ all satisfy the equation $\lambda=\lambda^2$, so only $0$ and $1$ are possible. Conversely,...


1

If $\psi$ is a state vector, meaning $\|\psi\|=1$, then the expected value of an observable $A$ for the system in the state $\psi$ is $$ (A\psi,\psi). $$ That's what you're trying to show. One must assume that $\psi$ is in the domain of the observable $A$. A selfadjoint operator $A : \mathcal{D}(A)\subseteq H\rightarrow H$ on a Hilbert space $H$ ...


1

$1\iff4$ : Assume $1$. Given $x\in N_+$, there exists nonzero $y\in N_\tau^+$ with $y\leq x$. Now use Zorn to find a maximal ordered family $\{y_j\}\subset N_\tau^+$ with $y_j\leq x$ for all $j$. As the net is bounded, it has a sup, say $y=\lim_{sot}y_j$. Then $y=x$, because otherwise a nonzero element of $N+\tau^+$ below $y-x$ contradicts the maximality. ...


2

Since $T - \lambda I$ is also normal, we have $$ \| T - \lambda I \| = \text{spr} (T - \lambda I) = 0, $$ showing that $T = \lambda I$. (I recently asked basically the same question (Self-adjoint operator with single point spectrum), but your formulation is more general so I thought it might be worth sharing the answer here.)


1

using spectral theorem since $T$ is normal it exist a spectral measure $E$ such that $$ Tx=\int_{\sigma(T)}tdE(x)=\int_{\{\lambda\}}tdE(x)=\lambda E(\{\lambda\})(x)=\lambda E(\sigma(T))(x)=\lambda I (x)=\lambda x $$


0

Note: There are (at least) two ways of indexing the Fibonacci series. Below I work with the $F_0 = F_1 = 1$ convention rather than the $F_0 = 0, F_1 = F_2 = 1$ convention. I assume you want $D$ to be a differential operator, given the tags. Then we have $$D = \sum_{k = 0}^\infty f_k(x) \frac{d^k}{dx^k}$$ for some functions $f_k(x)$, and we want $$F_n ...


1

Using Binet's formula: $$F_n=\frac{1}{\sqrt{5}}\left(\varphi^n-\psi^n\right)$$ where $\varphi=\frac{1+\sqrt{5}}{2}$ and $\psi = \frac{1-\sqrt{5}}{2}$. Then $$(Df)(x) = \frac{1}{\sqrt{5}}\left(f(\varphi x)-f(\psi x)\right)$$ If you write the scaling operators $(S_\alpha f)(x)=f(\alpha x)$, then you can write the above as: $$D=\frac{1}{\sqrt{5}}\left(S_\...


1

Consider $u\in B(H_1\oplus H_2, H_1\oplus H_2)$ and since $u^*u$ is projection, then $u$ is partial isometry, which means that $uu^*u=u.$


0

Using the dual basis, the matrix representation of $T^+$ is given by the transpose, $T^t$.


3

False. The closed span of $e_k$ for $k \ge n$ is invariant.


2

The result that most resembles what you might be looking for is the following result of E. Hille (see Theorem 4 of Page 54 here) : For an operator $T \in L(H)$, we say that $\sigma(T)$ is incongruent $\pmod{2\pi i}$ if $$ \sigma(T)\cap [\sigma(T) + 2k \pi i] = \emptyset\quad\forall k\in \mathbb{Z}\setminus \{0\} $$ In other words, if $\lambda_1,\lambda_2 \in ...


2

Start with $$ \overline{\mathcal{R}(T)}=\mathcal{N}(T)^\perp. $$ As you noted, $\mathcal{N}(T^{1/2})\subseteq\mathcal{N}(T)$. Because $T^{1/2}$ is selfadjoint, $$ \|T^{1/2}x\|^2=(T^{1/2}x,T^{1/2}x)=(Tx,x). $$ Therefore $\mathcal{N}(T)\subseteq\mathcal{N}(T^{1/2})$. So, $$ \overline{\mathcal{R}(T)}=\mathcal{N}(T)^{\perp}=\...


2

If $S=T^*$, and $f_j\to f$ weak$^*$ in $Y^*$, then for any $x\in X$ $$ Sf_j(x)=f_j(Tx)\to f(Tx)=T^*f(x)=Sf(x). $$ So $S$ is weak$^*$-continuous. Conversely, assume that $S$ is weak$^*$-continuous. We know that what our $T$ should satisfy if it exists: $Sf(x)=f(Tx)$. So let us use this to define $T$. For any $x\in X$, consider the functional on $Y^*$ ...


