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0

If $\alpha$ is a function of bounded variation on $[a,b]$, and if $f$ is a continuous function, then the classical definition of the Riemann-Stieljes integral of $f$ with respect to $\alpha$ is $$ \int_{a}^{b}f(t)\,d\alpha(t) = \lim_{\|\mathscr{P}\|\rightarrow 0} \sum_{n=1}^{k}f(t_{k}^{\star})(\alpha(t_{k})-\alpha(t_{k-1}), $$ where ...


2

The Dirac distribution is not the Dirac measure (point measure), but induced by the Dirac measure. Every Radon measure $\mu$ on $\mathbf{R}$ induces a distribution by $$\phi\mapsto \int_{\mathbf{R}}\phi \ d\mu.$$ Yes we can define the distribution $\delta_0\in\mathscr{D}'(\mathbf{R})$ to be the one induced by the Dirac measure, or simply by ...


0

To get this off the unanswered list (all credits go to Bryan): $f$ is not differentiable at $0$ since $$\lim_{h\downarrow 0}\frac{1}{h}(f(h)-f(0))\neq\lim_{h\uparrow 0}\frac{1}{h}(f(h)-f(0)).$$ You cannot show it to be rapidly decreasing if it is not differentiable (rapidly decreasing usually means to be an element of Schwartz space, but perhaps you are ...


1

Since the definition of weakly nuclear requires the codomain to be a von Neumann algebra, I will assume $\phi:A\to B^{**}$. Let $A=B(H)$, $B=K(H)$. Then $B^{**}=B(H)$. Then you can take $\phi$ to be the identity map, which is not nuclear (because $B(H)$ is not nuclear) but is weakly nuclear (by 2.1.4 in Brown-Ozawa).


4

The matrix $$ A= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$ is a counter-example: $$ \|A\|_2^2 = \|A^TA\|_2^2 = \|A^T\|_2\|A\|_2=1, $$ but $$ \|A^2\|_2=0 $$


1

The answers follow from the standard fact that $B$ has the metric $d(x,y) = \sum_{k=1}^\infty c_kp_k(x,y)/[1+p_k(x,y)]$ where $c_k$ is any sequence of scalars tending to $0$. E.g. $c_k=1/2^k$.


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 David 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 ...


3

Let us try to estimate $|B|$ from below. Every space $B_{n-1}^*$ ($n\geqslant 1)$ is of the form $C(K_n)$ for some compact Hausdorff space. For example, $K_1 = \beta\mathbb{N}$ and $|K_1| = \beth_1$. In particular, by the Riesz–Markov–Kakutani representation theorem each space $B_n$ is isometric to the space $M(K_n)$ of Radon measures on $K_n$. Moreover, we ...


2

Something more has to be said about the domain of $T$, even if you assume $T$ is closed. $T$ is unique if you assume the following: $$ \begin{align} 1.\;\; & T \mbox{ is linear with } Te_{n}=\lambda_{n}e_{n} \mbox{ for all } n; \\ 2.\;\; & T \mbox{ is closed};\\ 3.\;\; & \mathcal{D}(T)=\{ v \in H : \sum_{n}|\lambda_{n}(v,e_{n})|^{2} < ...


0

Because the set is equal to $\frac{1}{r} \cdot(C\cap (r\cdot B_n ))$ and hence it is compact.


0

Divergence operator is formally the adjoint of the gradient operator, in the sense that $$\int \varphi \operatorname{div}\vec F = - \int \nabla \varphi \cdot \vec F\tag1$$ when everything is smooth and $\varphi$ is compactly supported. Let's keep $\varphi$ being $ C^\infty$ and compactly supported, but allow $\vec F$ to be a vector-valued ...


0

@Ester, this isn't the best answer, but it will be enough for you to get the idea: consider $X$ to be an infinite-dimensional topological vector space. It is known from a theorem of Weil that it cannot be locally-compact. Can we have functions with compact support defined on it, then? Let $f \neq 0$ be such a function and $x$ such that $f(x) \neq 0$. Then ...


2

The point is that functions vanishing at infinity is a concept that is really only sensible on locally compact spaces. See http://en.wikipedia.org/wiki/Vanish_at_infinity and http://en.wikipedia.org/wiki/Locally_compact#The_point_at_infinity In particular, it turns out that the isomorphism class of the $C^*$-algebra $C_0(X)$ remebers the homeomorphism ...


