# Tag Info

1

There is no such operator. Your proof works when $T$ is injective, but you are right that it has a gap when it isn't since in that case "$T^{-1}$" is not a function but a one-to-many mapping, and none of our results about bounded operators apply to such mappings. Instead, use the original statement of the open mapping theorem: a surjective linear operator ...

1

For the first Q : You have a typo in your 3rd line, def'n of $(\eta_j)$ should be $$(\eta_j)=\sum_{k=1}^{\infty} \alpha_{j,k} \zeta_k$$ and also in what follows.As for the Q, let $\alpha_n=(\alpha_{n.k})_{k \in N}$ which belongs to the Hilbert space, otherwise $T$ is unbounded. Denote the inner product of vectors $x,y$ as $(x|y)$.We have, for $x \ne 0$, $$... 1 For the first part, use the fact that for every \epsilon > 0, \exists N_0 \in \mathbb{N} such that$$ \sum_{j=n+1}^{\infty} \sum_{k=1}^{\infty} |\alpha_{k,j}|^2 < \epsilon \quad\forall n\geq N_0 $$(because the tail of a convergent series goes to zero). For the second part, take$$ \alpha_{k,j} = \begin{cases} \frac{1}{k^{1/2}} &: k=j \\ 0 ...

1

Hint: since the operator $T$ is linear you can scale and translate any bounded set to be contained in the unit ball. If $T$ in not linear checking relative compactness only on the unit ball is generally not enough.

0

When $T\in B(H)$ with operator norm, yes $\|T^{-1}\|=\|T\|^{-1}$.

0

If there exists a continuous linear operator $T^{-1} : X\to X$ then $T$ will be a homeomorphism hence $B=T^{-1} (\overline{T(B)} )$ should be compact but this is impossible. $B$ - denotes the unit ball.

0

Hint: How do you prove that $y^*\colon H\to {\bf C}$ defined by $y^*(y'):=\langle y',y\rangle$ is bounded? Or, if you know what an adjoint operator is, just notice that $\Phi$ is just $T^*(y^*)$.

1

As the comment above indicates, there is no such inner product. One can deduce that this is the case by noting that the parallelogram identity fails to hold.

2

For (1), depending on whether you are a geometer or an analyst you may have a different sign on your Laplacian. However, if you locally write $$\Delta f = \sum \frac{\partial^{2} f}{\partial x_{j}^{2}}$$ then this is negative definite. This being the classic example of the sort of operator $L$ you are interested in, the author might be assuming $L$ to ...

1

The polynomial $p(\lambda)=\lambda(\lambda^{k-1}-1)$ has distinct roots $$0,\alpha,\alpha^{2},\cdots,\alpha^{k-1},\;\;\;\; \alpha=e^{2\pi i/(k-1)}.$$ Therefore, $X$ decomposes into the direct (not necessarily orthogonal) sum $$X = M_0\oplus M_1\oplus M_2\oplus \cdots \oplus M_{k-1},\\ ... 0 A counterexample for your EDIT question is i\frac{d}{dx} on \mathcal{L}^2(\mathbb R) with \mathcal{D}(C) = finite linear combinations of \mathcal{C^\infty}  functions of compact support. 3 You proved that T(t)x is a solution. So, it remains to show that the problem has only one solution. Let u and v be two solutions. Notice that the function w=u-v satisfies$$\left\{\begin{align} &\frac{dw}{dt}=Aw, & 0\le t \le t_{e} \\ &w(0)=0. \end{align}\right.$$Pick s\in[0,t_e] and define \phi:[0,s]\to X be setting ... 2 Notice that, the way you have defined T=MS_R, we have $$T^1(a_1,a_2,a_3,\cdots)=(0,c_2a_1,c_3a_2,c_4a_3,\cdots),$$ $$T^2(a_1,a_2,a_3,\cdots)=(0,0,c_3c_2a_1,c_4c_3a_2,c_5c_4a_3,\cdots),$$ and so on. In particular, if (e_i)_{i=1}^\infty is the canonical basis for \ell_p, p=2, then we have ... 0 Remember that if s=\sup X and x\leq t for all x\in X, then s\leq t. Also, if s=\inf X and t\in S, then s\leq t. Now, let us write$$\begin{align*} a&= \inf\{ k\;\colon\; \lVert Av\rVert\leq k\lVert v\rVert \text{ for all }v\in V\},\\ \\ b&=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lvert v\rVert\leq 1\},\\ \\ ...

