New answers tagged

2

No for $p=1$, yes for $1<p<\infty$. If $\phi\in \Delta C^\infty_c$ then $\int\phi=0$; this shows that $\Delta C^\infty_c$ is not dense in $L^1$. One might think at first that this shows the same thing for other $p$, but it doesn't, because the integral is not a bounded linear functional. Suppose from now on that $1<p<\infty$. Suppose that $K\...


1

a) Let $Z = \overline{T(Y)}$. Then $Z$ is a closed subspace of a Banach space and hence a Banach space. Define $i = T$. Then $i$ is an isometry and $\overline{i(Y)} = \overline{T(Y)} = Z$. So $(Z,i)$ is a completion of $Y$. b) Z is a closed subspace of a reflexive space, so it is reflexive.


0

The states on $\mathbb{C}G$ are given by positive definite functions in $F(G)$ such that $u(e)=1$. Therefore, states on $\mathbb{C} G$ are given by unitary representations $\rho:G\rightarrow \hom(V)$ and unit vectors $\xi\in V$ according to $$u(g)=\langle \rho(g)\xi,\xi\rangle.$$


1

When you say "then $f:L^{\infty}(\Omega)\rightarrow \mathbb{C}$ is bounded functional", it means $f\in (L^\infty)^*$, right? Then the convergence of $f_{n_k}\in L^1$ is going to be also as elements of $(L^\infty)^*$, and the limit may be not in $L^1$. (Note that $L^1$ is not closed in $(L^\infty)^*$ in the norm/weak topology.)


1

You cannot apply Banach-Alaoglu for $L^1(\Omega)$, since $L^1(\Omega)$ is not the dual space of a normed space. Rather you have to embed $L^1(\Omega)$ into larger spaces $L^\infty(\Omega)^*$ or $C(\bar\Omega)^*$ to obtain a weak-star convergent subsequence. To see that for $p=1$ the assertion is not true, consider the sequence $f_n(x)=n \chi_{0,1/n}(x)$ on $...


1

(You don't say how you got the second equality; since it is not trivial, I'm not sure how you did it and so it is done below) Since $A^*A$ is positive and compact, it is orthogonally diagonalizable (spectral theorem): $A^*A=U^*D^2U$ for some unitary $U$ and $D$ diagonal with diagonal $s_1(A),s_2(A),\ldots$ Assume $s_1(A)\geq s_2(A)\geq \cdots$ Since $U$ ...


2

(Huge) hint: Is it complete? Why or why not? (Even bigger hint below.) Follow-up:


2

Since $\|f - f_n \|_p \to 0$, we can extract a subsequence $f_{n_j}$ so that $\| f - f_{n_j} \|_p \le \frac{1}{2^j}$. Put $$g = \lvert f \rvert + \sum^\infty_{j=1} \lvert f - f_{n_j} \rvert.$$ Then $g \in L^p$ and by the triangle inequality we have $$\lvert f_{n_j} \rvert \le g \,\,\, (\text{almost everywhere}).$$ From this subsequence, you can extract a ...


0

You have the correct definition in Bogachev's book: to say that $X$ "is Gaussian with values in $C([a,b])$" is to say that $X$ is a random element of $C([a,b])$ with the property that $b^∗(X)$ is normally distributed for each $b^∗$ in the dual of $C([a,b])$.


2

Here is a simpler proof: $$ \sum_{n=1}^\infty| \langle x, e_n \rangle \langle y, e_n \rangle | \le \sum_{n=1}^\infty| \langle x, e_n \rangle|\cdot | \langle y, e_n \rangle | \\ \le \left(\sum_{n=1}^\infty| \langle x, e_n \rangle|^2\right)^{1/2}\left(\sum_{n=1}^\infty | \langle y, e_n \rangle |^2\right)^{1/2} \le \|x\| \cdot \|y\|. $$ First inequality is ...


