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(Reposted from comments at questioner's suggestion) Hint: You're looking for an operator which, if you differentiate it with respect to $\lambda$, gives you the original operator times $\lambda[A,B]$. You've got one operator which satisfies that already; can you find a second one?

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Consider e.g. the translation group on $L^2(\mathbb{R})$. These operators are isometries and thus the group is bounded, it is however not uniformly continuous. This and more examples can be found on pp. 34 in the book of Engel and Nagel.

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I do not know how to solve the equations to find the following, but for any $t_0$, an involute of $X, Y$ can be calculated as $$f(X,Y) = X - \frac{X'}{\def\abs#1{\left|#1\right|}\abs{(X,Y)'}}\cdot \int_{t_0}^t \abs{(X,Y)'}\, dt,\quad g(X,Y) = f(Y,X)$$ where $\abs\cdot$ denotes the standard absolute value on $\mathbb R^2$.

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Summary of comments Fix $a$ and let $T(x)=\sum_{k=1}^\infty a_kx_k$. For $n=1,2,\dots$ define linear functionals $T_nx = \sum_{k=1}^n a_kx_k$. These are bounded on $\ell^2$ and satisfy $\sup_n |T_n(x)|<\infty$ for every $x$, because the series $\sum_{k=1}^\infty a_kx_k$ converges. By the uniform boundedness principle, also known as the ...

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Suppose that $X$ is a complex inner-product space, and that $B(\cdot,\cdot)$ is a complex function on $X\times X$ such that $B$ is linear in the first coordinate and conjugate-linear in the second coordinate. Then $$B(x,y) = \frac{1}{4}\sum_{n=0}^{3}i^{n}B(x+i^{n}y,x+i^{n}y).$$ A special case is the inner-product $B(x,y)=(x,y)$, which gives ...

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If $T$ is normal on a Complex space $X$, then $$\|Tx\|^{2}=\|T^{\star}x\|^{2},\;\;\; x \in X.$$ Because $\lambda I$ is normal with adjoint $\overline{\lambda}I$, then $\|(T-\lambda I)x\|=\|(T^{\star}-\overline{\lambda}I)x\|$ also holds for all $x$ if $T$ is normal. That's the missing piece.

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Classical Solutions: First, assume $V\in L^{1}[a,b]$, and show the existence of classical solutions of $$-f''+Vf = \lambda f,\;\;\; f(a)=A,\;f'(a)=B.$$ This can be done by considering the equivalent integral equation $$f(x)=A+B(x-a)+\int_{a}^{x}\int_{a}^{t}(V(u)-\lambda)f(u)\,du\,dv.$$ This is a fixed point problem for $C[a,b]$ ...

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You have to show that $$\langle Sv,v\rangle =0 \quad \forall v\in V$$ implies $S=0$. To prove this one has to use polarization identities. In the complex case this works for a general mapping $S$. In the real case, this holds only for self-adjoint $S$. However, the mapping $T^*T-TT^*$ is self-adjoint.

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This is supposed to be a basis. If you only used a single element $x$, it would not be closed under intersections.

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The answer to the first question is simply the following: take $f(z)=z^4$ in the spectral mapping theorem and use $\mathfrak{F}^4=1$.

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The spectrum is not as stated. The spectrum of the Hamiltonian for the non-relativistic Hydrogen atom has eigenvalues corresponding to the bound states of Hydrogen, and these are negative. The positive spectrum corresponds to unbound states, and is a continuous spectrum. In spherical coordinates, for a function which depends on the radius $r$ only, one has ...

