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The operator $a$ has a polar decomposition $a=u|a|$, where $u$ is a partial isometry with initial space $\ker a^\perp$ and final space given by the closure of the range of $a$. Then $a=u|a|$ and $a^*=|a|u^*$. We can rewrite the given inequalities as $$\Vert |a|^2 - p \Vert < 1/4, \qquad \Vert u|a|^2 u^* -q \Vert < 1/4.$$ Looking at the second ...

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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 ...

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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 $$Obviously, all trajectories (invariant sets) are connected with the mapping x(t) = e^{At} x_0 regardless of the stability of the system. However, I don't know if all of the invariant sets are connected for an arbitrary ... 0 In short, yes. One thing that you need to clarify is, \epsilon is not a vector while the argument of \Psi is. I'll assume that \epsilon is multiplied by some constant vector v. Consider \Psi(\vec{r}+\epsilon\vec{v}) as a function of both \vec{r} and \epsilon. Name it \psi for clarity.$$ ...

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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 $$\|Tx\| \leq \sum_{i\in I}|y_i|\|Te_i\| \leq \sqrt{\sum_{i\in I} |y_i|^2}\sqrt{\sum_{i\in I}\|Te_i\|^2} = \|T\|_{HS}$$

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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 ...

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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 ...

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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 ...

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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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It is not necessarily true that $\pi(A)$ will have non-trivial projections. But $\pi(A)''$ will, because it is a von Neumann algebra. Now, the assertion you are looking at is basically trivial. You want to prove, for $A$ abelian, "if $\dim H\geq2$ then $\pi$ is not irreducible". This is the same as its contrapositive, which is "If $\pi$ is irreducible, ...

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The new operator $T$ is going to be also compact, as the product of a bounded and a compact and a bounded operator is also compact, and hence its spectrum, which could now contain even complex numbers, could only have as a limit point the $0$. The fact that $P$ is also self-adjoint does not change anything.

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Elements in the strong closure are only obviously limits of nets. The Corollary says that they are limits of sequences if the Hilbert space is separable.

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The strong operator topology on $\mathscr{B}(H)$ is not metrisable if the underlying Hilbert space $H$ is infinite-dimensional. Therefore it need not be sequential a priori. What the Kaplansky density theorem really tells you is that elements in the SOT-closure of a C*-algebra are limits of (norm) bounded nets. Moreover, if the Hilbert space is separable, ...

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As stated in the comments, you should specify what class of $\phi$ you are considering or what restrictions you impose. Anyway, proceeding a bit vaguely and assuming a "nice class" (for example, at least the inverse of $\phi(D)$ and $\phi(0)$ should be defined) you can transform your equation to (multiplying by $\phi(D)$ on the left and $\phi(0)$ on the ...

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If $\alpha_{nm}=\delta_{nm}$, the operator $A$ is the identity, which is not compact. So, there is a problem with your attempt... here: $A=\lim A_n$ This is true in the sense of pointwise convergence (strong operator topology). But to conclude compactness, you need norm convergence: $\|A-A_n\|\to 0$. This is not necessarily the case, as the example ...

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A basis $\{ e_{1},e_{2},\cdots,e_{N} \}$ of an $N$-dimensional vector space $X$ can be thought of as a way to transform a vector $x \in X$ into a coefficient function $\hat{x} : \{ 1,2,3,\cdots,N \}\rightarrow X$, where the inverse map is $(\hat{x})^{\vee} = \hat{x}(1)e_{1}+\cdots+\hat{x}(N)e_{N}=x$. A linear map $L$ on $X$ is diagonalized by the basis ...

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Restating the requirements, with $D = \dfrac{d}{dx}$ for convenience, we want $$\sum_k a_k(x) D^k(D(f)) = \sum_k b_k(x) D^k(x f) \tag 1$$ for any function $f$. Now by Leibniz's rule, $D(x f) = f + x D(f)$, and by induction $D^k(x f) = k D^{k-1}(f) + x D^k(f)$, so that (1) becomes \eqalign{ \sum_k a_k(x) D^{k+1}(f) &= \sum_k b_k(x) (k D^{k-1}(f) + x ... 1 For existence, one can often find an example. The book has given an example. Here is how it works: A(\frac{d}{dx}(f))=x^2f' and B(x(f))=(x\frac{d}{dx}-1)(xf)=x(\frac{d}{dx}xf) - xf=x(f+xf')-xf=x^2f'+xf-xf =x^2f' And so both are equal. If I have misunderstood you, please say. 1 In general, it is fully acceptable for a bounded self-adjoint operator to not have any eigenvalues. Consider for instance the operater M\in L^2([0,1]) defined on a representative by (Mf)(x)=xf(x). $$It is not hard to see that the spectrum of this operator is [0,1] (because it is a multiplication operator corresponding to a real continuous function ... 3 Rule of thumb: compactness should be expected when Tf is at least a little bit nicer than a generic element of the target space. Integration against a continuous kernel does this because Tf inherits modulus of continuity from the kernel, which is an improvement compared to having totally unknown modulus of continuity. But for compactness from C^0 to ... 1 The spectral theorem for normal operators on a Hilbert space gives you the result, but you don't get a unique selfadjoint T. Any unitary U can be written as$$ U = \int_{|\lambda|=1}\lambda dE(\lambda), $$where E is the unique spectral measure associated with U. Choose a branch of \log and define$$ ...

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Be careful that $T$ is not assumed to be bijective, so things like $T^{-1}x$ do not make sense. Maybe the following hint (in the spirit of the one you indicated for forward orbits) will be enough for you. As above, let $(y_n)_{n\geq 1}$ be dense in $X$, and let $x_0\in X$. Find $x_1$ close to $y_1$ and an integer $n_1$ such that $T^{n_1}x_1$ is close to ...

