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This can only happen if $f(\phi)=a\phi$ for some $a\in \mathbb{R}$. Indeed, differentiating the identity $$e^{i\phi X}=e^{if(\phi)Y}$$ at $\phi=0$ we get that $X=f'(0)Y$, so $e^{if(\phi) Y}=e^{i\phi f'(0) Y}$ for all $\phi\in\mathbb{R}$. Therefore $f(\phi)=f'(0)\phi$. In this case, trivially, $Y=X/{f'(0)}$ is the change of operators you are looking for.

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Yes, the answer is $w=\frac{Av}{\|Av\|}$

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You don't need open mapping theorem here. Assume $T^*$ is open: this means there is $C>0$ such that every unit norm functional $f\in X^*$ can be written as $g\circ T$ for some $g\in Y^*$ with $\|g\|\le C$. Following Etienne's hint, apply the above to the norming functional of an element $x\in X$; i.e., a unit functional $f$ such that $f(x)=\|x\|$. ...

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No. It is possible that an element is mapped to a compact in one irreducible representation and to a non-compact in another. For instance, let $I_n\in M_n(\mathbb C)$, $I\in B(\ell^2(\mathbb N))$ be the respectives identities. Let $\mathcal A=M_n(\mathbb C)\oplus B(\ell^2(\mathbb N))$ and $a=I_n\oplus I$. Then the irreducible representation $$... 0 Here are some thoughts - not sure if there is a clear-cut answer to your question even for matrices (I use r(T) to denote \text{rad}(T)) : T has this property iff r(T) \geq \|T\|, which happens iff$$ \|T(x)\| \leq r(T)\|x\| \quad\forall x  \Leftrightarrow \langle Tx,Tx\rangle \leq r(T)^2 \langle x,x\rangle  \Leftrightarrow r(T)^2I - ...