2

You can use the criterion that reflexivity is equivalent to the unit ball being compact in the weak topology. Then the intersection of the unit ball with the closed subspace (which is still closed in the weak topology) is also compact, hence $E\subseteq X$ is reflexive. Edit: If you want to do it directly, you can use the Hahn Banach theorem to do it ...


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 ...


1

No, it's not nearly that simple. The constant $C$ quantifies the connectivity of the manifold. It can be imagined as the severity of traffic jams that occur when all inhabitants of the manifold decide to drive to a random place at the same time. For example, let $M$ be two unit spheres $S^2$ joined by a thin cylinder of radius $r\ll 1$ and length $1$. There ...


2

Note that ${\rm Ker}\,U={\rm Ker}\, U^*U$. Thus $({\rm Ker}\,U)^\bot={\rm Im}\,(U^*U)$ because $U^*U$ is an orthogonal projection. So, for $x\in({\rm Ker}\,U)^\bot$ we have $U^*Ux=x$ that is $\langle x,U^*Ux\rangle=\langle x,x\rangle$ or equivalently $\Vert Ux\Vert=\Vert x\Vert$.


0

Hint. Try contradiction. If $E$ is not bounded, there is some $U$ such that $E \not\subseteq nU$ for all $n \in \mathbb N$. Choose $a_n \in E \setminus nU$. Is $A = \{a_n \mid n \in \mathbb N\}$ bounded?


1

Adding to martini's answer: From $$ \|f\|_p \le \|f\|_{p_0}^{1-\theta}\|f\|_{p_1}^\theta $$ one finds using Young's inequality in the form $$ ab \le \theta a^{\frac1\theta} + (1-\theta)b^{\frac1{1-\theta}} $$ the estimate $$ \|f\|_p \le \|f\|_{p_0}^{1-\theta}\|f\|_{p_1}^\theta \le (1-\theta)\|f\|_{p_0} + \theta\|f\|_{p_1}. $$ Using the definition of $p$ it ...


1

There is some convexity, don't know if this helps you, but: Let $p_0, p_1 \in [1,\infty]$, $\theta \in [0,1]$, $\frac 1p = \frac{1-\theta}{p_0} + \frac{\theta}{p_1}$. Then for $f \in L^{p_0}\cap L^{p_1}$, by Hölder \begin{align*}\def\norm#1#2{\left\|#1\right\|_{#2}}\def\abs#1{\left|#1\right|} \norm fp &= \norm{\abs{f}^{1-\theta}\abs{f}^\theta}p \\ ...


2

Suggestion: start with a symmetric operator $C$ that is not essentially self adjoint, and $A$ with ${\mathcal D}(A) \subset {\mathcal D}(C)$ that is self adjoint. Take $B = C - A$.


2

The question is closely related to Dual space of the space of finite measures, except that one asked "what" is the dual space, and you have a candidate for it. As the comment by t.b. (supported by a quote from Dunford and Schwartz) indicates, there is no satisfactory description of the dual, which sort of implies the negative answer to your question. But ...


0

The definition of a Besov space depends on a number of parameters. Some Besov spaces can boil down to certain Sobolev spaces and Sobolev spaces on nice enough domains tend to be isomorphic to $L_p$. In general, however, Besov spaces need not embed to $L_p$. A standard reference for completeness of Besov spaces is Chapter 3 of: J. Peetre, New thoughts on ...


1

Following the first comment above: the answer is yes. The definitions are equivalent, even in the case $\mathbb{L} = \mathbb{R}$. This is because there is also a real version of the polarization identity: $$(x,y) = \frac{1}{4}(||x+y||^2 - ||x-y||^2).$$


0

Zeidler's Nonlinear Functional Analysis... book, part II/A is good for this kind of information. I would also recommend the book about Navier Stokes by Boyer and Fabrie for a modern typesetting. Which also reminds that Temam's book (freely available on his website by Googling "Temam") also contains some basics of these kind of spaces.


1

It only makes sense to talk about the closure of a subset of a topological space; you need to specify which topological space $A$ is a subset of. Just saying the closure of $A$ without any further reference means (at least to me) the closure of $A$ as a subset of $A$, whereas you are referring to the closure of $A$ as a subset of $B$. To see why this ...


1

The example by David Mitra is perfectly valid, but it may be a little hard to see exactly what happens to $\sin nx$ as $n\to\infty$. Here's a slightly modified example: Rademacher functions $$r_n(x) = \operatorname{sign}\sin (2^n \pi x), \quad x\in [0,1]$$ These converge to $0$ weakly in $L^2$, but $|r_n|=1$ a.e. To prove weak convergence, first consider ...