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Let $X=L^2[0,1]$ and define $T:X \to X$ by $(Tf)(x) = x f(x)$. Then $T$ is bounded and $T^* =T$. Let $c\neq 0$ be a constant, and let $\gamma(x) = c$ for all $x$. Then we see that $\|Tf-\gamma\|>0$ for all $f \in X$ (since $x \mapsto { c \over x}$ is not in $X$). Hence $\gamma$ is not in the range of $T$. If we choose $g \in X$, we can define $f_n(x) = ... 1 For$y \in D(T)$, we have, due to the symmetry of$T$, $$x \mapsto \langle Tx,y\rangle = x \mapsto \langle x, Ty\rangle,$$ and the latter is easily seen to be continuous, hence$D(T) \subset D(T^\ast)$. 0 Since$T$is self-adjoint, you can use the spectral theorem (see e.g. Theorem 13.24 in Rudin, "Functional Analysis"). If$\text{Re}(z) > 0$,$f(\lambda) = \lambda^z$is a continuous function on$[0,\infty)$. If$\text{Re}(z) \le 0$,$f(\lambda)$is not defined at$\lambda=0$, but this is only a problem if$0$is an eigenvalue rather than just a point ... 3 A bounded linear operator$A$on a separable Hilbert space$X$with orthonormal basis$\{ e_{j} \}_{j=1}^{\infty}$is a Hilbert-Schmidt operator if $$\sum_{j=1}^{\infty}\|Ae_{j}\|^{2} < \infty.$$ This condition is true for one orthonormal basis iff it is true for every other orthonormal basis. If you're studying$X=L^{2}[a,b]$, then an ... 0 Consider$\alpha< 1/2 $$$||k(x,y)||_{L^2xL^2} = \int^b_a\int^b_a |k(x,y)|dxdy= \int^b_a\int^b_a \frac{1}{|x-y|^{2\alpha}} dxdy$$ So splitting this up, $$\int^b_a\int^y_a \frac{1}{(y-x)^{2\alpha}} dx+ \int^b_a \frac{1}{(x-y)^{2\alpha}} dx dy$$ Since$\alpha< 1/2 $,$1-2\alpha >0 , so integrating, $$=(2\alpha+1)^{-1}\int^b_a ... 0 An operator is a mapping from one vector space or module to another. where a function is a map from any arbitrary set to another 0 This was resolved in comments: \|T^*\|=\|T\| for every bounded linear operator between normed spaces. In fact, we only need the easier part here: \|T^*\|\le \|T\|, which comes directly from the fact that T^* acts on linear functionals by composing them with T. 1 Let \{ e_{j} \} be an orthonormal basis of L^{2}[0,1] consisting of real functions. Define f_{i,j}(x,y)=e_{i}(x)e_{j}(y). Then \{ f_{i,j} \}_{i,j} is a complete orthonormal basis of L^{2}([0,1]\times[0,1]); this can be verified by showing that (f,f_{i,j})=0 for all i, j for a continuous f(x,y) implies f=0. Notice that$$ \begin{align} ... 1 Can we accept as fact the following representation of the delta function? \begin{align} \delta(x-y) = \sum_{j\in \mathbb{Z}} e_{j}(x)\overline{e_{j}(y)} \end{align} If so, then \begin{align} \mathrm{tr}(K) &= \sum_{j\in \mathbb{Z}} \int_{0}^{1}Ke_{j}(x)\overline{e_{j}(x)} \, \mathrm{d}x \\ &= \sum_{j\in \mathbb{Z}} ... 0 Yes, the sequence\{f_n=e_n/n\}$is a Bessel sequence, but is not a frame, since for$v=e_k$we have$\sum_n|\langle v,f_n\rangle|^2 = 1/k^2$, which is not bounded below by$A\|v\|^2$for any fixed$A>0$. The above set is spanning. As far as I know, a Bessel sequence is not required to span the space. In this case,$\{0,0,0,\dots\}$is a trivial example ... 0 What would it mean to say ‘the solution of$u$depends on$F$continuously‘ It means that for every$F$and every$\epsilon>0$there exists$\delta>0$such that for every functional$G$with$\|F-G\|<\delta$we have$|\tilde u-u\|<\epsilon$, where$\tilde u$is the solution of the variational inequality with$G$instead of$F$. how is ... 0 In set theory, operator or operation is sometimes used for those operations that, unlike functions, cannot be modelled by sets of ordered pairs, but whose collection of ordered pairs instead constitute a proper class. The most common example of this is the power set operator$\mathcal{P}(X)$. 0 Same approach as previous answers, using limited operators: If$T\colon X \to X$is a compact operator and$S \colon X \to S$is a bounded operator, then both$TS$and$ST$are compact. Let$S: C^1[0,1] \to C^1[0,1]$such that$u(t) \mapsto u(\sqrt{t})$. Readily we see that$\|S\| = 1$, so$S$is bounded. But$TS = Id$. If$T$is compact so is$Id$, but ... 