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If $E_N$ is the spectral resolution for $N$, then the following is bounded and normal by the functional calculus: $$N(I+N^{\star}N)^{-1} = \int \frac{\lambda}{1+|\lambda|^{2}}dE_{N}(\lambda).$$ In line with your comment under this point that you are trying to derive the spectral measure, let's try a different approach. For any $x$, one has ...

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Start by showing $\|T(t)f\| \le \|f\|$ for all $f \in L^{2}[0,\infty)$ and $t \ge 0$: \begin{align} \|T(t)f\|^{2} & =\int_{0}^{\infty}|f(x+t)|^{2}dx \\ & = \int_{t}^{\infty}|f(x)|^{2}dx \\ & \le \int_{0}^{\infty}|f(x)|^{2}dx = \|f\|^{2}. \end{align} Because of this norm estimate, the problem of showing ...

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This is just a property of continuous functions. If $f(x)$ is continuous at $a$, and $\lim_{n\rightarrow\infty} a_n=a$, then $\lim_{n\rightarrow\infty} f(a_n)=f(\lim_{n\rightarrow\infty} a_n)=f(a)$. Be careful with your example as $\log(x)$ is not continuous at $x\leq 0$ so if $\lim_{n\rightarrow\infty} f(x)=0$ for some $x$, then the result doesn't ...

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It's not quite stated precisely. This should be more-or-less in Kato. The spectral projection for the isolated eigenvalue $\lambda$ is $$P = \dfrac{1}{2\pi i} \oint_\Gamma (z I-A)^{-1}\; dz$$ where $\Gamma$ is a small circle centred at $\lambda$. By assumption, this is a projection of finite rank. It is the limit (in operator norm) of the ...

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After thinking some more about it, I came up with this, but I am not sure this is right: Every probabilty measure $\mu$ on $B(S)$ can be written as: $\mu = \int_S \delta_y \mu(dy)$, then we have for $O\mu$ $O\mu = O\int_S \delta_y \mu(dy)=\int_S O\delta_y \mu(dy)$ (last equality holds because of the linearity of O) This cannot be defined, if ...

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For your second question the above answer is good.... For the first one let $x,y \in \overline{D(T)}$, i.e., there exist sequences $(x_n),(y_n)$ in $D(T)$ such that $x_n \longrightarrow x$ and $y_n \longrightarrow y$, then $(x_n+y_n)$ is a sequence in $D(T)$ (as $D(T)$ is a vector space) such that $x_n+y_n \longrightarrow x+y$ and if $\alpha$ is any scalar ...

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Suppose $$T(x_1)(t) = T(x_2)(t)$$for all $t \in[0,1]$. As you have already figured out that the range of the operator is the set of all continuously differentiable functions (this follows from Fundamental Theorem of Calculus). In light of your observation just differentiate the two sides of the equation to get $x_1(t) = x_2(t)$, ...