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The question is answered affirmative (and in a much simpler way) elsewhere: How to derive these Lie Series formulas Summary. First solve the differential equation: $$g(x) = \frac{1}{\phi'(x)} \quad \Longrightarrow \quad \phi(x) = \int \frac{dx}{g(x)}$$ Then we have (barring division by zero and other issues): $$e^{g(x)\partial} f(x) = ... 1 By the Banach-alaoglu theorem, the character space is a subset of a compact space (the closed unit ball of the dual space). Now suppose that A is not unital, and let B be the standard unitisation of A. Then the map which sends the unit in B to 1 and is identically zero on A is a character. This corresponds to the zero character on A. The addition ... 4 It does not follow that U 1_{B_n} \to U 1_B pointwise \mu-almost everywhere. Consider S = \mathbb{N},\Sigma = \mathcal{P}(\mathbb{N}) and \mu the counting measure (or any measure given by positive masses on every point, so \mu could even be a probability measure). Let \mathscr{U} be a free ultrafilter on \mathbb{N} and define ... 1 Start with a countable set D which is dense in the unit sphere of \Omega^*, say D=\{ x_i^*;\; i\in\mathbb N\}. For each i\in\mathbb N, one find a point x_i\in\Omega such that \Vert x_i\Vert=1 and \vert \langle x_i^*,x_i\rangle\vert>1/2. To show that \Omega is separable, it is enough to prove that the linear span of the vectors x_i is ... 0 Yes. This is the functional-analytic formulation of the study of linear PDEs, in which a linear differential operator L is viewed as a linear operator between two appropriate vector spaces. For example, L is a differential operator of order k and u is assumed to live on some domain U, then one might naturally think of considering L as an operator ... 7 The number after nine is represented in the usual base system by the sequence of digits 10, but it isn't a sequence of digits. Properties like "two digits long" are properties of the representation, while properties like "has four factors" are properties of the number. We could represent the number in binary, as 1010, and that would be just as valid. ... 4 The concept of matrix can be useful in other situation than for linear operators, and there exists linear operators that can not be represented by matrices. The matrix is just an array of elements and are added and multiplied according to certain rules. For this it is only required that there is a meaningful definition for addition and multiplication for ... 34 The operator T is a function from \Bbb R^n to \Bbb R^n; an n\times n matrix of real numbers is not a function. If one chooses the matrix A properly, one can say that the function T(x)=Ax for each x\in\Bbb R^n, but that is far from saying that the function T is the matrix A. Consider the more familiar case of real-valued functions on the ... 1 If T is continuous, then \ker(T) is closed and the quotient map$$ \pi : X \to X/\ker(T) $$is an open map. Furthermore, T induces an injective map$$ S : X/\ker(T) \to Y $$Since Y is finite dimensional, so is X/\ker(T), and so S (whose range is Y) is now a homeomorphism. In particular, S is an open map, so$$ T = S\circ \pi $$is also ... 0 Suppose that f(x)=0 for all f\in F. Let g\in X^*. Then there exists a net \{f_j\}\subset F with f_j\to g weak^*. So$$ g(x)=\lim f_j(x)=\lim 0=0. $$Thus g(x)=0 for all g\in X^*, and then x=0. 1 Nevanlinna-Pick interpolation (as well as Caratheodory-Fejer interpolation) is a special case of Nehari's problem. Indeed, NP interpolation is to find a function f from the unit ball in H^\infty that interpolates the given values f(\zeta_i)=\omega_i in the unit disc. If L(z) is the Lagrange interpolation polynomial with L(\zeta_i)=\omega_i then ... 0 Given the recursive relationship you found, the first three components of a vector in the null space define unambigously all the others. On the other hand, the first three components can be chosen arbitrarily, and completing the vector according to the recursive rule will still put it in the null space. Which is just a long way of saying that the null space ... 2 Note that \mathcal{A}\cap K is a closed ideal in \mathcal{A}, and we have a well-defined \ast-homomorphism$$ \varphi : \mathcal{A}/(\mathcal{A}\cap K) \to B(H)/K \text{ given by } a + (\mathcal{A}\cap K) \mapsto a+K $$The range of \varphi is precisely (\mathcal{A} + K)/K, and the range of a \ast-homomorphism is complete. Since K is ... 1 Since A is positive, its spectrum is contained in [0,\infty). Then the spectrum of A+I is contained in [1,\infty); thus 0 is not in the spectrum of A+I and A+I is invertible. To justify that the spectrum translates note that A+I-\lambda I=A-(\lambda-1)I. So A+I-\lambda is not invertible precisely when A-(\lambda -1)I is not invertible, ... 0 The spectral radius of T is \lim_n \|T^{n}\|^{1/n}. However, \|T^{2}\|=\|T\|^{2} for any selfadjoint T. This is because$$ \|T\|^{2}=\sup_{\|x\|=1}\|Tx\|^{2}=\sup_{\|x\|=1}(T^{2}x,x) \le \sup_{\|x\|=1}\|T^{2}x\|\|x\|=\|T^{2}\|. $$On the other hand,$$ \|T^{2}\|\le \|T\|\|T\|=\|T\|^{2}. $$You get the final result by noticing ... 3 It shouldn't be to hard to show that A is injective, for if Ax = 0 then$$\langle x,x \rangle \le \langle Ax,x\rangle =0.$$It will follow that A is invertible once we show that A is surjective: the range R(A) satisfies R(A) = H. Let A^* denote the adjoint of A. Suppose that y \in N(A^*) so that A^*y = 0. Then$$0 = \langle y,A^*y ...

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In context of the above comment, it suffices to show that the spectral radius $\rho(T)=\|T\|$ for a self-adjoint operator $T\in B(\mathcal{H})$. That $\rho(T)\leq \|T\|$ is a trivial observation. For the reverse inequality, we need to show that either $\lambda$ or $-\lambda$ is in $\sigma(T)$, where $\lambda= \|T\|$. This is same as showing that $\lambda ... 1$\phi(A)\neq\lbrace 0 \rbrace$is a (complex) subspace of$\mathbb C$and therefore equal to$\mathbb C$. The second part is a general fact about quotients: Whenever$\phi: X\to Y$is a surjective linear map between vector spaces then$\tilde T: X/\mathrm{ker} T \to Y$is a linear bijection. 3 You can deduce the convergence of the series$\sum_n a^{n^{2/d}}$with$0 < a < 1$by first applying Cauchy's condensation test. Thus the original series converges iff the series$\sum_n 2^n a^{2^{2n/d}}$converges. Now apply the root test: This latter series converges if $$\limsup_n 2 \cdot a^{2^{2n/d}/n} < 1 \, .$$ Since$2^{2n/d}/n = ...