0

This is true for any linear normed spaces. Let $T:X \to Y$ be a linear operator and $X,Y$ be two normed spaces, then 1- T is continuous iff T is bounded. 2- If T is continuous at a single point $x_0$, then it is continuous. the proof of second ostatement follows from the first statement, and here the details if you want: Assume $T$ is continuous at an ...

0

$||T|| := \inf\limits_{C \geq 0} \{ C: ||Tv|| \leq C||v||, \ \forall v \in V\} = \inf\limits_{C\geq 0}\{C: ||Tv|| \leq C, \forall ||v|| = 1\}$. We established the last equality in comments made above. So if $||T|| \gt \sup\limits_{||v|| = 1} ||Tv||$, then there exists $||Tv|| \gt \sup\limits_{||v|| = 1} ||Tv||$ with $||v|| = 1$, impossible!

1

If $M$ is a vector space, then we refer $A$ as an endomorphism. There are differences when a function takes an element from one space to an element in the same space. Take for example: $f:C[0,1]_{\|\cdot\|_1}\to C[0,1]_{\|\cdot\|_2}$, where $\|\cdot\|_1 \to$ integral norm $\|\cdot\|_2 \to$ maximum norm Note that $C[0,1]$ is complete under maximum ...

1

For the other direction, if $v \neq 0$, then $||Tv|| = ||v|| \cdot||T( \frac{v}{||v||})|| \le ||v|| \cdot \sup_{||v|| = 1} ||Tv||$. So, $||T|| \le \sup_{||v|| = 1} ||Tv||$

2

Your operator is $$Tf(x) = \int^\infty_{-\infty}1_{y\leq x}e^{-(x-y)}f(y)\,dy$$ which by change of variables $z=x-y$ becomes $$Tf(x) = \int^\infty_{-\infty}1_{z\geq 0}e^{-z} f(x-z)\,dz.$$ This is just the convolution operator $$f\mapsto \phi*f$$ with $$\phi(z) = 1_{z\geq 0}e^{-z}.$$ Now one of the fundamental inequalities (not hard to prove) about ...

2

$Tf=\phi*f$, where $\phi(t)=e^{-t}\chi_{(0,\infty)}(t)$. (I had $\phi$ backwards in the first version; noticed that guest's $\phi$ was different, then noticed his was right.) So $$\widehat{Tf}=\hat\phi\hat f.$$You can easily calculate $\hat\phi$; now the norm of $T$ is $||\hat\phi||_\infty$ and the spectrum of $T$ is the essential range of $\hat\phi$. (In ...

0

Using the Gram-Schmidt process, construct orthogonal vectors $f_n$ from the $A^nf$. Note that $\Vert f_n \Vert\le \Vert A^nf \Vert$. From the Gram-Schmidt equations it is apparent that, on the orthonormal vectors $\frac{f_n}{\Vert f_n\Vert}$, $A$ is a Jacobi matrix operator $J$ with $n,n+1$ elements $b_n := \frac{\Vert f_{n+1}\Vert}{\Vert f_n\Vert}$ and ...

1

The Spectral Mapping Theorem allows you to more easily compute the spectrum of some operators. If you know that you can write an operator $A$ as $A=f(a)$ for $f\in hol(a)$ and $a\in\mathcal{A}$, where you already know the spectrum of $a$, you can compute the spectrum of $A$, since $\sigma(A)=f(\sigma(a))$.

1

This is a "functional calculus". The point is that the function $f$, initially defined on complex numbers, can now be extended to suitable members of the Banach algebra. Moreover, this extension will turn out to have some useful properties. For a concrete example, consider the Banach algebra $\mathcal L(\mathbb C^n)$ of linear operators on $\mathbb C^n$ ...

0

This is not true !! to convince you, take $T$ the unilateral shift so : $$\sigma(T)=\bar{\mathbb{D}}\\ \sigma_p(T)=\mathbb{D}\\ \sigma_c(T)=\mathbb{T}$$ proof page 6-7.

2

In general there is no such surjection (the C*-case is quite special in that respect). Indeed, let $V=c_{00}$ be the vector space of finitely supported vectors. For $(\xi_n)$ in $V$ define $\|(\xi_n)\|_1 = \sup_n |\xi_n|$ and $\|(\xi_n)\|_2^p = \sum_{k=1}^\infty |\xi_k|^p$ for some fixed $p\in (1,\infty)$. Conspicuously, the respective completions are ...