0

Giving distinct values to $m$ you have each time $x_1+x_2=3m-1$ so your answer can not be independent of m. You have always as solution the (literal, not numerical) coefficient $3m-1$ according to the quite known Vieta's formulas.


0

If $(3m-1)^2-4(2m+3)\ge0$ then $x_1+x_2=3m-1$


1

In general, consider the polynomial $(x-r_1)(x-r_2)\cdots(x-r_n)$. Expanding, we have $$x^n - (r_1+r_2+\ldots+r_n)x^{n-1} + \ldots$$ So the sum of the roots appears as the negative of the $x^{n-1}$ term. Note that this applies to monic polynomials (i.e., the coefficient of $x^n$ is $1$); otherwise, first factor out the coefficient of the $x^n$ term.


1

A quadratic is of the form $x^2 - (\alpha + \beta)x + \alpha \beta$ where $\alpha$ and $\beta$ are its roots. This can be seen by expanding $(x-\alpha)(x-\beta)$. So the sum of your roots is given by the negative of the coefficient of $x$, i.e: $3m-1$.


6

There is no such operator. Note that $(C ([0,1]),\|\cdot\|_\infty ) $ is a Banach space, so that it suffices to show that $T $ has closed graph. Thus, assume $f_n \to f $ and $T f_n \to g $ (both with respect to $\|\cdot\|_\infty $). Then $f_n \to f $ and $T f_n \to g $ bith with respect to $\|\cdot\|_2$ (why?) and hence $T f_n \to Tf $ also with respect ...


0

I could solve the problems: We show that $\pi$ is open onto its image $\pi(S)$. Let $\{s\} \subseteq S$, then we have $\{\delta_s\} = \pi(s) \cap (\operatorname{ev}_{1_{\{s\}}})^{-1}(\mathbb C \setminus \{0\})$, where $\operatorname{ev}_{1_{\{s\}}}(\varphi) = \varphi(1_{\{s\}})$ for all $\varphi \in G_S$. Hence $\{\delta_s\}$ is open in $\pi(S)$ and $\pi(S)...


1

Note that the two equalities $A=AA^*A$ and $A^*=A^*AA^*$ are the same, since you can obtain one from the other by taking adjoints. Assume first that $A=AA^*A$. By multiplying by $A^*$ on the left, we get $$ A^*A=(A^*A)^2.$$ It follows that the eigenvalues of $A^*A$ all satisfy the equation $\lambda=\lambda^2$, so only $0$ and $1$ are possible. Conversely,...


1

Say $f\in C_c(X)$. Standard construction: For $n=1,2\dots$ and $j\in\Bbb Z$ define $$E_{n,j}=\{x:j/n<f(x)\le(j+1)/n\}$$and set $$\phi_n=\sum_{j\in\Bbb Z}\frac jn\chi_{E_{n,j}}.$$Then $\phi_n$ is $\sigma(\mathcal E)$-measurable and $\phi_n\to f$ (uniformly).


2

Any $*$-homomorphism between C$^*$-algebras is contractive. This is standard (i.e., it appears in every book on the subject) and is due to three things: The C $^*$-identity $\|a\|^2=\|a^*a\|$, which reduces the problem to norms of positives; The equality $\|a\|=\text {spr}\, (a) $ for $a $ positive; The fact that a $*$-homomorphism reduces the spectral ...


1

No, it's not a coincidence. The definition of "$\lim (T-T_k)=0$" is "$\lim||T-T_k||=0$". No, you can't do the same for any operator. You can't show that $||T-T_k||\to0$. (Let $Tx=x$. What is $||T-T_k||$?)


4

You could not find a proof because it is not true. In fact $BL(X)$ is separable in the topology of uniform convergence if and only if $X$ is totally bounded. If $X$ is totally bounded, every Lipschitz function can be extended to its completion $\tilde{X}$, which is compact. Thus we obtain an isometric embedding $BL(X) \to C(\tilde{X})$. If $X$ is not ...