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I assume you mean the transition $\|(I-T)S_n-I\|=\|I-T^{n+1}-I\|$? This is basically the geometric series: $$(I-T)S_n = S_n - T S_n = (I+T+...+T^n)-(T+T^2+...+T^{n+1}) = I-T^{n+1}$$

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It's not $I \cdot S_n=I$, it's a telescopic series: \begin{align} (I-T)S_n & = S_n - TS_n = \\ &= (T^0 + T^1 + \dots + T^n) - T(T^0 + T^1 + \dots + T^n)= \\ & = T^0 + T^1 + \dots + T^n - T^1 - T^2 - \cdots T^{n+1}= \\ & =T^0 - T^{n+1}=I-T^{n+1}. \end{align}

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Given $U=e^{iH}$, assume V diagonalizes H: $e^{V^{-1} iH V} = V^{-1} e^{iH} V = V^{-1} U V$ implying that V also diagonalizes U; hence, $V$ is easily found. Let $\alpha_i$ denote the $i^{th}$ diagonal element of $V^{-1} U V$, then $\alpha_i = e^{i \theta_i}$ where $e^{i \theta_i}$ is the $i^{th}$ diagonal element of $e^{V^{-1} iH V}$. Then, it easily ...

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Hint Let $p(t) = t^2 - t -1$. Note that $p(t) = (t-\lambda_1)(t-\lambda_2)$ where $\lambda_1,\lambda_2$ are the eigenvalues $$t = \frac{1\pm \sqrt{5}}{2}$$ Note that $\ker p(S) = \ker(S - \lambda_1 I) + \ker(S - \lambda_2 I)$.

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You have $(a_k)\in\ker(f)$ if and only if $a_{k+2}=a_{k+1}+a_k$. So any such sequence is determined by its first two terms. $$(a,b,a+b, a+2b,2a+3b, 3a+5b,....) = a(1,0,1,1,2,3,5,...) + b(0,1,1,2,3,5,...)$$

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You may have already had this proven, but I thought I would include it for completeness. The integral of $e^{-2\pi ix\cdot\xi}$ over a sphere of radius $r$ is \begin{align} 2\pi r\int_{-r}^re^{-2\pi it|\xi|}\,\mathrm{d}t &=2\pi r\int_{-r}^r\cos\left(-2\pi t|\xi|\right)\,\mathrm{d}t\\ &=\frac{2r}{|\xi|}\sin\left(2\pi r|\xi|\right)\tag{1} ... 1 Your result is true if T is a closed operator(ie: graph of T is a closed subspace of H \times H) because we can prove that the closure of T (ie: the smallest closed extension of T), \overline T = T^{**} (\overline T = T if T is a closed operator). For a proof(easy only) you may see any standard book which deals with unbounded operators for eg: ... 3 I had not noticed before that your indices on the matrix were restricted to positive entries. This matrix has the form \begin{pmatrix} \alpha_0 & \alpha_{-1} & \alpha_{-2} & \alpha_{-3} & \cdots \\ \alpha_1 & \alpha_{0} & \alpha_{-1} & \alpha_{-2} & \cdots \\ \alpha_2 & \alpha_{1} & ...

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The operator $Qf = \int_{0}^{1}f(x)dx$ is easily verified to satisfy $Q^{2}f=Qf$. Therefore $S= I-Q$ satisfies $S^{2}=I-2Q+Q^{2}=I-2Q+Q=I-Q=S$.

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\begin{align} S(Sg)(y)&=Sg(y)-\int_0^1Sg(x)dx \\ &=g(y)-\int_0^1g(x)dx-\int_0^1\left( g(x)-\int_0^1g(y)dy\right)dx \\ &=g(y)-\int_0^1g(x)dx-\int_0^1 g(x)dx+\int_0^1\int_0^1g(y)dydx \\ &=g(y)-\int_0^1g(x)dx-\int_0^1 g(x)dx+\int_0^1g(y)dy \\ &=g(y)-\int_0^1g(x)dx \\ &=Sg(y) \end{align}

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Historically, integral operators are the prototypical compact operators. Compactness came up in the late 1800's when studying differential operators by recasting them as integral operators. While differential operators are very discontinuous, integral operators are especially continuous on bounded regions because they typically map uniformly bounded ...

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It is not true in general. To see this, set $$S=\left(\begin{array}{cc} 0 & 0\\ \sqrt{2} & 0 \end{array}\right).$$ Then $$S^{\ast}S=\left(\begin{array}{cc} 0 & \sqrt{2}\\ 0 & 0 \end{array}\right)\left(\begin{array}{cc} 0 & 0\\ \sqrt{2} & 0 \end{array}\right)=\left(\begin{array}{cc} 2 & 0\\ 0 & 0 \end{array}\right),$$ so ...

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Sorry I got the answer it was simple, Consider $l^2(\mathbf R)$ with the orthonorman basis$\{(1,0,0\cdots),(0,1,0,\cdots) \}$and consider the infinite diagonal matrix $diag (1, 1/2, 1/3,\cdots)$ which is a compact operator. This is a strictly positive compact operator.

