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0

You shouldn't open parantheses for Leibnitz rule, use the boundary conditions with integration by parts instead. Take $y$ that satisfies the same boundary data: $$y^{(j)}(a)=y^{(j)}(b)=0,\, j=0,1,\dots,n-1,$$then $$\int_a^b (px^{(j)})^{(j)}y\ dt =(px^{(j)})^{(j-1)}y\big|_{t=a}^{t=b} - \int_a^b (px^{(j)})^{(j-1)}y^{(1)}\ dt $$ $$=\dots= (-1)^j\int_a^b ...


1

The general fact, that a continuous map between Hausdorff topological spaces has closed graph implies that $G(T)$ is closed in $(X,\sigma(X,X^*))\times (Y,\sigma(Y,Y^*) = (X\times Y, \sigma(X\times Y,(X\times Y)^*))$. You do not need Mazur's theorem to conclude that $G(T)$ is strongly closed: The strong (or norm-) topology on $X\times Y$ is finer than the ...


0

The spectral integral reduces to $$ (e^{-itA}f)(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-its}e^{isx}\int_{-\infty}^{\infty}e^{-isv}f(v)\,dv \\ = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{is(x-t)}\int_{-\infty}^{\infty}e^{-isv}f(v)\,dv = f(x-t). $$ This is translation semigroup. You can verify: $$ \frac{d}{dt}f(x-t)|_{t=0}=-f'(x) ...


0

It depends a lot on what you call "explicit". In the case of the exponential, you have $$ e^{-itH}=\sum_{k=0}^\infty\frac{(-i)^kt^k}{k!}\,H^k=\sum_{k=0}^\infty\frac{(-1)^kt^k}{k!}\,\partial_x^k. $$


0

The exact same idea as in the answer to your other question works. That is, now take a Hamel basis of $B(H)$ that extends a Hamel basis of $\mathbb RI$, and you can still get a $\mathbb Q$-linear map (so additive) such that $\mathbb RI\subsetneq \Phi(\mathbb RI)$.


0

Let $A=B=\mathbb C$. Fix a Hamel basis $X$ of $\mathbb R$ as a vector space over $\mathbb Q$. Then $X\cup Xi\ $ is a Hamel basis of $\mathbb C$ over $\mathbb Q$. Since $X$ and $X\cup\{i\}$ have the same cardinality, there exists a bijection $\gamma:X\to X\cup\{i\}$ with $\gamma(1)=i$. Let $\eta:iX\to iX\setminus\{i\}$ be a bijection. These bijections induce ...


0

Hint: $T-\lambda I$ is invertible with inverse $A \in \mathcal{B}(H)$ iff $$ I = (T-\lambda I)A = A(T-\lambda I) $$ The above holds iff $$ I = A^{\star}(T^{\star}-\overline{\lambda}I) = (T^{\star}-\overline{\lambda}I)A^{\star}. $$


1

For the proof of the author's claim, note that $c \geq 1 = c ^{-1} c$ if and only if $1 \geq c^{-1}$. This example doesn't quite work since it doesn't meet the hypotheses set out by the author. Let's consider the C*-algebra $A := C([2,10])$ and the same function $f$, which you defined above. There are various notions of invertibility in the C*-algebra ...


0

With some manipulation with integration by parts I found the solution \begin{eqnarray} \int^{+\infty}_{-\infty} (L\psi) \psi_1 dx &=& \int^{+\infty}_{-\infty} \left(\left[\frac{d^2}{dx^2} + f(x)\right]\psi \right)\psi_1 dx \\ &=& \frac{d\psi}{dx}\psi_1\Bigg|^{+\infty}_{-\infty} -\int^{+\infty}_{-\infty}\frac{d\psi}{dx}\frac{d\psi_1}{dx} dx + ...


2

First thing: self-adjoint in which space? Second thing: try to apply the Fredholm alternative - if you have an operator $A$ satisfying the Fredholm alternative, then $$b\in im(A)\iff b \in (\ker A^*)^\bot.$$


0

The product of positive operators is not even selfadjoint in general. For instance, $$ \begin{bmatrix}1&0\\0&0\end{bmatrix}\,\begin{bmatrix}1&1\\1&1\end{bmatrix}=\begin{bmatrix}1&1\\0&0\end{bmatrix}, $$ while $$ ...


0

The spectrum is invariant under unitary transformations, thus it is rotational invariant: $$\sigma(W(f\neq0))=\sigma(W(g)W(f\neq0)W(g)^*)=e^{-i\sigma(g,f\neq0)}\sigma(W(f\neq0))$$ Note that this excludes the special case: $$\sigma(W(0))=\sigma(1)=\{1\}$$


1

No. Counterexample: $$ A = \pmatrix{ 1 & 2\cr 2 & 5\cr},\ B = \pmatrix{1 & -1\cr -1 & 2\cr}\\ AB = \pmatrix{-1 & 3\cr -3 & 8\cr}\\ \left\langle AB\pmatrix{1\\0},\pmatrix{1\\0} \right\rangle = \pmatrix{1&0} A B \pmatrix{1\cr 0\cr} = -1 $$ (example taken from here).


