New answers tagged

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 $n\in\mathbb N$ $F:[0,t]\times H\to E$ 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 ...


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Be careful: the definition of the Laplace transform is \begin{equation} \mathcal{L}\{f(t) \} = \int_{0}^{\infty}f(t)e^{-st}dt, \end{equation} so when you substitute $ct$ in, the $e^{-st}$ factor needs to remain as is.


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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\\ ...



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