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0

A (probably to) simple example is $T:D\to X:x\mapsto 0$ for some unclosed dense $D\subsetneq\overline{D}=X$. There $T$ is bounded densely defined but not closed, however, closable. A more sophisticated bounded nonclosable example would require a little more work since one needs to consider an incomplete space and a nice fitted operator since: "A bounded ...


1

For Example: $A:H_0^1(I) \to L^2(I)$ $Af=df/dx$ but be careful $A:L^2(I) \to L^2(I)$, with the same definition, is not bounded.


0

$$ \begin{align} \nabla \cdot (fv) &= \sum \partial_i (f v_i) \\ &= \sum (f \partial_i v_i + v_i \partial_i f) \\ &= f \sum \partial_i v_i + \sum v_i \partial_i f \\ &= f \nabla \cdot v + v \cdot \nabla f \end{align} $$


1

Write out both expressions: $$\nabla\cdot(fv) = \nabla\cdot(fv_1,fv_2,fv_3) = \partial_1(fv_1) + \partial_2(fv_2) + \partial_3(fv_3) = (\partial_1f)v_1 + f(\partial_1 v_1) + \dots$$ and $$ \nabla f \cdot v = (\partial_1 f, \partial_2 f, \partial_3 f)\cdot(v_1,v_2,v_3) = (\partial_1 f) v_1 + \dots $$ $$ f \nabla \cdot v = f(\partial_1 v_1 + \dots) $$ then ...


0

For the closeness you have to pay attention carefully to domains, an operator can be turned into closed operator only by changing the domain of definition.


1

I found the answer, in a book written by robinson, 2001, "infinite dimensional spaces and systems" A $k$th order differential operator in the Sobolev space $H^k_0$ a closed operator.


1

Exactly, you are right, it requires $D(A)=D(A^*)$; but I emphasized on densely defined operator. In fact, is it possible to say that when an operator is densely defined on a suitable Hilbert space and also symmetric then $D(A)=D(A^*)$? what more condition does it need to give $D(A)=D(A^*)$?


-1

no of course not. en effet, you must require $A=A^\ast$, you are missing $D(A)=D(A^\ast)$.


1

if an operator is the generator of a $C_0$ semi-group then we will have the following properties (refering to Pazy, "semi-group Theory"): 1) For $z_0\in D(A)$ then $ T(t)z_0 \in D(A), \forall t>0$ 2) $\bigcup\limits_{n=1}^{\infty}D(A^n)$ is dense in X Note that the first property does not guarantee that $Az_0\in D(A)$, However, from the property 2, it ...


0

First, note that the equation you wrote is incorrect. Composition of linear operators (and multiplication of matrices) is not, in general, a commutative binary operation. What you want to prove is $$(A+B)C = AB+AC = A(B+C).$$ It then follows that $$(A+B)(A+B)=AA+AB+BA+BB$$. Note that $AB \neq BA$ in general since we are dealing with linear operators. I ...


2

I don't know of a theorem to help you in this case. However your operator is a sum of an inclusion from $L^2$ to $H_0^k$ in the first coordinate and a $k$th derivative into $L^2$ from $H^k$ in the second. The differential part is in fact a bounded operator. Therefore you only need to consider the inclusion. I think you can verify directly that the ...


1

Spectal Theory evolved with the thought in mind of expanding functions in eigenfunctions of some operator. Eigenfunctions and eigenvalues make no sense if you consider $L : X\rightarrow Y$ because $Lf=\lambda f$ makes no sense if $X\ne Y$. You can't have discrete eigenvalues and discrete eigenfunction expansions, or approximate eigenvvalues and continuous ...


2

Considering the spectrum of unbounded operators as a very special definition that doesn't quite fit into the general definition of the spectrum as follows... The basic ingredient for spectral theory in general is a unital algebra $1\in\mathcal{A}$. Denote the set of invertibles by $\mathcal{A}^\ast$. Then, the spectrum of an element is nothing but: ...


3

I think you have already done the hard part. Assume $h \neq 0$ and fix some $a \in \mathcal{A}$ with $h(a) \neq 0$. For each $i$, we have $h(e_i) h(a) = h(e_i a)$, so $$ h(e_i) = \frac{h(e_ia)}{h(a)} \to 1.$$ Thus there is a net of elements of norm $\leq 1$ whose images tend to $1$. It follows that $\|h\| \geq 1$.