4 Since the spectrum does not depend on the (C$^*$) algebra, we can assume that$\mathcal A=C^*(x)$. Using the Gelfand transform, we can identify$C^*(x)$with$C(\sigma(x))$, with$x$mapped to the function$z\mapsto z$. The point evaluation states$f\mapsto f(t)$are precisely the pure states The pure states are the extremal points of the set of states of ... 5 If$A^*$denotes the adjoint matrix (complex conjugate) of$A$that is not true. Take: $$A = \left( \begin{array}{c c} 1 & 0 \\ 0 & 0 \end{array} \right) \qquad B = \left( \begin{array}{c c} 1 & 1 \\ 0 & 0 \end{array} \right)$$ Then: $$A^* = \left( \begin{array}{c c} 1 & 0 \\ 0 & 0 \end{array} \right) \qquad B^* = ... 2 The point in doing so in general is, that x^* = \left<\cdot, x\right>\colon X \to \mathbb R is a continuous linear functional. Let \xi \colon \Omega \to X be a random variable, we will show that \def\E{\mathbb E}\E\def\<#1>{\left<#1\right>}x^*(\xi) = x^*(\E \xi) holds if \E\|\xi\| exists. We start with a characteristic \xi = ... 0 The definition you give of pure point spectrum is not the usual one. Usually the pure point spectrum is the spectrum associated with the pure point part of the spectral measure (hence it is usually associated with self-adjoint or normal operators, for which the spectral theorem holds). Therefore it could correspond to finite or infinite multiplicity ... 3 You can probably start with the smooth case f \in C_0(\mathbb{R}), and write down explicitly$$\int_{-1/2}^{1/2} f(x-y)\, dy =0 \quad\hbox{for (almost) every$x \in \mathbb{R}$}.$$This should imply rather easily that f vanishes identically. Then remember that a generic f \in L^2(\mathbb{R}) can be approximated by continuous functions with compact ... 2 Here is how you advance.$$ u_{np} = \dfrac{1}{E - 2}n^22^n = -\frac{1}{2} \left( 1-\frac{E}{2} \right)^{-1}n^22^n  =-\frac{1}{2} \sum_{k=0}^{\infty}\frac{E^k}{2^k}n^22^n = -\frac{1}{2} \sum_{k=0}^{\infty}\frac{1}{2^k}(n+k)^2\,2^{n+k} = -\frac{2^n}{2}\sum_{k=0}^{\infty}(n+k)^2=-\frac{2^n}{2}\sum_{m=n}^{\infty} m^2  = -\frac{2^n}{2}\left( ... 1 I'm not familiar with operator theory, but this is an answer using generating functions. First, define $$U(x)=\sum^{\infty}_{n=0}u_nx^n$$ Multiply the equation throughout by$x^n$and sum up the terms from$0$to$\infty$. We get $$\sum^{\infty}_{n=0}u_{n+1}x^n- 2\sum^{\infty}_{n=0}u_nx^n=\sum^{\infty}_{n=0}n^2(2x)^n$$ The sum on the right hand side is ... 1 Perhaps define $$v_n=\frac{u_n}{2^{n-1}}.$$ Then the given relation becomes $$v_{n+1}-v_{n}=n^2.$$ In your notation this means $$(E-I)v_n=n^2.$$ Thus $$v_n=(E-I)^{-1}n^2.$$ Since$E$is the difference operator then its inverse would be to increase the degree (like integration is inverse of differentiation). Thus$v_n$should be an appropriate cubic ... 0 ? As you've suggested already yourself, is the answer not simply this:$T = S + D$, with $$S = \frac{1}{2} \left( T + T^* \right) \qquad ; \qquad D = \frac{1}{2} \left(T - T^* \right)$$ Where$S$is self-adjoint and$D$is a so-called anti self-adjoint operator:$D^* = -D$. In three dimensional space, the anti self-adjoint operator ($3 \times 3$matrix) ... 1 Here's Daniel Fischer's comment in more explicit terms: any compact operator on an infinite dimensional Banach space (e.g.$C([0,1])$) cannot be invertible. However the map$x(t)\mapsto x(\sqrt{t})$is the inverse map for$T$so$T$cannot be compact. 1 Here is a much less intuitive answer. Note: Unfortunately this was not as simple as I had originally thought (thanks to Davide for catching my oversight). The result depends on the fact that$C[0,1]$has the 'approximation property', that is, any compact operator is the limit (in the operator norm) of a sequence of finite rank operators. (See Remark 1.1.15 ... 1 If$A=A^{\star}$is a densely-defined selfadjoint linear operator on a complex Hilbert space$H$, and if there is a complete orthonormal basis of$H$consisting of eigenvectors of$A$, then it is true that the point spectrum$\sigma_{p}(A)$of$A$is dense in$\sigma(A)$. The converse is not true. To show this, let$H$be a Complex Hilbert space and suppose ... 