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If $\parallel x\parallel =M$ then $$|T(x)(t)|=|\int_0^t x(s)ds | \leq \int_0^t M \leq M$$ Hence bounded. And $\frac{d}{dt} T(x)(t)=x(t)$ is continuous. And $T(x)(0)=0$. If $T(x)=T(y)$ then $\parallel T(x)- T(y)\parallel =0$ So $$\forall t,\ \int_0^t (x-y)(s) ds =0$$ Assume that $t_0\in (0,1)$ with $(x-y)(t_0) >0$. Then $x-y \geq c> 0$ on $... 0 Yes. Your answer about range is correct. The operator is injective. To see it, since$x(t)$is continuous at$[0,1]$, you can apply Fundamental theorem of calculus, given$F(t)\in R(T)$,$x(t)$is uniquely determined by$x(t)=F'(t)$. Or you can argue directly that if$x_1(t_0)\not =x_2(t_0)$, for some$t_0\in[0,1]$, by continuity$T(x_1)(1)\not =T(x_2)(1)$... 1 You can define$G$as follows: if$x_n \in D(T)$and$w \in W$with$x_n \to w$, then$G(y) = \lim_{n \to \infty} T(x_n)$(use boundedness of$T$to show that this is well-defined). Then if$x_n \to w$and$y_n \to v$, and$a,b$are scalars, you want to show that$G(aw + bv) = a G(w) + b G(v)$. Well, what sequence (defined in terms of$x_n$,$y_n$,$a$, ... 2 Yes, the closure of a vector space is a vector space. Yes,$D(T)$might not be a closed set. For example, polynomials form a non-closed subspace of$C[0,1]$(the continuous functions on$[0,1]$). 1 First, suppose$T$is bounded. Let$\{x_n\}$be a sequence in$X$converging to zero. Then,$\{\lVert x_n\rVert\}$also converges to zero and hence is bounded (by some$M$). Thus,$\lVert Tx_n \rVert \leq \lVert T \rVert \lVert x_n \rVert \leq \lVert T \rVert M < \infty$. Conversely, suppose$T$maps zero-convergent sequences to bounded sequences. If$T$... 1 If$1_A$is the characteristic function of some set, then $$1_A \circ T=1_{T^{-1}(A)}$$ Therefore, if$f =\sum a_i 1_{A_i}$you have $$f \circ T = \sum a_i ( 1_{A_i} \circ T)= \sum a_i 1_{T^{-1}(A_i)}$$ Thus $$\int\limits_X|f\circ T|^p\mathrm{d}m=\int\limits_X|\sum a_i 1_{T^{-1}(A_i)}|^p\mathrm{d}m=\int\limits_X|\sum a_i 1_{A_i}|^p\mathrm{d} T^* m$$ 1 Imagine that the vector field in question describes the velocity of fluid at a given point in a giant tank of fluid. In this instance, a net positive divergence over a solid region means that there is fluid flowing out of that region or, equivalently, that fluid is being produced within the region, a 'source' if you like. A net negative divergence, on the ... 2 Since $$f(0)=f(0+0)=f(0)+f(0)=2f(0),$$ we have$f(0)=0$. It follows that $$f(-x)=f(-x)+f(x)-f(x)=f(-x+x)-f(x)=f(0)-f(x)=-f(x) \quad \forall x\in E.$$ Also for every$x\in E$we have \begin{eqnarray} f(2x)&=&f(x)+f(x)=2f(x)\\ f(3x)&=&f(2x)+f(x)=3f(x)\\ &\vdots&\\ f(nx)&=&nx \quad \forall n\in \mathbb{N}. \end{eqnarray} ... 2 For example, note that$f(2x) = f(x + x) = f(x) + f(x) = 2 f(x)$, and hence$f(x) = \frac{1}{2} f(2x)$, which means$f(\frac{1}{2} x) = \frac{1}{2} f(x)$. Similarly one can prove that, for every$q \in \mathbb{Q}$, we have$f(qx) = q f(x)$. Now suppose$\lambda \in \mathbb{R}$. Choose a sequence$q_n \in \mathbb{Q}$such that$q_n \rightarrow \lambda$. This ... 0 I don't think so. According to your definition, every bounded linear operator is a relatively bounded perturbation. 1 We are talking about a flow field$v$which does not change in time; but the field vectors change from point to point. Consider a fixed point$p$within this flow field and a cube$C$of side length$s\ll1$centered at$p$. The net flux$\Phi$of$v$through the surface$\partial C$of this cube represents the amount of fluid produced within$C$per ... 