0

I guess that "positive" here means that the quadratic form is positive, right? In this case, I guess that the proposition is false unless $[A, B]=0$. You can find counterexamples even in finite dimensional spaces, so the compactness assumption does not help you much.


0

Goal The pure-point-space agrees with the eigenspace: $$\mathcal{H}_\text{pp}(E)=\overline{\langle\mathcal{E}(T)\rangle}$$ with pure-point-space and eigenspace being: $$\mathcal{H}_\text{pp}(E)=\{\varphi:\exists\#\Lambda_0\leq\aleph_0:\nu_\varphi(\Lambda_0)=\nu_\varphi(\Omega)\}$$ ...


0

Goal The spectral subspaces decompose the Hilbert space: $$\mathcal{H}=\mathcal{H}_\text{ac}\oplus\mathcal{H}_\text{sc}\oplus\mathcal{H}_\text{pp}$$ with the spectral subspaces being: $$\mathcal{H}_\alpha=\{\varphi:\nu_{\varphi,\alpha}=\nu_\varphi\}$$ Preparation Check the equivalences: $$\nu_\varphi(A)=0\iff E(A)\varphi=0$$ ...


1

If $\phi$ is an unbounded linear functional, take $u$ so that $\phi(u) = 1$, and define $P x = \phi(x) u$. You won't get an explicit example though, because the existence of an unbounded linear operator on an infinite-dimensional Hilbert space requires some form of the Axiom of Choice.


1

Self-adjointness is used to show that $\langle Ax,x\rangle$ is always real. Recall that $\langle u,v\rangle = \overline{\langle v,u\rangle} $ and use the definition of self-adjointness: $$ \langle Ax,x\rangle=\langle x,Ax\rangle = \overline{\langle Ax,x\rangle} $$ The boundedness of $A$ implies that the set $\{\langle Ax,x\rangle:\|x\|=1\}$ is bounded. ...


0

The authors may be using some notion of monotonicity different from what you expect (they never defined it). The inequality $$\langle (A + G)(u_{1})-(A+G)(u_{2}),u_{1}-u_{2}\rangle \geq 0\quad \forall u_{1},u_{2} \in C^{\infty}_{c}(\Omega) \tag{1}$$ does not follow from their assumptions. Indeed, since both $A$ and $G$ could be scaled independently from one ...


0

Here's a proof outline that directly uses the uniform boundedness principle rather than the closed graph theorem. This is probably more or less equivalent to Mizar's idea. Consider the family of linear functionals $$ T_x(y) = \frac{1}{\Vert x \Vert}(Ax,y), $$ indexed by $x\in H$. Define $S_x$ analogously, except with the $S$ on the other side of the inner ...


7

Show that $T$ and $S$ are closed. Then apply the closed graph theorem to conclude that $T$ and $S$ are continuous. To show that $T$ is closed, suppose $\{ x_{n} \}$ converges to $x$ and suppose $\{ Tx_{n} \}$ converges to $y$, and show that $Tx=y$; to do this observe that $$ (Tx,z) = (x,Sz) = \lim_{n}(x_{n},Sz) = \lim_{n}(Tx_{n},z)=(y,z),\;\;\; z \in ...


2

If you know the uniform boundedness principle (Banach-Steinhaus theorem), you could first show that the set $C=\{Tx:\|x\|\le 1\}$ is weakly bounded. Then, if you apply Banach-Steinhaus theorem properly (in a subtle way), you get that $C$ is bounded (which tells you that $T$ is continuous).


0

A convenient to show that an operator is compact is to show that it is the limit (for the operator norm) of operators of finite rank. Define the operator $T_n$ by $T_n(\phi_k) = \frac 1k \phi_{k+1}$ if $k \leq n$ and $T_n(\phi_k) = 0$ for $k > n$. Let $x = \sum x_k \phi_k$ be an element in $H$. We want to evaluate $\|(T-T_n)x\|$. This is given by $$ ...


0

For $T \ge 0$, and orthonormal basis $\{ e_{n}\}$, your trace is $$ \mbox{tr}(T)=\sum_{n} \|T^{1/2}e_{n}\|^{2}. $$ Assume your trace is finite. If $\{ f_{m} \}$ is another orthonormal basis, then Parseval's equality gives $$ \begin{align} \mbox{tr}(T) & =\sum_{n} \sum_{m}|(T^{1/2}e_{n},f_{m})|^{2} \\ & = ...


