# Tag Info

## New answers tagged operator-theory

0

Using the Spectral Theorem, $$Ax=\int_{0}^{\infty}\lambda dE(\lambda)x \\ \mathcal{D}(A) = \left\{ x : \int_{0}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty \right\}.$$ Then the positive square root $\sqrt{A}$ is $$\sqrt{A}x = \int_{0}^{\infty}\sqrt{\lambda}dE(\lambda)x \\ \mathcal{D}(\sqrt{A}) = \left\{ x : ... 2 The Spectral Theorem for A is given in terms of a Borel Spectral measure E$$ Ax = \int_{-\infty}^{\infty}\lambda dE(\lambda)x, $$and x \in \mathcal{D}(A) iff$$ \int_{-\infty}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty. $$The operator e^{iA^2} is defined through the functional calculus as$$ e^{iA^2}x = ...

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 ...

2

Example, $$A = \frac{1}{i}\frac{d}{dx}$$ on the domain $\mathcal{D}(A)$ of absolutely continuous functions $f \in L^2[0,1]$ for which $f' \in L^2[0,1]$ and $f(0)=0$. Then $A^*$ is the same as $A$ except that the condition $f(0)=0$ is replaced by $f(1)=0$. Then $A^{\star\star}=A$ because $A$ is closed and densely-defined. However, ...

1

Yes, it is fine if you interpret $\langle D^2 f(x) , e_n \rangle$ as the bilinear form $$(y,z) \mapsto \langle D^2 f(x)[y,z], e_n \rangle.$$ As you already said, this follows simply from the chain rule and the linearity of $L_n$.

1

Let $T$ be the Fréchet derivative of $f$ at $e\in E$. Then $$\frac{|f_n(e+h)-f_n(e)-\langle Th,e_n\rangle|}{\|h\|} =\frac{|\langle f(e+h),e_n\rangle-\langle f(e),e_n\rangle-\langle Th,e_n\rangle}{\|h\|} =\frac{|\langle f(e+h)- f(e)- Th,e_n\rangle}{\|h\|} \leq\frac{\| f(e+h)- f(e)- Th\|}{\|h\|}\to0.$$ So the derivative of $f_n$ is $h\longmapsto \langle ... 3 You can write the similarity as$NS=SM $. As$N $and$M $are normal, the Fuglede-Putnam theorem guarantees that$N^*S=SM^*$. Taking adjoints,$S^*N=MS^*$. Then $$S^*SM=S^*NS=MS^*S.$$ Using this identity repeteadly,$p (S^*S)M=Mp (S^*S ) $for all polynomials; taking limits,$f (S^*S)M=Mf (S^*S) $for all continuous functions$f $. In particular, if ... 1 They are not equivalent on an infinite-dimensional Hilbert space. The weak convergence for operators in this case is in the usually called "weak operator topology": $$A_n\xrightarrow{wot} A\ \ \iff\ \ \langle A_nx,y\rangle\to\langle Ax,y\rangle,\ \ \forall x,y\in H.$$ The weak operator topology is known to be coarser than the$\sigma$-weak operator ... 1 Let's try to prove this rigorously. Lemma 1 Let$E$be a topological space$t\ge 0$and$t_0^{(n)},\ldots,t_n^{(n)}\ge 0$with $$0=t_0^{(n)}<\cdots<t_n^{(n)}=t$$ for some$n\in\mathbb NF:[0,t]\times H\to E$be continuous$(X_t)_{t\ge 0}$be a left-continuous$H$-valued stochastic process on$(\Omega,\mathcal A,\operatorname P)$... 2 Showing that$T$is well-defined amounts to observing that, for any orthonormal set$\{ e_n \}_{n=1}^{\infty}$in a Hilbert space, the sum$\sum_{n=1}^{\infty}\alpha_n e_n$converges in$H$to a vector$y$iff$\sum_{n=1}^{\infty}|\alpha_n|^2 < \infty$, and, in that case,$\|y\|^2 = \sum_{n=1}^{\infty}|\alpha_n|^2$. Because$\sum_{n=1}^{\infty}|(x,e_n)|^2 ...