8

If such a function exists, then you may add any function whose integral between $0$ and $1$ is zero. Now you should find one particular solution. For this, a constant (with respect to $x$) function suffices.


0

Endow $R(A)$ with the quotient norm having the unit ball $A(K_X)$ where $K_X$ is the unit ball of $X$. This is a Banach space and $\tilde B:X\to R(A)$, $x\mapsto B(x)$ is well-defined by assumption, linear, and has closed graph (because the norm of $Y$ gives a coarser norm on $R(A)$ which makes $B$ continuous). The closed graph theorem implies the continuity ...


0

It depends on what exactly you want to achieve. The reason most of the programs contain a lot of geometry is that they try to process images from a 3D point of view, and the reality of 3D arrangement and motion of objects is best described in a geometric way. However, there might be problems where there is no need to use geometry. One example that comes to ...


1

If $\psi$ is a state vector, meaning $\|\psi\|=1$, then the expected value of an observable $A$ for the system in the state $\psi$ is $$ (A\psi,\psi). $$ That's what you're trying to show. One must assume that $\psi$ is in the domain of the observable $A$. A selfadjoint operator $A : \mathcal{D}(A)\subseteq H\rightarrow H$ on a Hilbert space $H$ ...


1

I recently wrote up a solution to this very problem, which I've copied below. Note that in this context, $\mathfrak A$ is the space of bounded operators on $B$. I hope you find this helpful. $ \newcommand{\f}{\mathfrak} \DeclareMathOperator{\rad}{r} \newcommand{\eps}{\varepsilon} \DeclareMathOperator{\spec}{spec} \newcommand{\lp}{\left(} \newcommand{\rp}{\...


0

Since $E^*$ is reflexive you got that $E^{**}$ is reflexive (you showed that). Since $E$ is isometric isomorphic to a closed, complete subset of $E^{**}$ you know that $E$ is also reflexive because every closed subspace of a reflexive space is also reflexive. Hope that it helps you :)


1

$1\iff4$ : Assume $1$. Given $x\in N_+$, there exists nonzero $y\in N_\tau^+$ with $y\leq x$. Now use Zorn to find a maximal ordered family $\{y_j\}\subset N_\tau^+$ with $y_j\leq x$ for all $j$. As the net is bounded, it has a sup, say $y=\lim_{sot}y_j$. Then $y=x$, because otherwise a nonzero element of $N+\tau^+$ below $y-x$ contradicts the maximality. ...


2

Since $T - \lambda I$ is also normal, we have $$ \| T - \lambda I \| = \text{spr} (T - \lambda I) = 0, $$ showing that $T = \lambda I$. (I recently asked basically the same question (Self-adjoint operator with single point spectrum), but your formulation is more general so I thought it might be worth sharing the answer here.)


1

using spectral theorem since $T$ is normal it exist a spectral measure $E$ such that $$ Tx=\int_{\sigma(T)}tdE(x)=\int_{\{\lambda\}}tdE(x)=\lambda E(\{\lambda\})(x)=\lambda E(\sigma(T))(x)=\lambda I (x)=\lambda x $$


0

Once you have the Parseval (Plancherel) identity for Schwartz class, then the Fourier transform extends to $L^2$ as you noted in the first paragraph. For $f\in L^2\cap L^1$, it's not hard to show that this continuous extension to $L^2$ is the same as the classical integral definition: $$ \hat{f}(s)=\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\...


2

As @AlexanderFrei pointed out, it should read $$\forall x: \lim_{n \to \infty} T_n(x) = T(x) \iff \forall K \subseteq X \, \text{compact}: \lim_{n \to \infty} \sup_{x \in K} \|T_n(x)-T(x)\| = 0.$$ The implication "$\Rightarrow$" is trivial, just choose $K= \{x\}$ for fixed $x \in X$. It remains to prove "$\Leftarrow$". Suppose that $T_n(x) \to T(x)$ for ...