2

I don't understand why you cannot get an answer from the eigenvalues of $A^*A$? Evaluating directly the maximal eigenvalue of this matrix gives $$ \|A\|_2^2=|a|^2+\frac{|b|^2}{2}+\frac{|b|}{2}\sqrt{4|a|^2+|b|^2}. $$ Now put the assumption $|b|\leq 1-|a|^2$ (with the one that $|a|\leq 1$ to make it possible) in there to verify that $\|A\|_2\leq 1$. The ...


2

The right naming here would not be "Heine-Borel" but "Banach-Alaoglu".


1

I prefer sticking to the classical context for the first round of dealing with the spectral theorem; in particular, I would use Riemann-Stieltjes integrals instead of Borel measures. Once you have the Riemann-Stieltjes version, it is a fairly trivial matter to extend to the measure theoretic, when you want it. I highly recommend this text for self-study at ...


1

Start with an example of the translation semigroup: $(T(t)f)(x)=f(x+t)$ for $f \in L^{1}[0,\infty)$. The generator $A$ of this semigroup is differentiation. So you do not have $\mathcal{R}(A)\subseteq\mathcal{D}(A)$ because that would force every $f \in \mathcal{D}(A)$ to be infinitely differentiable. If the generator $A$ is bounded, then you do have ...


5

In a unital Banach algebra the set of invertible elements is open and nonvoid, hence the set of noninvertible elements is closed ( and also nonvoid, since it contains zero). Any proper ideal is contained in the set of noninvertible elements. Hence, no dense proper ideals in a unital Banach algebra.


0

I'm not sure that we are talking about the same Green's theorem, I think that there are many version of it. In my experience Green's theorem is formulated in $\mathbb{R}^2$ and it is equivalent to the divergence theorem in dimension 2. http://en.wikipedia.org/wiki/Green's_theorem#Relationship_to_the_divergence_theorem


2

Let $A=\sum_k \lambda_k P_k$ be the orthogonal decomposition of $A$ where $u_k$ are its eigenvectors. Let $P=1-\sum_{k=1}^{N-1} P_k$ be the projection on the orthogonal complement of the space spanned by the first $N-1$ eigenvectors. We then have $$\begin{align}\langle Ae_i,e_i\rangle&= \langle \sum_{k=1}^\infty\lambda_kP_k e_i,e_i\rangle\\ &=\langle ...


0

Hints: $|f(x)| \le \|f\|$. Approximate $|g|$ by a linear combination of indicator functions for disjoint closed sets.


0

Every linear functional $x^{\star}$ on $C[a,b]$ has the form $$ x^{\star}(f) = \int_{a}^{b}f(t)\,d\mu(t) $$ for a unique finite complex Borel measure $\mu$ on $[a,b]$. Furthermore, $$ \|x^{\star}\| = \|\mu\|, $$ where $\|\mu\|$ is the total variation of $\mu$ on $[a,b]$. If $K : C[a,b]\rightarrow C[a,b]$ is a ...


1

Observe that \begin{align*} \langle (\mathcal A \mathcal P+\mathcal P\mathcal A)x,x\rangle&=\langle \mathcal A \mathcal Px,x\rangle+\langle\mathcal P\mathcal Ax,x\rangle\\ &=\langle \mathcal Px,\mathcal Ax\rangle+\langle\mathcal Ax,\mathcal Px\rangle\\ &=2\mbox{Re }\langle\mathcal Px,\mathcal Ax\rangle=2\mbox{Re }\langle\mathcal A\mathcal ...


1

Since everything is taking place in the orthogonal complement of the nullspace of $A$, we can assume that the Hilbert space is seperable. Pick then an orthonormal basis of the range of $P$, say $\varphi_n$, such that $$ Px=\sum_n \langle x,\varphi_n\rangle\varphi_n. $$ From this form of $P$, it is not hard to see that $$ |\langle Px,y\rangle|\leq|\langle ...


0

Yes, more or less. Ordinary integrals do not give a general representation of bounded operators between spaces of continuous functions, the identity operator on $C(a,b)$ is not of this form, at least if you want $E(u,t)$ to actually be a function. However, there is the Schwartz kernel theorem that represents every operator on the space of test functions ...


2

You can always define an unbounded operator on the whole space $X$, as long as $X$ is infinite dimensional. Simply take any unbounded linear functional $\varphi : X \to \Bbb{K}$ (with $\Bbb{K} \in \{\Bbb{R}, \Bbb{C}\}$) and some $x_0 \in X \setminus \{0\}$ and define $T : X \to X, x \mapsto \varphi(x) \cdot x_0$. For the existence of $\varphi$, see On ...