4

For your first question, I think you are misunderstanding what the theorems say. When you restrict your $\sigma$-weakly continuous functional to the unit ball, you don't get a wot functional on the whole space: so 4.6.4 does not apply. For your second question, here is an example: fix an orthonormal basis $\{e_n\}$ and let $$ ...


1

Another way to calculate that commutator using $[x,p]=i\hbar$ and introducing intermediate commutators: $$[x,p^2]=xp^2-px^2=xp^2-ppx+pxp-pxp=xp^2-pxp+p(xp-px)=xp^2-pxp+p[x,p]=xp^2 -pxp+i\hbar p= xp^2-pxp+xpp-xpp+i\hbar p=xp^2-xp^2+(xp-px)p+i\hbar p=2i\hbar$$


1

I'll use $X$, $U$ instead of $x$, $u$ because I want to use vectors, and it just looks wrong otherwise. I'm assuming you have some sort of functional calculus in order to form $|X|=(X^{\star}X)^{1/2}$. And that means you can verify that $|X|(|X|+\epsilon I)^{-1}$ is uniformly bounded in norm by $1$ for $\epsilon > 0$. I'll show you that $$ ...


3

niki, he keeps the assumption that that $\mathfrak{m}$ is $\sigma$-weakly closed. So the third line should read: If, in addition, $\mathfrak{m}$ is a two-sided ideal, then.... Of course, when you deal with norm-closed ideals it is very easy to come up with a counter-example: take the compact operators inside $B(H)$. This is also an illustration to ...


0

Hint: We want functions $x(t)$ such that $Tx(t) = ax(t)$ for some scalar $a$. But, $Tx(t) = \displaystyle\int_{-1}^{1}(1-3tz)x(z)\,dz = \left[\int_{-1}^{1}x(z)\,dz\right]-\left[\int_{-1}^{1}3zx(z)\,dz\right]t$. Thus, $Tx(t)$ is always a polynomial in $t$ with degree at most $1$. So if $x(t)$ is an eigenfunction, then $x(t)$ must also be a polynomial in ...


1

You can do it when the function $\boldsymbol B$, and the function $\dfrac{\partial \boldsymbol B}{\partial t}$ are continuous. Why? Here is the answer.


2

As for (b), let $m = \liminf_n x_n$ and $M= \limsup_n x_n$. For all $\varepsilon >0$ you have eventually $x_n - m + \varepsilon \geq 0$. Then, shifting enough the sequence, you get that $$l(x +(-m + \varepsilon)e) \geq 0$$ i.e. $$l(x) + (-m + \varepsilon) l(e) = l(x) + -m + \varepsilon \geq 0$$ by arbitrarity of $\varepsilon$ you get $l(x) \geq m$. The ...


1

For example: ${\rm dom}(T)$ is the set of polynomials on $(-1,1)$ with sup norm, and $T(p) = p'(0)$ is the derivative.


1

No. For example, unbounded linear functionals do exist.


1

Yes, the two closures are equal. Let $x\in \bar A^{wot}$. We have $\bar A^{wot}=\bar A^{sot}$, so there is a net $\{x_j\}\subset A$ such that $x_j\to x$ sot. By Kaplansky, there is a bounded net $\{x_j'\}\subset A$ with $ x_j'\to x$ sot, and so wot. That is, $x$ belongs to the wot closure of a ball in $A$. But on balls, the wot and ultraweak topologies ...


0

Here is an example of a net that is weakly convergent, but not ultraweakly convergent. I will use the fact that $X_j\to0$ ultraweakly precisely when $\text{Tr}(AX_j)\to0$ for all trace-class operators $A$. Fix $A$ to be an injective trace-class operator, so that $AP\ne0$ for all projections $P$. For example, you could take $A=\sum_k\frac1{k^2}\,\langle\, ...


2

Typically, one has a locally compact or compact topological Hausdorff space $\Omega$, and the projection measure is defined on the Borel subsets of $\Omega$. In the context of the Spectral Theorem, $\Omega$ is the closed subset of $\mathbb{C}$ which is the spectrum of a normal operator $N$, but the abstraction to a more general space can be useful, ...


0

We assume the linear operator $\hat{D}$ is defined on the Hilbert space $H = \mathcal{L}^2(\mathbb{R})$ equipped with the usual inner product $(f,g)=\int_{\mathbb{R}} f(x)g(x) \, \textrm{d}x$ since it is not specified. The linearity just follows from definition. Indeed, for $f, g \in H$, $\alpha, \beta \in \mathbb{C}$, $$(\hat{D}(\alpha f + \beta g))(x) = ...