2

There is a more general fact: A C*-algebra $A$ is reflexive if and only if it is finite-dimensional. In the infinite-dimensional case, there is a normal element $x\in A$ with infinite spectrum (actually, one can find a positive element with infinite spectrum). Therefore, by the spectral theorem, $C^*(x)$ is an infinite-dimensional commutative C*-algebra, ...


0

No. Consider $H(x) = x^3$ and $$ f_n(x) = \begin{cases} 1 & \frac in \le x < \frac{i+1/2}n \quad i = 0,\ldots, (n-1)\\ 0 & \text{otherwise} \end{cases} $$ Then $H(f_n) = f_n$, $f_n \rightharpoonup \frac 12$, $H(f_n) = f_n \rightharpoonup \frac 12 \ne H(\frac 12) = \frac 18$. Note this trick works for any $H$ which has $H(\frac 12) \ne \frac{H(0) ...


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 ...


3

Clearly this is not answer you requested, but may be this will be helpful for someone else. We may regard $c_0$ as a closed subspace of $\mathcal{B}(H)$. Indeed, take any orthnormal basis of vectors $(e_i)_{i\in I}\subset H$ and for each $s\in c_0(I)$ consider bounded linear operator $T_s:H\to H$ well defined by $T(e_i)=s_ie_i$. Consider map ...


1

In the weak$^*$-topology of $X^*$, you can linearly and continuously inject $X$ into $X^{**}$. Rudin considers this injection an identification, which eases notation. In this relation, see page 95 for a more thorough discussion of the special case of Banach spaces.


2

Hint: For the step $(*)$ use $$\left|\int_0^sf(x,t)dx - \lim_{L\to\infty}\int_0^L f(x,t)dx\right| = \left|\int_s^\infty e^{-xt}{\sin(x)\over x} dx\right| < \int_s^\infty e^{-xt}{dx \over x} < \frac{e^{-st}}{st}$$ instead. Now take limsup.


6

Not only is $f$ constant, that constant is either $0$ or $1$. $$ \begin{align} \int_a^b\left[f(x)^2-f(x)\right]^2\,\mathrm{d}x &=\int_a^b\left[f(x)^4-2f(x)^3+f(x)^2\right]\,\mathrm{d}x\\ &=0 \end{align} $$ Thus, $(f(x)-1)f(x)=0$ for almost all $x\in[a,b]$. Since $f$ is continuous, we have either $f(x)=0$ for $x\in[a,b]$ or $f(x)=1$ for $x\in[a,b]$.


8

By the Cauchy Schwarz Inequality, for any integrable function $f(x)$: $\displaystyle\left(\int_a^b f(x) \cdot f(x)^2\,dx\right)^2 \le \left(\int_a^b f(x)^2\,dx\right) \left(\int_a^b (f(x)^2)^2\,dx\right)$ $\displaystyle\left(\int_a^b f(x)^3\,dx\right)^2 \le \left(\int_a^b f(x)^2\,dx\right) \left(\int_a^b f(x)^4\,dx\right)$ But by the given conditions, we ...


2

One can forget $p(s)$, $r(s)$ and the rest and simply try to show that, for every $a\lt b$, $$\|u\|_a\leqslant\|u\|_b.$$ To wit, considering $v=|u|^a$ and $p=b/a\gt1$, note that Hölder inequality yields $$ \int |u|^a=\int v\leqslant\left(\int v^p\right)^{1/p}=\left(\int |u|^b\right)^{a/b}, $$ that is, $$ \left(\int |u|^a\right)^{1/a}\leqslant\left(\int ...


1

The book does not say that $(a)\iff (b)\iff (c)$ because of $$\overline{G+L}=(G^{\perp}\cap L^{\perp})^{\perp} \tag{19}$$ It says that $(a)\iff (c)$ because of (19). The implication $(b)\implies (a)$ is proved later, and takes about a page. $(a)\implies (c)$. If (a) holds, then $G+L=\overline{G+L}=(G^{\perp}\cap L^{\perp})^{\perp}$, so (c) holds. ...


5

No. Nonlinear transformations and weak convergence go together like drinking and driving. For example, let $r_k$ be the $k$th Rademacher function on $[0,1]$, that is $r_k = \operatorname{sign}\sin ( 2^k \pi x) $. Then $2^p r_k \rightharpoonup 2^{p-1}\mathbf {1}$ in $L^1$, where $\mathbf{1}$ is the constant function equal to $1$. On the other hand, ...