1

If $A$ is a bounded linear operator on a complex Hilbert space $H$ such that $\|A\| \le 1$, then there exists a positive operator measure $P$ on the unit circle $T$ such that $$ A^n = \int_{T} z^n dP(z),\;\;\; n=1,2,3,\cdots. $$ Note that $P(T)=I$. This becomes an abstract moment problem. If $A : \mathcal{D}(A)\subseteq H\rightarrow H$ is the ...


3

We have that $T^{\times}(b_j^{\times})(e_i)=b_j^{\times}(T(e_i))=b_j^{\times}(\sum\limits_k a_{k,i}b_k )=a_{j,i}.$ Therefore, $$T^{\times}(b_j^{\times})=\sum_{i} a_{j,i}b_i^{\times}.$$ It follows that the matrix representation of the adjoint is the transpose.


2

One can even have $T|_{c_0}=0$. Indeed, first note that if $X\subseteq \ell_\infty$ is isomorphic to $\ell_2$, then $X\cap c_0$ is finite-dimensional so there is a subspace $X\subseteq \ell_\infty$ isometric to $\ell_2$ with $X\cap c_0=\{0\}$. (Actually, by simple repetition of the coordinates in the standard way of embedding separable spaces into $\ell_\...


0

1) Your reasoning about continuity is wrong. Best is to read a bit about continuity. 2) Yes, the map is a bijection


4

There are two things that will help: First, whenever $c_0$ is isomorphic to a subspace of a separable Banach space $X$ then that subspace is complemented in $X$. Second, whenever $Y$ is a subspace of a normed space $X$ and $T:Y\to\ell_\infty$ is a continuous linear operator then there exists a continuous linear extension $\widetilde{T}:X\to\ell_\infty$ ...


0

Recall Gelfand's formula for the spectral radius of a bounded operator $T$: $$r(T) = \lim_{n\to\infty} \|T^n\|^{\frac1n}. $$ If $T$ is a self-adjoint operator and $\|f\|=1$ then $$\|Tf\|^2 = \langle Tf, Tf\rangle = \langle T^2f, f\rangle\leqslant\|T^2f\|\|f\|=\|T^2f\| $$ which implies $\|T^2\|=\|T\|^2$. By induction it follows that $\|T^{2^n}\|=\|T\|^{2^n}$ ...


1

Notice that $\|Lx\|^2 = \langle x, L^2 x\rangle$, and by assumption, this is $\le \|Ax\|^2=\langle x, A^2 x\rangle$. In other words, $L^2\le A^2$, and since the square root function is operator monotone, this implies that $L\le A$ and thus also $L+t\le A+t$. Now we obtain in the same way that $$ \|(L+t)^{1/2}x\|^2 = \langle x, (L+t) x\rangle\le \|(A+t)^{1/2}...


3

Yes. The Hahn-Banach theorem shows that if $Y$ is a normed vector space and $y\in Y$ then $||y||_Y$ is equal to the norm of $y$ regarded as a linear functional on $Y'$. Say $X_1$ and $X_2$ are normed spaces consisting of the same space with two different norms, and $X_1'=X_2'$. We need to show that if $||x_n||_1=1$ for $n=1,2\dots$ then $||x_n||_2$ is ...


1

Yes. The necessary and sufficient condition is that $A$ is bounded. Indeed, the domain of $AE_I$ is the whole Hilbert space, and the domain of $E_IA$ is the domain $\mathcal D(A)$ of $A$.


1

First, note that $X_2$ is an $A$-invariant subspace, so that $A_2:X_2\to X_2$. We can show formally that the adjoint of $A_2$ should be the restriction of $A^*$ to $X_2$, which is again $A^*$. It then suffices to note that the image of the closed unit ball under $A_2$ is a closed subset of the image of the closed unit ball under $A$ and is therefore ...


1

It follows from the definition of $\sup$. If $$ M=\sup\left( f(x)\ : x \in B\right), $$ then there exists a sequence $x_n\in B$ such that $$ M=\lim_{n\to \infty} f(x_n).$$ Apply this observation with $B=\text{unit sphere}$, $f(x)=|(Ax_n, x_n)|$ and $M=\|A\|$.


0

Hint You have to find a vector $x$ such that $||Ax||_1 = \max_\limits{j=1,...,d} \sum_\limits{i=1}^d |a_{ij}|$ Let $j_0$ such that: $\max_\limits{j=1,...,d} \sum_\limits{i=1}^d |a_{ij}| = \sum_\limits{i=1}^d |a_{ij_0}|$ Let $e$ the vector such that $e_j = 1$ if $j=j_0$ and $e_j = 0$ otherwise. Now compute $||Ae||_1$.