1 Starting with the comment, $$E(S)\left(\int f(\mu)dE(\mu)\right)=\left(\int f(\mu)dE(\mu)\right) E(S)=\int_{S}f(\mu)dE(\mu).$$ Let$f(\mu)=\mu-\lambda$and let$S$be the singleton set$\{\lambda\}$. Then the second equality gives $$(u-\lambda I)E\{\lambda\} = 0.$$ So, as you deduced,$E\{\lambda\}H \subseteq \mbox{ker}(u-\lambda I)$. ... 1 If$\operatorname{range}(S)$is not closed, then we can find a sequence$(y_n)_n\subset \operatorname{range}(S) $such that$y_n\to y$and$y\notin \operatorname{range}(S)$. We may write$y_n$as$x_n-Kx_n$, where$x_n\in Z$. Claim. The sequence$(x_n)_{n\geqslant 1}$is not bounded. Indeed, if it was, then the sequence$(Kx_n)_{n\geqslant 1}$would ... 0 I suppose that what you are claiming is that$\langle\psi|\rho_i|\psi\rangle\neq 0$for just one$i\in\{1,2\}$. Hence the criterion you are looking for. 1 There's an implicit hypothesis that$T$is self-adjoint (otherwise the desired formula might be false - take antysymetric$T$). Let $$a=\sup_{\|x\|=1}|(Tx,x)|.$$ You can write that $$(Tx,y) = \frac14(T(x+y),(x+y))-\frac14(T(x-y),(x-y)),$$ hence $$|(Tx,y)|\le \frac a4 (\|x+y\|^2+\|x-y\|^2) =\frac a2 (\|x\|^2+\|y\|^2)$$ We also know that ... 3 (1) Define$F \colon c_0 \to \def\K{\mathbf K}\K$by $$F(x) = \sum_{n=0}^\infty x_n^n$$ Then$F$is continuous, as the series converges locally uniform, but$F$is unbounded, as the elements$x^{(n)} = (1, \ldots, 1,0, \ldots) \in \bar B_{c_0}$have $$\def\norm#1{\left\|#1\right\|} \norm{x^{(n)}} = 1,\qquad \def\abs#1{\left|#1\right|}\abs{F(x^{(n)})} = ... 0 What about the following generalisation (it works, but is it useful?) ? Let's deal with polynomials of the form P(A,B,z)=A-Bz first and let's give a modified definition to the "generalised resolvent set" \rho(A,B): \left\{z\in \mathbb{C}\left| \right. \left(A-zB\right)^{-1}\text{,}B\left(A-zB \right)^{-1}\text{exist on a dense (common) ... 1 Given that \|f(x)-f(y)\|=\|x-y\|, define g(x)=f(x)-f(0). Then \|g(x)\|=\|x\|. By the Parallelogram law,$$ \|g(x)+g(-x)\|^{2}+\|g(x)-g(-x)\|^{2}=2\|g(x)\|^{2}+2\|g(-x)\|^{2} $$Hence,$$ \|g(x)+g(-x)\|^{2}+\|x-(-x)\|^{2}=2\|x\|^{2}+2\|-x\|^{2} \\ \|g(x)+g(-x)\|^{2} = 0 \\ \implies ... 2 For$n=1$the structure of the graph of a maximal monotone operator is known. First the domain and range of such an operator are (possibly degenerate) intervals. If the domain is just one value then the graph is a vertical line passing through that value (similarly for the range being degenerate). If the domain is a non-degenerate interval ($(a,b)$or ... 1 You can find the source of maximal monotone mappings in the following and the references therein: Variational Analysis, Rockafellar, R. Tyrrell, Wets, Roger J.-B. http://www.springer.com/gp/book/9783540627722 Maximal Monotone Operators in Banach Spaces, Viorel Barbu http://link.springer.com/chapter/10.1007%2F978-1-4419-5542-5_2 Lectures on Maximal Monotone ... 3 More generally, for any self-adjoint operator$A$in a Hilbert space$H$and any Borel function$f$on$\sigma(A)$, we define$f(A)$by the functional calculus associated to the Spectral theorem. If$E$is the resolution of the identity associated to$A$by$A = \int_{\mathbb R} \lambda \; dE(\lambda)$, then$f(A) = \int_{\mathbb R} f(\lambda)\; ...

