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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 \begin{equation}T^1(a_1,a_2,a_3,\cdots)=(0,c_2a_1,c_3a_2,c_4a_3,\cdots),\end{equation} \begin{equation}T^2(a_1,a_2,a_3,\cdots)=(0,0,c_3c_2a_1,c_4c_3a_2,c_5c_4a_3,\cdots),\end{equation} 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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