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Diagonalization is synonymous with finding a basis of eigenvectors. If you have a diagonal matrix, the elements of the standard basis are all clearly eigenvectors, and conversely, if you write the matrix in a basis of eigenvectors it will be diagonal. The exponentials form a basis of eigenvectors,so in this sense they diagonalize convolution. Here follows a ...

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I obtained the following upper bound. Let $D_z$ denote a diagonal matrix whose diagonal entries are given by vector $z$. Furthermore, write the SVD of $X$ as $X=UD_\sigma V^\mathrm{T}=\sum _{i=1}^r\sigma_i u_iv_i^\mathrm{T}$ where $r$ denotes the rank of $X$. Then we have \begin{align*} F\odot X &=\sum_{i=1}^r F\odot \left(\sigma_i u_i ...

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First observe that the proof given in Bounded linear operator maps norm-bounded, closed sets to closed sets. Implies closed range? does not use the fact that $X,Y$ are assumed to be complete. It hence remains to show that we can reduce to the case of an injective operator. To this end, we "factor out the kernel": As $T$ is bounded, the kernel $Z := ... 1 This is just a thought which might be helpful but is a bit long to be a comment. Consider a$N\times N$matrix$A$whose$(1,1)$entry is$|z|^2$. Let$e_1=[1,0,\dots,0]^T$denote first column of$N\times N$Identity matrix. Then note that, $$\lambda_{min}(A) = \min_{x^Tx=1}~x^TAx\leq e_1^TAe_1=|z|^2$$ 2 A common way to prove that some operator$T$is not compact is to exhibit an infinite-dimensional subspace$M$on which$T$has a lower bound: that is, there exists$c>0$such that $$\|Tx\|\ge c\|x\|,\quad \forall\ x\in M \tag{1}$$ If (1) holds, then the image of unit ball under$T$contains a ball of radius$c$in the infinite-dimensional subspace$TM$, ... 1 Jonas Meyer's comment gives an example of a self-adjoint$T$whose spectrum is$[-1,1]$. Then we may take the absolute value function, for instance, or even any continuous function whose zeros accumulate in$(-1,1)$to see that that the space of continuous functions on the spectrum contains functions which are not power series. Since$C(\sigma (T))$is ... 0 I'll assume the underlying space of interest is$L^{2}[0,L]$because any discussion of 'selfadjoint' requires a Hilbert space. And I'm assuming you're starting with a domain$C^{2}[0,L]$. There are variations where you assume$f \in C^{2}(0,L)$with$f''\in L^{2}[0,L]$, and I'll leave those to you. Define$\Delta f = f''$on$\mathcal{D}(\Delta)=\{ f \in ...

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In the following, I will assume that $\mu$ is $\sigma$-finite for simplicity. One can probably drop that assumption with a bit of extra work. Also, I do not claim that my proof is one of the simpler ones. Let us define $$\left\Vert k\right\Vert _{L^{p,q}}:=\left(\int\left(\int\left|k\left(x,y\right)\right|^{p}d\mu\left(x\right)\right)^{q/p}\, ... 3$$\int_0^1 \underbrace{|k(s,t)-k(s',t)|}_{\leq \sup_{r \in [0,1]} |k(s,r)-k(s',r)|} \underbrace{|f(t)|}_{\leq \sup_{r \in [0,1]} |f(r)|} \, dt \leq \sup_{r \in [0,1]} |k(s,r)-k(s',r)| \cdot \|f\|_{\infty} \cdot \int_0^1 \, dt.$$1 Nota Bene: In my haste and arrogance I misread the question as written, blithely and blindly assuming that the only case of interest was t \to \infty; when my error was graciously pointed out by our colleague PhoemueX (see comments), I deleted my answer until such time as I could correct it. Having done so, I present my rectified answer below; however, I ... 2 You can find P so that PGP^{-1}=T=\begin{pmatrix}\lambda_1 & * & *\\&\ddots & *\\&&\lambda_n\end{pmatrix} where the \lambda_is are the eigenvalues and * means "whatever". Then, \exp(tG)=\exp(tP^{-1}TP)=P^{-1}\exp(tT)P. \exp(tT)=\begin{pmatrix}e^{t\lambda_1} & * & *\\&\ddots & ... 2 If you have a bounded operator A, then the holomorphic functional calculus is always an option, and it is based on Cauchy's integral representation:$$ f(A) = \frac{1}{2\pi i} \oint_{C} f(\lambda)\frac{1}{\lambda I-A}"\,d\lambda = \frac{1}{2\pi i} \oint_{C} f(\lambda)(\lambda I-A)^{-1}\,d\lambda.  The contour $C$ is any simple ...

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The nearest point projection onto a closed subset $E\subset \mathbb R^n$ is single-valued if and only if $E$ is convex*. In this case, the Lipschitz constant is equal to $1$. If $E$ is not convex, there is at least one point $x\in \mathbb R^n$ for which $\min_{y\in E}\|x-y\|$ is attained at more than one point. We could try to discuss the continuity of ...

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This is a comment to @ABC 's answer. I seem to have difficulties to understand things correctly, maybe it's simply too late. Here is a list of a: the alternating series of units, b: the partial sums (=using Cesaro-order 0), c: the partial sums using Cesaro-order 1, d: and the squares of $c$, (where it is focused that the ...

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We have the following characterization of adjoint operators: Suppose that $X$ and $Y$ are normed spaces. If $T:X\rightarrow Y$ is a bounded linear operator, then $T^*$ is weak*-weak* continuous. Conversely, if $S$ is a weak*-weak* continuous linear operator from $Y^*$ to $X^*$, then there is a bounded linear operator $T:X\rightarrow Y$ such that $T^*=S$. ...

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