0

Both assumptions imply convergence in the sense of distributions: that is, for every smooth compactly supported function $\varphi$ we have $$ \int \varphi u_n\to \int \varphi u,\qquad \int \varphi u_n\to \int \varphi v \tag1$$ So, $$\int \varphi u = \int \varphi v\tag2$$ for every such $\varphi$. This means exactly that $u=v$ in the sense of distributions. ...


1

Let's assume that the functions in $PC[a,b]$ and in $S[a,b]$ are normalized so that $$ f(x+0)=f(x),\;\;\; a \le x < b. $$ All of the functions are required to be continuous at $a$ using this normalization. But these funtions may be discontinuous at $b$. If $y \in (a,b]$ is a point of discontinuity of $f \in PC[a,b]$, then $$ s(x) = ...


0

So i am going to answer my own question; I hope it will be helpful for others . Basically being $\psi$ a distribution it is defined as acting on a test function through a scalar product or analogously integrated over its spatial variable $<\psi|\phi>=\int dr \psi\phi$. Moreover one has to consider $\psi$ and $\psi^*$ independend one from the other ...


0

Instead of $\mu <x, y> $ I have used $ <x, y>.$ Let $ <\mathcal{X}, .>$ be semi inner product. Let $x $, $y $ be fixed vectors in $\mathcal{X}$ and $\gamma$ be scalar. Consider $$ <x - \gamma y, x - \gamma y> = <x, x> - \gamma<y, x> - \bar\gamma<x, y> + |\gamma|^2<y,y> $$ Put $<y,x> = b ...


1

No continuity is needed. Only linearity, positivity, and traciality. We can write $I=P+Q$ for two infinite projections, both equivalent to $I$. Explicitly, for matrix units $E_{kj} $ we can take $$ P=\sum E_{2n-1,2n-1},\ \ \ Q=\sum E_{2n,2n},\ \ \ V=\sum E_{2n,2n-1}, \ \ \ W=\sum E_{2n-1,n}. $$ Then $V^*V=P$, $VV^*=Q$, $W^*W=I$, $WW^*=P$. So $$ ...


0

NOTE: There is no such thing as a boundary condition $f(0)=0$ or $f(\pi)=0$ on $L^{2}[0,\pi]$. Functions in $L^{2}[0,\pi]$ are equivalence classes of functions which are equal to each other a.e.. That part should be tossed out of your question. What is true is that $$\left\{ \sqrt{\frac{2}{\pi}}\sin(nx)\right\}_{n=1}^{\infty}$$ is a complete orthonormal ...


0

Try the following, if you want to avoid any Fourier series methods: first throw in the function $1$ into your set $\{\sin{kx}\}$. The resulting linear span is a subalgebra of $C(T,\mathbb{R})$, where $T$ is the one-dimensional torus. Then apply Stone-Weierstrass to obtain density in the $C$-norm. Now use the fact that $C(T,\mathbb{R})$ is dense in $L^2(T, ...


1

Hints: Let's use the max norm: $|| x||=\max |x_i|$. The $(i,j)$ entry of the Jacobian: $Jf(x)_{ij} =-\dfrac{d_{ij}}{(1+D_{i*}x)^2}$. Now $\forall y$ such that $||y||=1$, we have $|| Jf(x) \cdot y ||\le \max_{i,j} d_{ij}<1$. Why? The are gaps here for you to fill. You need to use the fact $x\in [0,1]^n$. So the norm of the Jacobian ...


1

Your interpretation of $(x^\ast \otimes y)(x) = x^\ast(x)y$ is the correct one. Ryan explains several different interpretations of tensors in section 1.3 of his book, among which the one intended here. With this interpretation, the algebraic tensor product $X^\ast \otimes Y$ is simply the space of finite rank operators from $X$ to $Y$. In particular, ...


1

I don't like to introduce $\lambda = \beta/\alpha$ because this destroys the symmetry between $x$ and $y$. I'd rather begin with an algebraic observation: the polynomial (aka quadratic form) $P(t,s)=At^2+2Bts+Cs^2 $ is: is strictly positive on $\mathbb R^2\setminus \{(0,0)\}$ iff $AC>B^2$ and $A>0$ is nonnegative on $\mathbb R^2\setminus ...



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