3

Take an $x\in H\setminus \{0\}$. You need to show that $\langle x,Tx\rangle > 0$. Can you see that a certain projection with one-dimensional range could be helpful?


1

Since $e_1,\ldots,e_n$ are fixed, the vector $c^n$ is simply the vector of the components of $u_n$ in the basis $\{e_1,\ldots,e_n\}$. Thus, considering the vector $u_n$ or $c^n$ is completely equivalent.


0

Here's a more elaborate example to give you a bounded group that is not $C_{0}$. To do this, I'll define a function $F : \mathbb{R}\rightarrow\mathbb{R}$ such that $F(x+y)=F(x)+F(y)$. Then $E(x)=e^{iF(x)}$ satisfies $E(x+y)=e^{iF(x+y)}=e^{iF(x)}e^{iF(y)}=E(x)E(y)$. Automatically $F(0)=0$ because $F(0)=F(0+0)=F(0)+F(0)$, which gives $E(0)=1$. I'll describe ...


1

Let $P_{k}$ be the orthogonal projection onto the closure of the range of $T_{k}$. Then $P_{k}P_{k'}=P_{k'}P_{k}=0$ for $k\ne k'$ because $(T_{k}x,T_{k'}y)=(T_{k'}^{\star}T_{k}x,y)=0$ for all $x,y \in H$. Similarly, if $Q_{k}$ is the orthogonal projection onto the closure of the range of $T_{k}^{\star}$, then $Q_{k}Q_{k'}=0$ for $k\ne k'$. Furthermore, $$ ...


1

First observation: For all $u,v\in\mathcal H$ and $j\ne k$ $$ \langle T_ju,T_k v\rangle=\langle T_j^*u,T_k^* v\rangle=0. $$ Thus the linear subspaces $T_j^*\mathcal H$, $j\in\mathbb Z$, are perpendicular to each other, and so are their closures. Set $$ Y=\bigoplus_{j\in\mathbb Z}\overline{T_j^*\mathcal H}. $$ This infinite direct sum contains elements of ...


3

Yes, let $X=L^{2}[0,1]$, and let $(Tx)(t)=tx(t)$, i.e., multiplication by $t$. This is the prototypical example in many regards. You don't have any eigenfunctions because $Tx=\lambda x$ would require $(t-\lambda)x(t)=0$ for a.e. $t\in[0,1]$ which means that $x=0$ a.e.. However, you have approximate eigenfunctions. For example, if $\lambda\in(0,1)$ and ...


0

Sorry, but I don't have the rights to add a comment. Can you explain the line $$ C_\pm(w)(A\pm i\mu I)=(A\pm i\mu I) +w B $$ I cannot see that this is true.


0

Let $H = K = \ell_2$. Define a function $\Phi$ by $\Phi( x,y ) = \tfrac{1}{2}\|x\|\|y\|I_{H\otimes K}$. If there were an extenstion of $\Phi$ to a bounded linear operator $\ell_2(\mathbb{N}\times \mathbb{N})\to B(H\otimes K)$ (still dented by $\Phi$), it would map the weakly null sequence $(e_n\otimes e_n)_{n=1}^\infty$ in the domain to a weakly null ...


0

An idea: maybe you can show that if $x_1M \subset M$ then $(x_1M \cap H^2) \subset (M\cap H^2)$ (+ remaining a CLOSED subspace, which I have no idea how to show). Once there, you could apply Beurling's Thm and you know that $M \cap H^2$ is of the form $\phi H^2$. By density of $H^2$ as a subspace of $H^1$, you can probably show that the space $M$ is also of ...


1

Compact operators on Banach spaces can in general be upper-triangularized---generalizing the Jordan canonical form. This means that the space is Schauder-spanned by the generalized eigenvectors. So what you would need to show is that, given any non-zero $\lambda \in \sigma(L)$, the finite dimensional subspace $\mbox{ker}(L - \lambda)$ is spanned by ...


1

You are given that $$ (-2\pi i)Tf = \int_{-\pi}^{\pi}K(x-y)f(y)\,dy \\ =\int_{-\pi}^{\pi}(\pi\,\mbox{sgn}(x-y)-(x-y))f(y)\,dy\\ =\int_{-\pi}^{x}(\pi\,-(x-y))f(y)\,dy\\ +\int_{x}^{\pi}(-\pi-(x-y))f(y)\,dy\\ =(\pi-x)\int_{-\pi}^{x}f(y)\,dy+\int_{-\pi}^{x}yf(y)\,dy \\ -(\pi+x)\int_{x}^{\pi}f(y)\,dy+\int_{x}^{\pi}yf(y)\,dy. $$ ...