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The first part is a standard result (easy to prove also!) available in most books on operator theory for eg: Proposition 4.6 in Banach Algebra Techniques in Operator Theory by R.G. Douglas, 2nd edition states it as follows If $T$ is an operator on the Hilbert space $H$, then $ker T = (ran T^*)^\perp$ and $(ran T)^\perp = ker T^*$ This proposition ...

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The comment by Geoff Robinson was really an answer: You have enough knowledge from class to know that $I-A^{N}$ is invertible, and $$I - A^{N}= (I-A) ( I + A + A^{2} + \ldots + A^{N-2} + A^{N-1})$$

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Expanding on John Ma's comment: consider the bilinear form on $W^{2,2}$ defined by $$B(u,v)=\int_\Omega \Delta u\Delta v$$ Only two derivatives of each function are involved, so this form is bounded on $W^{2,2}$: $$|B(u,v)|\le C\|u\|_{W^{2,2}}\|v\|_{W^{2,2}}\tag{1}$$ Hence, for every $u\in W^{2,2}$ the map $v\mapsto B(u,v)$ is a bounded linear functional ...

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Write $$f = \sum_{n=-\infty}^{\infty}\hat{f}(n)e^{inx}$$ What happens if you mutliply $f(x)$ by $a(x)=\sum_{n=-\infty}^{\infty}\alpha_ne^{inx}$? The function $a$ is continuous, periodic and bounded. So this multiplication operator $M_{a}$ is a bounded linear operator on $L^{2}$. $$M_{a}f = ... 1 Here the answer again: p isn't minimal in M_4(\mathbb{C}), because consider q= \begin{pmatrix} 1 & 0&0&0 \\ 0 & 0&0&0\\0 & 0&0&0\\ 0 & 0&0&0 \end{pmatrix}. It is q=q^*=q^2 and 0\le q\le p, but it isn't p=q, because p has a smaller image than p. 3 Counterexample to the first question: A nilpotent matrix. The second part is true, however: Since A is bounded linear there is a uniform bound c on ||A^1||,\dots, ||A^n||, and then we have |A^{kn+i}v|\le c||A^n||^k|v|\to 0. 1 Let A be your first statement "For all \varphi \in L^p, if (\lambda - m)\varphi = 0 a.e. then \varphi = 0 a.e.", and B your second statement "\lambda - m \ne 0 a.e.". To show A implies B, we would like to simply say: suppose A holds, and let C = \{x : m(x) = \lambda\}, and take \varphi = 1_C. Then certainly (\lambda - m) 1_C = 0 everywhere, ... 2 Test it yourself, with A=\pmatrix{0&1\\ 1&1},\ B=\pmatrix{1&2\\ 0&2} and the operator norm. The LHS of your inequality is about 1.44, but the RHS is \sqrt{2}. 1 Solutions can be found, thanks to the separation of variables. Not really a smart method, but effective anyways. In addition, the boundary condition \left(\frac{\partial(f(r,t)}{\partial t} \right)_{r=R}=v_0 \cos(\omega t) leads to look for a solution made with real sinusoidal functions of \omega t . This implies a particular form of the exponential ... 3 Let us assume that you are working with a reproducing kernel Hilbert space (otherwise multiplier algebras don't make all that much sense, and H^2 is certainly a RKHS). This means that the space H consists of functions on some space X and point evaluations (i.e. H \ni h \mapsto h(x)) are continuous functionals for all x \in X. If \phi \in M(H), ... 0 The idea is that on finite dimensional vector spaces, all norms are equivalent. In particular you can find \alpha >0 with$$\Vert A \Vert_\infty \le \alpha \Vert A \Vert$$for all A \in \mathbb R^{n \times n}. Where$$\Vert A \Vert_\infty = \sup\limits_{1 \le i,j \le n} \vert a_{i,j} \vert.$$And it is easy to prove that \mathbb R^{n \times n} is ... 0 You just have to verify that x \mapsto Ax is Lipschitz, which is not difficult considering the definition of continuity for a linear map. And in finite dimension, all linear maps are continuous. Then you'll prove that x = \exp(At)x_o is solution of the equation. According to Picard–Lindelöf theorem, the solution is unique. 1 After reading the Luiro's paper again more carefully, I realize that he answers why \mathcal{M} is not sublinear in the sense that$$\left\|\mathcal{M}f-\mathcal{M}g\right\|_{W^{1,p}}\leq\left\|\mathcal{M}(f-g)\right\|_{W^{1,p}},\quad\forall f,g\in W^{1,p}(\mathbb{R}^{n}) in the concluding remark of the paper. The failure of sublinearity is a ...

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