2 The first equality is just the parallelogram identity, valid for all norms arising from an inner product. Next, for the purported equality $$2[\Vert u_{j} \Vert^{2}+ \Vert u_{l} \Vert^{2}] - \Vert u_{j} + u_{l} \Vert^{2} = 2[E(u_{j}) + E(u_{l})] - 4E(\frac{u_{j}+u_{l}}{2}),$$ we observe that the linear part of$E$cancels on the right hand side, so we are ... 1 You need to show that$B=i(B(0,1))$is relatively compact. Suppose$y_n \in B$. Then since$\|y_n\|_0 + \|y_n'\|_0 < 1$, we see that$\|y_n\|_0 < 1$for all$n$, hence$y_n$are uniformly bounded. Furthermore, since$\|y_n'\|_0 <1$, we see that the$y_n$are Lipschitz with rank at most one, hence equicontinuous. Applying the Arzelà–Ascoli theorem, ... 1 Suggestion: Apply the Arzelà–Ascoli theorem. 1 Because nobody is answering, I thought I'd offer a start. Your equation for the resolvent of$A$is $$(\lambda I -A)\left(\begin{array}{c}f \\ g\end{array}\right)= \left(\begin{array}{cc}\lambda & -I \\ I-\Delta & \lambda\end{array}\right) \left(\begin{array}{c}f \\ ... 2 It is clear that |\langle Tx,x\rangle|\leq \|T\| for \|x\|=1. For the converse, it suffices to show that |\langle Tx,y\rangle|\leq \alpha for all \|x\|=\|y\|=1, with \alpha=\sup|\langle Tx,x\rangle|. We can clearly assume \langle Tx,y\rangle \in\mathbb R. Then$$ \langle Tx,y\rangle = (\langle T(x+y),x+y\rangle - \langle T(x-y),x-y\rangle)/4. $$... 1 Why are PDE's classified in this way where it relates to conic sections? They are not. The parallel with conic sections is an artifact of second-order PDE in two dimensions. It is not a classification of PDE in general, as one quickly discovers when encountering higher order equations and higher dimensions. The properties recognized in two dimensions ... 1 First of all, your requirements imply that for each x \in H, the sequence \iota(T_n x) of bounded linear functionals, where \iota denotes the canonical embedding of H into the dual H' by \iota(x)(y) := \langle y, x\rangle is pointwise bounded. By the uniform boundedness principle, this implies that \Vert T_n x \Vert is a bounded sequence for ... 1 Hint: Let L(x,y) = \lim_{n \to \infty} \langle y, T_n x \rangle. You want to define T so that \langle y, T x \rangle = L(x,y). The Principle of Uniform Boundedness will be useful. 0 The Bogoliubov transformation is well defined since it respects the antilinearity:$$a(U(\lambda f))+a(V(\lambda f))^*=\gamma[a(\lambda f)]=\gamma[\overline{\lambda}a(f)]=\overline{\lambda}\gamma[a(f)] \\=\overline{\lambda}a(Uf)+\overline{\lambda}a(Vf)^*=a(U(\lambda f))+a(V(\lambda f))^*\\ ... 1 Assume there is a non-zero$v \in \mathcal{D}(Q)\setminus \mathcal{D}(A)$, and let$K$be the one-dimensional subspace spanned by$v$. Then$K\cap \mathcal{D}(Q)=K$, and$K$is closed in$H$. The restriction of the form$Q$to$K$is finite-dimensional, and positive-semidefinite.$B$is trivially defined, but$\mathcal{D}(B)\ne\mathcal{D}(A)\cap K=\{0\}$. ... 2 The Fourier transform is a unitary operator on your space. This means that its transpose is its inverse,$\mathcal F ^* = \mathcal F^{-1}$. The typical thing to do is to replace$T$with$\mathcal F T \mathcal F^*\$. Observe that with this convention, you have $$(\mathcal F T \mathcal F^*)\hat f=(\mathcal F T \mathcal F^*)\mathcal F f=\mathcal F ... 0 Suppose that the ideal I is projectionless. If a partial isometry v\in A maps to a unitary, it means that 1-v^*v\in I, but 1-v^*v is a projection, impossible. This situation can be achieved for instance when A=C(X), I=C_0(X) for some compact subset of \mathbb C. 0 Here is an alternative proof for the first statement. Suppose that x_n\to x in E and Tx_n\to f in E'. By hypothesis, we have that$$\langle Tx_n-Ty,x_n-y\rangle\ge 0,\ \forall \ y\in E\tag{1}$$If we pass the limit in (1) we get that$$\langle f-Ty,x-y\rangle\ge 0,\ \forall\ y\in E\tag{2}$$Now take y=x+tv where t\in \mathbb{R} and v\in X. ... 1 Note that,$$ \frac{\partial}{\partial x} \delta(x) = - \delta(x) \frac{\partial}{\partial x}, $$which implies that,$$ \frac{\partial^2}{\partial x^2} \delta(x) = \frac{\partial}{\partial x} \frac{\partial}{\partial x} \delta(x) = - \frac{\partial}{\partial x} \delta(x) \frac{\partial}{\partial x} = + \delta(x) \frac{\partial}{\partial x} ...

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