0

Notice that for symmetric matrices $A$ and $B$, eigenvalue of $A\otimes B$ coincide with multiplication of eigenvalue of $A$ and $B$. For this you can see page 708 of Abstract harmonic analysis Hewitt&Ross(same method). About your question, Suppose $x\otimes w$ is an eigenvector of $A\otimes I$(thus $x$&$w$ arenot zero), $$(A\otimes I)(x\otimes ... 1 Observe that F^\perp = \ker P because P is an orthogonal projection and F = {\rm im} P. If PT = TP, then we have T(F) = T(P(H)) = P(T(H)) \subseteq P(H) = F so F is T-invariant. Similarly, P(T(F^\perp)) = T(P(F^\perp)) = T(P(\ker P)) = 0, hence T(F^\perp) \subseteq \ker P = F^\perp and F^\perp is T-invariant. Conversely, if F and ... 3 We know T is linear. Let x be any vector with norm less than or equal to 1. Then we have ||Tx||=||x||*||\frac{Tx}{||x||}||\leq||\frac{Tx}{||x||}||=||T(\frac{x}{||x||})||. What does this mean? Well, it means that if ||x||\leq 1, there is some other point vector, given by x'=x/||x||, with norm 1, such that ||Tx||\leq||Tx'||. Thus, if we're ... 2 Because if \;0<\lVert x\rVert\le 1, then \Biggl\lVert\dfrac{x}{\lVert x\rVert}\Biggr\rVert=1, and because \lVert\lambda x\rVert=\lvert \lambda\rvert\lVert x\rVert. 1 The operator$$ Tf = f'-\frac{x}{\sqrt{1+x^{2}}}f $$is a bounded operator from X to Y because$$ \begin{align} |Tf| & \le |f'|+|f|,\\ |Tf|^{2} & \le |f'|^{2}+|f|^{2}+2|f'||f| \\ & \le 2|f'|^{2}+2|f|^{2} \\ \|Tf\|_{Y}^{2} & \le 2\|f\|_{X}^{2}. \end{align} $$To ... 1 Argyros and Haydon constructed a separable Banach space X such that each bounded linear operator T\colon X\to X is of the form T=cI_X + K where K is compact. Let Y be an infinite-dimensional subspace of X that has infinite codimension. Then Y is not the range of any operator on X. Indeed, if T\colon X\to X were an operator with {\rm ... 2 Let L denote the matrix$$ L = \pmatrix{u \mathrm{I} + i S_3 && i S_- \\ i S_+ && u \mathrm{I} - i S_3} $$My best guess is that whatever the author is getting at has something to do with the fact that$$ \operatorname{tr}(L^N) = 2I\,u^N + q_{2}u^{N-2} + \cdots + q_N $$or, if N is odd,$$ \operatorname{tr}(L^N) = 2I\,u^N + q_{2}u^{N-2} + ...

0

I think $A$ considered as an operator in $H \bigoplus H'$ (where $H'$ is another Hilbert space) satisfies the same conditions with respect to $H \bigoplus H'$. But $D(A)$ is not dense in $H \bigoplus H'$

1

Yes if and only if $H$ is separable. Yes if and only if $H$ is separable. If I remember correctly yes. Please check this survey paper.

1

No. Kakutani proved that a Banach space $X$ is isometric to a Hilbert space if and only if each two-dimensional subspace $Y$ is complemented by a norm-one projection. Now suppose that $X$ is not isometric to a Hilbert space and take a two-dimensional subspace $Y\subset X$ such that each projection $P\colon X\to X$ with ${\rm im}\,P = Y$ has norm $>1$. ...

3

This doesn't necessarily use the theorems you refer to (I don't own that book), but this is the one proof I used to know. Hope it helps you anyway. Remember that in a complete metric space, totally bounded (having finite $\varepsilon$-net for any $\varepsilon$) and relatively compact are the same. I also assumed that your $T_n$ are continuous (i.e. bounded ...

1

Let $A=-\frac{d^{2}}{dx^{2}}$ on the domain $\mathcal{D}(A)=W^{2}_{2}(\mathbb{R})$. The restriction $A_{0}$ of $A$ to $\mathcal{C}_{0}^{\infty}(\mathbb{R})$ has a closure $\overline{A_{0}}=A$. In other words, the closure of the graph of $A_{0}$ is equal to the graph of $A$. If $V$ is any bounded measurable function on $\mathbb{R}$, then $V$ defines a ...