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You can of course try to apply the binomial series, $$(1+A)^r=\sum_{k=0}^\infty\binom{r}{k}A^k$$ with $A=\frac d{dx}$ and $r=-2n$. Note that $\binom{-m}{k}=(-1)^k\binom{m+k-1}k=(-1)^k\binom{m+k-1}{m-1}$. You can also transform $V(x)=e^{-x}U(x)$ with $$U'(x)=e^x\left(I+\frac d{dx}\right)V(x),\quad U^{(k)}(x)=e^x\left(I+\frac d{dx}\right)^kV(x).$$

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$S$ is bijective and continuous, so if $S$ is also open, it's a homeomorphism from $Y$ to $X$. Let $f_i\in X$ denotes the function which sends $j\in\mathbb N$ to $1$ if $j=i$ and $0$ if $j\neq i$. Note that $S$ sends the non-convergent sequence $\{\sqrt nf_n\}$ to convergent sequence $\{\frac 1{\sqrt n}f_n\}$, so it cannot be a homeomorphism.

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I found this notation in Simon's paper "Hamiltonians defined as quadratic forms", Commun. Math. Phys. 21 (1971), 192-210. (PDF available on his web page). A footnote at the bottom of the first page reads: $$X + (L^\infty)_\varepsilon = \{ f \mid (\forall \varepsilon) f = x_\varepsilon + g_\varepsilon \text{ with } x_\varepsilon \in X; ... 0 That I - A is both injective and surjective follow from the fact that it is invertible. Indeed, let B:X \to X be any invertible operator; then we have an operator B^{-1} such that BB^{-1} = B^{-1}B = I. \tag{1} To see that (1) implies B is injective, suppose that Bx_1 = Bx_2 \tag{2} for some x_1, x_2 \in X. Then from (1) x_1 = Ix_1 = ... 2 Hint Prove that Id+A+A^2+..+A^n+... defines an operator which is the inverse of Id-A. 1 You know that B(X), the space of bounded linear operators on X is a Banach space. Consider the series$$ \sum_{n=0}^{\infty} A^n $$Since \|A\| < 1, it converges absolutely, and thus converges in B(X) to an operator B. Now check that B(I-A) = (I-A)B = I and so (I-A) must be surjective. 3 I am assuming all vector spaces are over \mathbb{C}. To show that \alpha I - A is not injective for any \alpha \in \mathbb{C}, it is sufficient to find (given \alpha \in \mathbb{C}) an f \in L^2(0,1;\mathbb{C}) such that \alpha f - Af = 0. Set f(x) = e^{\alpha x}. Then |f(x)| \le e^{\max(0,\Re \alpha)} for all x \in (0,1) and hence f ... 1 When you have to differentiate a series, it often pays to look at the series of derivatives, integrate, and justify changing the order of integration and summation. It's a trick that helps you avoid interchanging differentiation and summation, which seems always to be more difficult to justify. For example, you know the derivative should be$$ ...

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Nevermind, I answered it myself using an elementary argument. Thanks anyway! EDIT: Here is the argument, in brief. Consider the real Banach spaces $\widetilde{X}=X\oplus_{\ell_1}X$ and $\widetilde{Y}=Y\oplus_{\ell_1}Y$. It is a simple exercise to show that $T\oplus 0$ and $0\oplus T$ both lie in $\mathcal{FSS}(\widetilde{X},\widetilde{Y})$, and hence so ...

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If $B$ denotes the infinitesimal generator of $T$ then we know that: If $f \in \mathcal{D}(B)$ and $t \geq 0$, then $T(t)f \in \mathcal{D}(B)$ and $$B T(t) f = \frac{d}{dt}T(t)f .$$ (c.f. Ethier and Kurtz page $9$). Here $\mathcal{D}(B)$ is the domain of the operator $B$. So you just need to check the domain is the whole space, and justify the ...

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SOT-limit of a net of projections is a projection. This is not true for WOT-limit (as your example shows).