0

Well, \begin{align} S_n&:=\sum_{i=1}^n\int_{t_{i-1}}^{t_i}L_i\Phi_0\;{\rm d}W_s\\ &=\sum_{i=1}^n\int_{t_{i-1}}^{t_i}F_x(t_{i-1},X_{t_{i-1}})\Phi_0\;{\rm d}W_s\\ &=\int_0^t\bigg[\sum_{i=1}^nF_x(t_{i-1},X_{t_{i-1}})\Phi_0\mathbf{1}_{(t_{i-1},t_‌​i]}(s)\bigg]\;{\rm d}W_s. \end{align} Now if $F(s,x)$ is continuous for $(s,x)\in[0,t]\times H$ and ...

0

Yes. The continuity of all $f\circ T$ implies that $T$ has closed graph: If $x_n\to 0$ and $Tx_n\to y$ then we have to show $y=0$. But the continuity of $f$ and $f\circ T$ imply $f(y)=\lim\limits_{n\to\infty} f(T(x_n))=f(T(0))=0$; as this holds for all $f\in Y^*$ we get $y=0$ from Hahn-Banach. The closed graph theorem thus implies that $T$ is continuous.

1

You can also finish your proof by noting that $k \le 2 \, \|x\|^2$ (by applying Hölder's inequality). Hence, $$\langle L \, x , x \rangle \ge -4 \, k + 17 \, \|x\|^2 \ge 9 \, \|x\|^2.$$

1

Your isometries are not such. Consider $$\rho_1=\begin{bmatrix}1/2&0\\0&1/2\end{bmatrix},\ \ \rho_2=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$$ Then $$\|\rho_1\|=1/2,\ \ \|\rho_2\|=1,$$ while $$\|\rho_1\|_1=\|\rho_2\|_1=1.\ \$$ For the proof, if $\rho=\sum_k\lambda_kP_k$ with $\sum_k\lambda_k=1$ and $0\leq\lambda_k$ for all $k$ and at ...

1

Let $e_1,\ldots,e_k$ be an orthonormal basis of $\text{Coker}\,P$. Note that $P(H)$ is closed, so there exists $f_1\in\mathfrak A$ with $$\text{dist}\,(f_1,e_1)<\frac{\text{dist}\,(e_1,P(H))}2.$$ The triangle inequality guarantees that $f_1\not\in P(H)$. Now $\text{span}\,\{P(H),f_1\}$ is closed, and we can repeat the process to obtain ...

2

If $V$ is the bilateral shift, we have $$L=17 I - 4(V+V^*).$$ From $\|V\|=1$, we get that $V+V^*$ is a selfadjoint with $\|V+V^*\|\leq2$. Then, for a unit vector $x$, $$\langle Lx,x\rangle=17-4\langle(V+V^*)x,x\rangle\geq17-4\|V+V^*\|\geq 17-8=9.$$ In other words, you can take $c=9$.

0

Let me split this answer into two parts: Part 1 Let $U$ and $H$ be arbitrary Hilbert spaces, $L\in\mathcal L(U,H)$ and $x\in H$. As Q. Huang noted in his answer, the authors of Stochastic Differential Equations in Infinite Dimensions$^3$ "define" $$(L^\ast x)u:=\langle Lu,x\rangle\;\;\;\text{for }u\in U\;.\tag 7$$ I hate that, it's awful. Why? Well, cause ...

1

There is exactly a definition of the term $\int_0^t\langle\Phi_s{\rm d}W_s,F_x(s,X_s)\rangle$. For $\Phi_s$ taking values in $\mathfrak L_2(Q^\frac{1}{2}U,H)$ and satisfying the condition that the integral of $\Phi_s$-'s square-norm in $\mathfrak L_2(Q^\frac{1}{2}U,H)$ is a.s. finite (just called the "Energe Condition" privately), and for $\Psi_s$ a ...