0

Lemma: Let $A$ be a unital Banach algebra and $\{a_n\} \subset A$ such that $a_n \to a\in A$. Suppose $\lambda_n \in \sigma(a_n)$ are such that $\lambda_n \to \lambda$ in $\mathbb{C}$, then $\lambda \in \sigma(a)$ Proof: Suppose $\lambda \notin \sigma(a)$, then $(a-\lambda 1) \in GL(A)$, which is open. So $\exists \epsilon > 0$ such that \|y - ... 1 Absent any relation between the measure m and the manifold/flow structure, you can't do this. For example, let M=S^1, and let f_t be the rotation of S^1 by angle t, and choose m to be the restriction of the arclength measure to the upper semi-circle. For any t\in (0,\pi) there exists a function supported in the lower semi-circle such that after ... 1 First off, I think your statement of the theorem is a bit off. It should go something like: Given A:\mathbb{X}\rightarrow\mathbb{R} (not \mathbb{L}\rightarrow\mathbb{R}) sublinear and B:\mathbb{L}\rightarrow\mathbb{R} linear such that B(u)\le A(u) for all u\in\mathbb{L}, there exists a linear extension of B (call it C) from ... 1 Yes and you don't even have to take the closure. For instance if A is a non-simple C*-algebra with trivial cetnre that has unique (faithful) trace (for instance A=C^*(G) for a sufficiently non-commutative amenable group such as the group of permutations of integers that move at most finitely many entries), then A\otimes \mathcal{Z}, where \mathcal{Z} ... 2 A function is said to be an extension of another one if, given the original function f_1 : A \to X, there is a function f_2:B \to X such that A\subset B and f_1(x) = f_2(x) \,\,\,\forall x \in A. Clearly, an extension doesn't have to be unique, it just has to be defined on a larger domain that contains the original one and agree with the first ... 0 In addition to Stefan Perko's answer.. The square matrices are not linear maps. They rather form an algebra: The simplest examples of unital C*-algebras. Now, there is the Gelfand-Naimark theorem: It says that they can be interpreted as continuous linear operators over some Hilbert space. If I'm not mistaken, this turns out to be the canonical bijection ... 0 Assume that J=JJ^*J. If \varphi\in\mathcal R=(\ker J)^\perp=\overline{\mathcal R J^*}, then for any \xi\in\mathcal H we can write \xi=J^*\eta+\nu with \nu\in(\mathcal RJ^*)^\perp=\ker J, so \nu\perp\varphi and \begin{align} \langle J^*J\varphi,\xi\rangle=\langle J^*J\varphi,J^*\eta+\nu\rangle=\langle J^*J\varphi,J^*\eta\rangle=\langle ... 2 One has to specify a space of functions (or some linear space, if you are not interpreting the elements as functions) on which D acts, and some topology with respect to which \sum D^n f might converge, or not converge, to an element of the space. For example, the series (1-D)^{-1}, interpreted as a power series, does always converge and give a ... 2 Simply note that |A|, |A^\ast| are positive semi-definite, self-adjoint operators which both have operator norm \Vert A \Vert. Hence, \sigma(|A|) \cup \sigma(|A^\ast|) \subset [0, \Vert A \Vert]. Since the operator norm equals the spectral-radius for self-adjoint operators, we get \Vert A \Vert \in \sigma(|A|) \cap \sigma(|A^\ast|). Now, since ... 4 They do not satisfy the definition of a linear map. Instead, there is a canonical bijection:\text{Vect}(\mathbb{K}^m,\mathbb{K}^n) \to \mathbb{K}^{n\times m}$$where \text{Vect}(\mathbb{K}^m,\mathbb{K}^n) is the set of linear maps from \mathbb{K}^m to \mathbb{K}^n. It is given by mapping a linear map f : \mathbb{K}^m \to \mathbb{K}^n to the ... 0 First note that$$W(e_n)=W(\underbrace{0,\ldots,0}_{n-1 \text{ terms}},1,0,\ldots)=(\underbrace{0,\ldots,0}_{n \text{ terms}},\frac 1n,0,\ldots)$$so that$$\sum_{n=1}^\infty \|We_n\|^2=\sum_{n=1}^\infty \frac 1{n^2} < \infty.$$2 Of course, there are plenty: first the identity. A second idea is to take f\in X^* such that f(x)\ne 0. Then define$$ Tv = \frac{1}{f(x)}f(v) x.  Both of these operators are linear, and can be easily shown to be bounded.

1

Hint: Since $\|x_n\|$ is bounded, there is a subsequence $(x_{n(k)})_{k\geq 1}$ that converges weakly to some $x\in H$. Show that this $x$ is a solution.

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