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One can show that $T$ is a compact operator: define $S_n:\ell^2(\mathbb{C}) \to \ell^2(\mathbb{C})$ by $$(S_nx)_m = \left\{ \begin{matrix} \frac{x_m}{m} & m \leq n \\ 0 & m > n \end{matrix} \right.$$ Note that the $S_n$ are finite-rank and that $$(T-S_nx)_m = \left\{ \begin{matrix} 0 & m \leq n \\ \frac{x_m}{m} & m > n \end{matrix} ... 0 For \lambda \notin \{0\} \cup \sigma_p(T), you can explicitly write down (\lambda I - T)^{-1}. 2 Without loss of generality, assume A = 0. Let \epsilon > 0. Since \langle A_n\xi,\eta\rangle \to 0 uniformly for \|\eta\| = 1, there exists an index N = N(\epsilon) such that if n\ge N, |\langle A_n\xi,\eta\rangle| < \epsilon/2 for all \eta \in H with \|\eta\| = 1. Thus, if n \ge N,$$\|A_n\xi\| = \sup_{\|\eta\| = 1}|\langle ...

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Hint: Let $e_n$ $(n\in \mathbb Z)$ be the standard basis in $\ell^2(\mathbb Z)$. For $\lambda \in {\mathbb T}$, let $$\xi_n=\frac{1}{2n+1}\bigl( \lambda^{-n} e_{-n}+\lambda^{-n+1} e_{-n+1}+\cdots+e_0+\cdots+\lambda^{n} e_n\bigr)\qquad (n\in \mathbb Z).$$ Note that $(\xi_n)$ is a sequnce of vectors with norm $1$. Consider $$u\xi_n-\lambda \xi_n\quad ... 1 The operator Af=(\frac{d}{dx}+a)f is, in a classical sense, similar to the differentiation operator because$$ Af = e^{-\int a(x)dx}\frac{d}{dx}e^{\int a(x)dx}f = M^{-1}\frac{d}{dx}Mf. $$Because of this, any function of the operator A is a corresponding function of \frac{d}{dx}. In particular,$$ e^{tA}f = e^{-\int ...

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\begin{align} 2\Phi(x+h)-2\Phi(x) & = (T(x+h),x+h)-(Tx,x) \\ & = (Tx,x)+(Tx,h)+(Th,x)+(Th,h)-(Tx,x) \\ & = (Tx,h)+(Th,x)+(Th,h) \\ & = (Tx,h)+(h,Tx)+(Th,h) \end{align} I assume you must be working on a real space, and working with an operator $T$ defined on the whole space, which makes it ...

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Your calculation for $A\mapsto A^2$ is essentially right. It should be $L(X)H=XH+HX$ and $o(H^2)\neq o(H)$, however, $H^2\in o(H)$ and that suffices. The inverse map is only defined on the open subset $GL_2(\mathbb R)$ of invertible matrices. To calculate the differential, you can use Neumann series: \begin{align}(X+H)^{-1}&=(X(1+X^{-1}H))^{-1}\\ ...

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This operator is self-adjoint, i.e. $A^*=A$. Indeed, consider the "minimal" operator, $$A_0u=u'',\ D(A_0) = \{u \in H^2([0, 1]) : u(0) = u(1) = u'(0) = u'(1) = 0\}.$$ Then $A_0^*u=u''$ and $D(A_0^*)=H^2([0, 1])$. Since $A_0\subset A\subset A_0^*$, we also have $A_0=A_0^{**}\subset A^*\subset A_0^*$, i.e. $A^*$ is a restriction of $A_0^*$, in particular ...

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Everything starts from knowing what a stochastic process is. I feel usefull to intuitively grab the meaning of a stochastic process to start thinking about the deterministic description of a fired cannon ball. You can calculate the trajectory of the ball. In every moment you knows where the ball is. From laws of physics you can build a function of time that ...

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