1

You haven't written the Laplace transform correctly in several ways: Notice that the left depends on $f$ while the integrals do not. Furthermore, the kernel is $e^{-st}$, not $e^{-sct}$. But once you have $$\mathcal{L}\left\{f(ct)\right\} = \int_0^{\infty} e^{-st} f(ct) \, dt$$ Now make the change of variables $u = ct$ to rewrite this as $$\frac 1 c ... 0 Be careful: the definition of the Laplace transform is $$\mathcal{L}\{f(t) \} = \int_{0}^{\infty}f(t)e^{-st}dt,$$ so when you substitute ct in, the e^{-st} factor needs to remain as is. 0 The inclusion i is surjective, because both B(\mathbb C^2)\otimes B(\mathbb C^2) and B(\mathbb C^4) have the same dimension (concretely, 16). The issue, and what you showed, is that if you restrict i to the subset of elementary tensors, then i is not surjective. Your element p is not a sot limit of elementary tensors. For starters, because we ... 1 This seems false. Take U=H, Z=Id, and identify L(H,\mathbb{R}) with H. Take and orthogonal base e_1,e_2\ldots, and define f(e_1)=e_2, f(e_i)=0 for i>1, extend by linearity. Then$$ \langle e_2,f(e_1)\rangle=1\neq0\langle e_1,f(e_2)\rangle. $$2 Your two ideas would have been my first attempts; but I have no idea how to make them work (well, for the first one, I would try with T the flip, but I still wouldn't know how to do it). Let K\subset B(\ell^2) be the compact operators. On \overline {B(\ell^2)\odot B(\ell^2)}, consider the ideals \overline{K\odot B(\ell^2)} and \overline{K\odot ... 0 I figured out what is wrong in my example. Of course D as I defined it is not dense. I mixed up two spaces and didn't saw it later on. Thanks for any comment - they helped alot to rethink my example! 0 Integration by parts \int_{-1}^1 f'g=[fg]_{-1}^1-\int_{-1}^1 fg' is a good idea. But you have to be in a context where the term [fg]_{-1}^1 is zero. In order that this condition is always fulfilled, you must restrict your working space to functions that are zero at both bounds -1 and 1 (it is no longer a vector space). In this framework, we have ... 0 Here is a counterexample. Let X=Y=H be an infinite-dimensional Hilbert space with orthonormal basis \{e_j\}. Let T_k be the projection onto the span of \{e_1,\ldots,e_k\}. Let D be the span of \{e_1,e_2,\ldots\}. For each x\in D, for k big enough T_kx=x, so the sequence \{T_kx\} is Cauchy. As we have T_kx\to x, we have T_k\to I ... 1 The inequality does not hold in general. Let$$ T=\begin{bmatrix}1&1\\0&0\end{bmatrix},\ \ Q_1=\begin{bmatrix}0&0\\0&1\end{bmatrix}, \ \ Q_2=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$Note that$$ ...

2

This is what in finite dimension is called the gradient of $f$, if it exists (and which you may call gradient in this case as well). It's the same idea as in the finite dimensional case. In order for this to work you need a natural isomorphism between the vector space and it's dual (which you usually don't have but) which is induced by the scalar product in ...

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 $T=\lambda_1\,E$.

3

The trick is to show the contrapositive. If an operator $T$ has infinite-dimensional and closed image, then it is not compact. Indeed, by restriction we get a bijective bounded linear operator $T:(\ker T)^\perp\to\text{Im}\,T$. By the open mapping theorem, $T$ maps open sets to open sets. So the image of the unit ball is open, and this is not a compact ...