0

You haven't specified the domain and range of this function. If you mean the operator over the $C^\infty$ functions, note that it will not be invertible. Taking for example $f(x) = \sin(x)$, we have $$ \left(I + \frac{\partial^2}{\partial x^2}\right) f(x) = 0 $$ so your function is certainly not injective. Suppose we are considering the operator over the ...


2

Yiorgos S. Smyrlis and Davide Giraudo have already explained that the result does not hold whenever the operator is indefinite. However, it is fruitful to think about how the concept could be extended to the indefinite case. One might consider the following generalization that accounts for the possiblity of negative eigenvalues. Replace $$\inf_x ...


3

The answer is NO. Simpest possible example $H=\mathbb R^2$ and $$ T=\left( \begin{matrix} 0& 1 \\ -1 & 0\end{matrix} \right) $$ Then, for $x=(x_1,x_2)$, we have that $Tx=(x_2,-x_1)$, and hence $$ \|Tx\|=\|x\|, $$ while $$ \langle Tx,x\rangle=0. $$


0

The question reduces to the following: if $S\colon H\to H$ has a closed range, then the following are equivalent: $\mathrm{Range}(S)=H$; $\ker(S^*)=\{0\}$. Assume that 1. is satisfied. If $S^*x=0$, then $\langle x,Sy\rangle=0$ for any $y\in H$ and by 1., there is some $y$ for which $Sy=x$ hence $\langle x,x\rangle =0$ and 2. is satisfied. Assume that ...


2

In your remarks, you asked about two specific issues. Question 1: Why is the closure of $\mathcal{C}^{\infty}(S)$ under the graph norm of the Laplacian the same as $H^{2}(S)$, even though the first derivative terms are not present in this norm $\|f\|+\|\Delta f\|$? The Laplacian in spherical coordinates is $$ \frac{1}{r^{2}}\frac{\partial}{\partial ...


2

Let $\{E_{kj}\}$ be the canonical matrix units and $P_n=\sum_1^nE_{kk}$. Take $S$ the unilateral shift, $S=\sum_kE_{k+1,k}$, and $T_1=S^*$, $T_2=S$. Then $$ P_nT_1T_2P_n=P_nS^*SP_n=P_n, $$ while $$ P_nS^*P_nSP_n=P_{n-1}. $$ Thus $$ \|P_nT_1T_2P_n-P_nT_1P_nT_2P_n\|=\|P_n-P_{n-1}\|=1 $$ for all $n$.


2

The spectrum of $A$ is $\{0\}$ because there has to be something in the spectrum, and resolvent series converges everywhere except at $\lambda=0$: $$ (\lambda I-A)^{-1}=\sum_{n=0}^{\infty} \frac{1}{\lambda^{n+1}}A^{n} $$


4

Or you can proceed directly by algebraic methods. The spectrum of $A$ is $\{ \lambda \in \mathbb{C} : A- \lambda I$ is not invertible $\}$. If $A$ is nilpotent and $\lambda \neq 0$, then there is a positive integer $n$ such that $A^{n} - \lambda^{n}I = -\lambda^{n}I,$ which is an invertible operator. Since $A^{n} - \lambda^{n} I = (A-\lambda I) (A^{n-1} + ...


4

If $A$ is a bounded linear operator on a Banach space, then the spectral radius $r(A) = \sup\{|\lambda|\ |\lambda\in\sigma(A)\}$ satisfies $$ r(A) = \lim_{n\to\infty} \|A^n\|^{1/n}.$$ So the spectrum of a nilpotent operator $A$ is not only countable and finite, it contains only zero. (c.f. these notes, proposition 9.5, pp 220-1 --- the theorem is for ...


0

In case $G$ is self-adjoint $P$, and hence $Q=1-P$, reduces $G$ so $${G_0}=QGQ+P.$$ Then ${G_0}f=0$ implies that $QGQf=0$ and $Pf=0$ separately. Since $QGQ$ has a vanishing kernel on the subspace projected upon by $Q$, also $Qf=0$ so $f=0$ and hence $0$ is in the resolvent set of $G_0$.


1

Your proof can be generalised easily to arbitrary unital Banach algebras. Let $A$ be a unital Banach algebra and consider the left regular representation of $A$ on itself, that is, the map $\lambda_a x = ax\;(a,x\in A)$. Since $A$ is unital, $\lambda\colon A\to B(A)$ is isometric. Now, let $(a_n)_{n=1}^\infty$ be a sequence of invertible elements in $A$ ...


2

Let $A : \mathcal{D}(A) \subset \mathcal{H}\rightarrow \mathcal{H}$ be densely defined and selfadjoint. Choose any $x \notin\mathcal{D}(A)$ and let $S$ be the span of the single vector $x$. Then $\mathcal{D}(A)\cap S=\{0\}$, which means that the domain of your proposed operator is $\{0\}\oplus S^{\perp}$ which is not dense in $\mathcal{H}$.



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