1

Note that $$ Tx \otimes \Phi(T)^*f = \Phi(T)x \otimes T^*f \implies\\ \langle (Tx \otimes \Phi(T)^*f)(y),g \rangle = \langle (\Phi(T)x \otimes T^*f)(y),g \rangle \quad \forall y \in X, g \in X^* \implies\\ g([\Phi(T)^*f(y)]Tx) = g([T^*f(y)]\Phi(T)x) \quad \forall y \in X, g \in X^* \implies\\ g([f(\Phi(T)y)]Tx) = g([f(Ty)]\Phi(T)x) \quad \forall y \in X, g \...


0

Your analysis is correct. Also your $\tilde f$ is the only such extension. By the Riesz representation theorem, every bounded linear functional g on $\mathbb{R}^3$ (with Euclidean norm) is of the form $g(x) = \left<x, y \right>$ for some $y \in \mathbb{R}^3$. If $g$ is defined by $y = (\alpha_1,\alpha_2,\alpha_3)$ for some $\alpha_3 \neq 0$, then $...


4

Definition [$C_0$ Semigroup]: Let $X$ be a Banach space, and let $T : [0,\infty)\rightarrow\mathcal{B}(X)$ be a function into the bounded linear operators on $X$. Then $T$ is a semigroup if $T(0)=I$ and $T(t)T(t')=T(t+t')$ for all $t,t' \ge 0$. $T$ is a $C_0$ semigroup if $\lim_{t\downarrow 0}T(t)x=x$ for all $x\in X$. Suppose $X$ is a Banach space, and ...


0

The contour $\Gamma$ doesn't matter, as long as all $\lambda_j$ are inside. You have, with $D=\text{Diag}\,(\lambda_1,\ldots,\lambda_n)$, $$ \lambda I-A=\lambda V^* V-V^*DV=V^*(\lambda I-D)V=V^*\,\begin{bmatrix}\lambda-\lambda_1&0&\cdots&0\\ 0&\lambda-\lambda_2&\cdots&0\\ \vdots& &\ddots&\vdots \\0&0&\cdots&\...


0

Case 1. You know approximations of the eigenvalues of $A$; then you know also (the calculation is in $O(n^3)$) a convenient matrix $V$ and finally $f(A)=V^*diag(f(\lambda_1),\cdots,f(\lambda_n))V$. Of course, you can also use the mickep's method. Yet, the calculation of $(\lambda I_n-A)^{-1}$ becomes complicated when $n$ grows. Case 2. You know only bounds ...


1

I don't know enough to give you a very authoritative answer. But, as far as I can tell, there is no "general theory" of non-selfadjoint subalgebras of $B(H)$ the way that there is a theory for c$^*$-algebras or von Neumann algebras. There is a rather complete classification of nest algebras, and some generalizations. The original source for non-selfadjoint ...


0

Any bounded linear functional defined on a subspace of a Hilbert space admits a unique norm preserving extension. The proof is given here. More on Banach spaces with unique extension property for functionals you can find in this discussion


2

For every $x$ we have $x=Px+(x-Px)$ where $Px\in \operatorname{im}P$ and $(x-Px)\in \ker P$. So, $\ker P$ is a complement of $\operatorname{im} P$. Suppose it is not an orthogonal complement; then there exist $u\in \operatorname{im}P$ and $v\in \ker P$ such that $\operatorname{Re}\langle u, v \rangle < 0$. For sufficiently small $t>0$ we have $\|u+tv\...


3

You have, since $0\leq q\leq I$ and $0\leq p\leq I$, $$ -I\leq -q\leq p-q\leq I-q\leq I. $$ So, as you mentioned, it follows that $\sigma(p-q)\subset[-1,1]$. Note also that the argument does not use that $p,q$ are projections, only that they are positive elements of the unit ball.


1

As mentioned in the comments, you are assuming the underlying space $H$ is a Hilbert space, and you're assuming the operators $A_j$, $B$ are defined everywhere on $H$. I'll assume $H$ is also complex. Then any symmetric operator on $H$ is bounded because it is closed, as noted in the comments. I guess you were able to show that $a(x)=\lim_j (A_jx,x)$ exists ...