0

We start from $$Df(t) = DA\bigl(B(t)\bigr)U$$ By the product rule, we have (note that the product rule holds for composition of linear maps, as the product is bilinear and continuous): $$D^2 f(t) = D\Bigl(DA\bigl(B(t)\bigr)\Bigr)U + DA\bigl(B(t)\bigr)DU$$ As $U$ is a constant, and does not depend on $t$, $DU = 0$, we have, by the chain rule $$D^2 ... 1 No. Let$$ T=\begin{bmatrix}0&1\\0&0\end{bmatrix},$$and take P(x)=x. You have \sigma(T)=\{0\}, P(0)=0, but P(T)\ne 0. If we are talking operators on a Hilbert space and T is normal, then the answer is yes, because the spectrum of P(T) is P(\sigma(T)) (always) and for a normal operator if the spectrum is \{0\} then the operator is ... 1 You are implicitly assuming that A_0 is injective. Let y\in H. Then there exists x\in D(A_0) with y=A_0x. Then$$ \langle A_0^{-1}y,y\rangle=\langle x,A_0x\rangle\geq0. $$Thus A_0^{-1} is positive, and so there exists an orthonormal basis \{e_n\} of eigenvectors. Since A_0^{-1} is compact, its eigenvalues \lambda_n satisfy ... 1 Yes, it is enough. Because A_0 is positive, symmetric and surjective, then A_0 is densely-defined, injective and selfadjoint. Therefore, A_0^{-1} is compact, selfadjoint with trivial null space. So A_{0}^{-1} has an orthnormal basis of eigenfunctions \{e_n \} with corresponding eigenvalue sequence of positive numbers$$ \lambda_1 \ge ...

1

Yes. Since $Z=|PTP|=(PTPTP)^{1/2}$, it is a limit of polynomials of the form $PX_nP$, and so $PZP=Z$. The spectral projections of an operator always belong to the von Neumann algebra generated by the operator. If $\mathcal M=W^*(|PTP|)=\{|PTP|\}''$, then $E^{|PTP|}(\Delta)\in\mathcal M$ for any Borel set $\Delta$. From the first paragraph we now that ...