1

I think you are thinking about it backwards. Every Banach space is a vector space. Every vector space has a basis. If your eigenfunctions are: (1) linearly independent under the inner product (2) have the same dimension as the space. They must be an equivalent basis. However, neither (1) nor (2) are necessary conditions of eigenfunctions as far as I know. ...


1

See if you can extend your operator to some $L^2$ space containing your functions. Maybe your operator is even self adjoint and still compact. Use the result for compact self adjoint operators on Hilbert spaces to find a basis of eigenvectors in that $L^2$ space. Show that in fact your eigenfunctions are already in your original space. Show that any ...


1

The following is a solution to part (c). One direction was proved in part (a). Therefore, I will only prove the other direction: assuming $\sum_{n = 1}^\infty |\lambda_n|^2 < \infty$, I will show that $T$ is a Hilbert-Schmidt operator (where $\lambda_n$ are the eigenvalues corresponding to some orthonormal basis of eigenvectors, such that $T(f) = \sum_{n ...


2

Let $H$ be a separable Hilbert space as desired. For simplicity assuming we are working over the real field. Then being an integral operator, we have $$ K: H\rightarrow H, (Kf)(x)=\int K(x,y)f(y)dy $$ Since $K$ is a symmetric compact operator, it is normal and can be diagonalized. Let us write its eigenvectors as $\phi_{j}$ with eigenvalue $\lambda_{j}$. ...


1

There is no any relation between stability and connectivity. Take two non intersecting invariant sets ( it might be any 2 trajectories of the system, which in regular autonomous case do not intersect). The union of these trajectories is also an invariant set, which is not connected. I assume you question is more relevant when the invariant set is minimal. ...


0

I don't think stability and connectedness of an invariant set is related. Consider the linear system $$ \dot{x} = A x, ~~ x(0) = x_0 $$ Obviously, all trajectories (invariant sets) are connected with the mapping $x(t) = e^{At} x_0$ regardless of the stability of the system. However, I don't know if all of the invariant sets are connected for an arbitrary ...


0

In short, yes. One thing that you need to clarify is, $\epsilon$ is not a vector while the argument of $\Psi$ is. I'll assume that $\epsilon$ is multiplied by some constant vector $v$. Consider $\Psi(\vec{r}+\epsilon\vec{v})$ as a function of both $\vec{r}$ and $\epsilon$. Name it $\psi$ for clarity. $$ ...


2

Let $(e_i)_{i\in I}$ be an an orthonormal basis, and $x =\sum_{i\in I} y_i e_i$ with norm $1$, i.e. $\sum_{i\in I} |y_i|^2 = 1$, then $$\|Tx\| \leq \sum_{i\in I}|y_i|\|Te_i\| \leq \sqrt{\sum_{i\in I} |y_i|^2}\sqrt{\sum_{i\in I}\|Te_i\|^2} = \|T\|_{HS}$$


0

Slightly different approach: Assume to the contrary that $T$ is compact. Clearly $\|T\|=1$, in fact $\|T^n\|=1,$ so spectral radius is equal to 1. Since the ambient space in infinite dimensional then there exists a sequence of eigenvalues that converges to zero . in particular we can find eigenvalue $\lambda$ with $|\lambda|<1$ and $g\in C[0,1]$ such ...


0

I assume we're talking about normed linear spaces here, and (for convenience) real scalars. 1) If $Y$ is not complete, there is a Cauchy sequence $y_n$ in $Y$ that has no limit there. Take linear operators $T_n$ on $\mathbb R$ such that $T_n 1 = y_n$. Then these form a Cauchy sequence in $B(\mathbb R, Y)$ with no limit. 2) I'm not sure I understand your ...


1

Suppose that $Ax$ is not a scalar multiple of $x$ for some $x \in X$. Then $x\ne 0$, and there is a linear functional $\phi$ defined on the linear span of $\{x,Ax\}$ such that $\phi(x)=1$ and $\phi(Ax)=0$. Such a linear functional is continuous on the two-dimensional subspace spanned by $Ax,x$ and, so, by the Hahn-Banach Theorem, extends to a continuous ...


2

I think we can just play around with some special compact operators, unless I'm missing some subtlety. If $x$ and $y$ are linearly independent, then we can choose a basis for $X$ which contains $x$ and $y$. Let $K_{x, y}$ be the linear operator which swaps $x$ and $y$ and which sends all of our other basis vectors to zero. Then $K_{x, y}$ is compact, ...



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