1

In the real case it is especially easy. Expand the right hand side of the following: $$ 0=\langle A(x+y), x+y\rangle, $$ you get $2\langle Ax, y\rangle=0$ for all $x, y\in H$. In the complex case you have to play a little bit more because you arrive at $2\Re \langle Ax, y\rangle=0$, but that's not a serious issue.


3

You can use the trick which lets you recover a scalar product from the associated norm. Let $A$ be as in your question, and $x$, $y$ be any elements of the Hilbert space. Then: $$0 = \langle A(x+y)|x+y\rangle = \langle Ax|x\rangle + \langle Ay|y\rangle+\langle Ax|y\rangle+\langle Ay|x\rangle = \langle Ax|y\rangle+\langle Ay|x\rangle.$$ Since $A$ is ...


0

Suppose that $a$ is positive, that $\|a\|=1$ and that $a-1$ is invertible. This means that $0\not\in\sigma(a-1)$, so $1\not\in\sigma(a)$. As the spectrum is closed, there exists $\varepsilon>0$ such that $(1-\varepsilon,1+\varepsilon)\cap\sigma(a)=\varnothing$. Consider the function $$f(t)=\begin{cases}t,&\ 0\leq t\leq 1-\varepsilon,\\ 1-\varepsilon,...


1

This operator is the orthogonal projection onto $U$, it holds $C^2=C$: First we find that $$ \langle Cv,e_n\rangle = \langle v,e_n\rangle, $$ since $(e_n)$ is orthonormal. This implies $$ C^2v = \sum_n \langle Cv,e_n\rangle Ce_n = \sum_n \langle v,e_n\rangle e_n =Cv. $$ Hence, $C^2 = C$, which implies $C^{1/2}=C$. Edit: I assumed that the scalar product ...


1

The fact that $\iota$ is HS means that if $\{e_n\}$ is an orthonormal basis of $U$, then $$\tag{1} \sum_n\langle e_n,e_n\rangle_V=\sum_n\|\iota(e_n)\|_V^2<\infty. $$ After identifying $U$ and $V$ with their respective duals, we have $C:V\to U$ given by $$\tag{2} \langle Cv,u\rangle_U=\langle v,\iota (u)\rangle_V=\langle v,u\rangle_V. $$ In particular, ...


2

The following formula for $\|T^{-1}\|$ is relevant for the question posted. Let $(\mathcal E, \|\cdot\|_{\mathcal E})$ and $(\mathcal F, \|\cdot\|_{\mathcal F})$ be Banach spaces and let $\mathcal L(\mathcal E,\mathcal F)$ be the space of all bounded operators from $\mathcal E$ into $\mathcal F$. Let $T \in \mathcal L(\mathcal E,\mathcal F)$. The following ...


1

Starting with the half of the question that no longer exists: $M$ invertible does not imply $T$ invertible. Say $\phi=e_{-2}$. Then $M$ is invertible. But if $f=e_1$ then $Tf=P(e_{-1})=0$, so $T$ is not invertible. On the other hand $T$ invertible does imply $M$ invertible. Suppose $M$ is not invertible. Let $\epsilon>0$, and set $$E=\{x:|\phi(x)|<\...


2

$W_C = \{u + iv \in X_C \mid u, v \in W\}$ is invariant under $T_C$, so we may as well forget about the real Banach spaces and consider an operator $T$ on a complex Banach space $X$ with a finite-codimensional invariant subspace $W$, such that $\sigma(T) \subseteq \mathbb R$. The claim is that $\sigma(T|_W) \subseteq \mathbb R$. If $\lambda \notin \sigma(T)...


0

I think you are right. If $Tf=\lambda f$, we have $ \lambda f = f + f(1) - f(0), $ or $$\tag{1} (\lambda - 1) \, f= f(1)-f(0). $$ If $\lambda=1$, then any $f$ with $f(1)=f(0)$ satisfies the equation, so $1\in\sigma_p(T)$. In particular, as you mention, constant functions are eigenfunctions for the eigenvalue $1$. When $\lambda\ne1$, the equation $(1)$ has ...


1

Say $x \in V$. Since the $e_n$ are a basis for $H$ and $x \in H$, $x = \sum_n c_n e_n$ for some coefficients $c_n \in \mathbb{K}$. Since $Q$ and $1-Q$ are orthogonal projections whose sum is 1, $x = Qx + (1-Q)x = (Q \sum_n c_n e_n) + ((1-Q) \sum_n c_n e_n)$. But the second term is zero since $x \in V$ implies $Qx = x$. So we have $x = Q \sum_n c_n e_n$, and ...



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