0

They key observation is that if $\lambda_n=0$ then $Qe_n=0$, and that $$\ker Q=\overline{\text{span}}\,\{e_n:\ \lambda_n=0\},\ \ \ \ (\ker Q)^\perp=\overline{\text{span}}\,\{e_n:\ \lambda_n>0\}.$$ To check this, let $x\in U$. We can write, since $\{e_n\}$ is an orthonormal basis, $$x=\sum_n\alpha_n\,e_n.$$ Then $$... 1 In a complex Hilbert space, every positive operator is selfadjoint. But in a real Hilbert space, one can find positive operators which are not selfadjoint; for instance,$$ T=\begin{bmatrix}1&-1\\ 0&1\end{bmatrix} $$as an operator on \mathbb R^2 satisfies \langle Tx,x\rangle\geq0 for all x. This ambiguity is common in the definition of real ... 2 This is not true. Let$$ T=\begin{bmatrix}1&2\\2&4\end{bmatrix},\ \ P=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$Then T is positive (selfadjoint, with eigenvalues 0 and 5), but$$ T-PTP=\begin{bmatrix}0&2\\2&4\end{bmatrix} $$is not positive (selfadjoint with eigenvalues 2\pm2\sqrt2). 0 Take H=\mathbb{C}^2, T=\begin{pmatrix}-1&-1\\6&4\\ \end{pmatrix}, P=\begin{pmatrix}0&0\\0&1\\ \end{pmatrix}. Then T is positive with eigenvalues 1,2, but T-PTP=\begin{pmatrix}-1&-1\\6&0\\ \end{pmatrix} is not positive. 1 Let X = L^2[-\pi,\pi], and let A=\frac{d^2}{dx^2} on the domain consisting of all polynomials. Let B=\frac{d^2}{dx^2} on the domain \mathcal{D}(B) of all linear combinations of \{ \sin(t),\sin(2t),\ldots \}. Both A and B are densely-defined, and they're both closable. However \mathcal{D}(A)\cap\mathcal{D}(B)=\{0\}, which forces ... 1 Assuming you mean the closure in the natural topology on AC, that being the one given by the norm |f(0)|+||f'||_1, then yes. This is clear because any L^1 function (of mean zero) can be approximated in L^1 by continuous functions (of mean zero). In detail: Choose a sequence g_n of continuous functions such that \int_0^{2\pi}g_n=0 and ... 1 Well, self adjoint implies \langle Ax,x\rangle is real for all x in the Hilbert space, and that better be the case if \langle Ax,x\rangle \geq 0 for all x. In fact, positive operator implies self adjoint. Indeed, since \langle Ax,x\rangle =\langle x,Ax\rangle=\langle A^*x,x\rangle for every x, we have \langle (A^*-A)x,x \rangle =0 for every ... 1 You have a linear bijection$$ Q^{1/2} : U \rightarrow U_0 $$Q^{1/2} is an isometric isomorphism because, by definition, Q^{1/2} is surjective, and is injective because \lambda_n > 0 for all n, and$$ \|Q^{1/2}y\|_{U_0}=\|Q^{-1/2}Q^{1/2}y\|_{U}=\|y\|_{U}. $$That also implies that \{ f_n=Q^{1/2}e_n \} is an orthonormal ... 3 When you say K is continuous I assume it is with respect the norm on C([0,1]), that is, |f\|_\infty. The two norms are not comparable. Let K be the identity. Then$$\sup_{\|f\|_\infty\leq 1} \|Kf\|_\infty=1$$but$$\sup_{\|f\|_2\leq 1} \|Kf\|_\infty=\infty.To see it consider a sequence \{f_n\} of functions such that f_n is supported on ... 1 The answer by TrialandError already explained the given characterization of the domain of the closure. Just some remarks on your first question when D(\bar A) and X coincide: The operator \bar A is closed by definition. If D(\bar A)=X, then \bar A is bounded by the closed graph theorem, and it follows that A is bounded as well. Conversely, if ... 1 The graph is a subspace of X\times Y, and it's never going to equal X\times Y unless X=\{0\}, Y=\{0\}, which keeps you from applying whatever result you were thinking about for A\times B. Here's a simple way to look at the graph of a linear operator: Theorem [Linear Operator Graph] Let X and Y be vector spaces over the same field. Let ... 1 Define S=V+V^* then S:L^2(0,1) \to L^2(0,1) is a bounded operator. First we show that S is idempotent (S^2=S). Let f \in L^2(0,1): \begin{align*} S^2f = S(Sf) &= S(\int_0^x f(t)dt + \int_x^1 f(t)dt) \\& = S\left(\int_0^1 f(t)dt\right) \\ &= \int_0^x\int_0^1 f(t) dt + \int_x^1 \int_0^1 f(t)dt)\\ &=\int_0^1f(t)dt=Sf\end{align*} So S ... 1 I will assume that the norm \|AB\| in the definition of \|A\|_{L(X)} is \|AB\|_{L(\mathbb C^n)}. The two norms are equal: you have, by definition, \|A\|_{L(X)}\leq \|A\|_{L(\mathbb C^n)}, $$since \|AB\|_{L(\mathbb C^n)}\leq \|A\|_{L(\mathbb C^n)}\,\|B\|_{L(\mathbb C^n)}=\|A\|_{L(\mathbb C^n)}. Conversely,$$ ...

1

The closed graph theorem indeed suggests itself: Suppose that $f_n\to f$ and $\phi f_n\to g$ in norm. We must show that then $g=\phi f$. This follows from $$g(x)=\langle K_x, g\rangle = \lim\, \langle K_x, \phi f_n\rangle =\phi(x) \lim\, \langle K_x, f_n\rangle =\phi(x)f(x) .$$

1

We will use the following results of my first answer: $y\ge \frac x2(3x-1)$. The bound given by the OP is attained for every $n$ and every admissible $x$. Set  A_0=\begin{pmatrix} 1&0\\ 0&0 \end{pmatrix},\quad B_0=\begin{pmatrix} 1/4&-\sqrt{3}/4\\ -\sqrt{3}/4&3/4 \end{pmatrix}\quad\text{and}\quad C_0=\begin{pmatrix} 1/4&\sqrt{3}/4\\ ...

Top 50 recent answers are included