1

Define $x(t) = w(t) e^{i \zeta t}$. We want to show that $x \in C$, i.e, $\lim_{|t| \rightarrow \infty} x(t) = 0$ according to (2), or, equivalently, $|x(t)| \rightarrow 0$ as $|t| \rightarrow \infty$. Note that $e^{i \zeta t} = e^{i (Re \zeta + i Im \zeta) t} = e^{-(Im \zeta) t} e^{i (Re \zeta) t}$, and note also that the text requires $|Im \zeta| < c$, ...


3

If your example $\mathcal{B}$ were a real Banach algebra instead of a complex Banach algebra, then you would be right that there are four connected components, since $\mathcal{B}$ can be identified with $\mathbb{R}^2$ and the invertible elements split into four quadrants. But over $\mathbb{C}$, you have the complement in $\mathbb{C}^2$ of two (complex) one-...


2

Hah! This is actually a specific example of something in my research! (My work attacks a more general set of integral equations, in some sense.) Let's go for something nontrivial (unlike previous answers/comments). If you consider what I like to call a diagonal kernel, i.e. $g(x,t) = f(xt)$ for some $f$ and assume $g$ is real analytic, then this is very ...


2

An easy solution can be obtained by making $g(x,t)$ degenerate: $$ g(x,t)={1\over2}\exp(-bx^4)\exp(at^4-|t|) $$ which is susceptible to the generalization $$ g(x,t)={1\over C}\exp(-bx^4)\exp(at^4)f(t) $$ where $f$ is integrable over $\mathbb R$ and $$ \int_{-\infty}^{\infty}f(t)dt=C\neq0 $$


1

Well, $$||x||^{m^2}=(||x||^m)^m\le(K||x^m||)^m=K^m||x^m||^m\le K^mK||(x^m)^m||=K^{m+1}||x^{m^2}||.$$Now your argument for $n\le m$ also works for $n\le m^2$...


4

Triangle inequality $$|\, ||f_n ||_2 -|| f ||_2\, |\leq ||f_n-f||_2$$


4

since $\langle T(x+y),x+y\rangle =0$ that implies : \begin{eqnarray} \langle T(x+y),x+y\rangle &=&\langle Tx+Ty,x+y\rangle \\ &=&\langle Tx,x+y\rangle+\langle Ty,x+y\rangle\\ &=& \langle Tx,x\rangle+\langle Tx,y\rangle+\langle Ty,x\rangle+\langle Ty,y\rangle\\ \end{eqnarray} Then $$ \langle T x,y\rangle +\langle Ty,x\rangle=0 \qquad ...


3

A counterexample for $d=2$: let $\Omega$ be the disk $\{x:\|x\|<\exp(-\exp(\pi))\}$, and $$ f(x) = \sin \log \log \frac{1}{\|x\|} $$ This function is in $W_0^{1,2}(\Omega)\cap L^\infty(\Omega)$ (relevant calculations here) but has a discontinuity at $0$, and moreover cannot be made continuous by redefining it on a set of measure zero. On the other ...


2

According to Daniel Fischer: The left hand side cries for an application of the Cauchy-Schwarz inequality. And according to siminore: $$ x=\sum_n x_n e_n = \sum_n \langle x,e_n \rangle e_n \quad ; \quad y=\sum_n y_n e_n = \sum_n \langle y,e_n \rangle e_n $$ But we give it a twist: $$ x'=\sum_n |x_n| e_n = \sum_n \left|\langle x,e_n \rangle\right| e_n \quad ; ...


0

This is well known exercise many times solved on MSE. Here is one example. $\phantom{}\phantom{}\phantom{}\phantom{}\phantom{}\phantom{}\phantom{}\phantom{}$


0

the norm of $f$ is given by : $$ \|f\|=\inf\left\{k\in \mathbb{R}^+ \; ; \forall (x,y) \; |f(x,y)|\leq k\|(x,y)\|_\infty \right\} $$ Let $(x,y)\in \mathbb{R^2}$ then we have $|x|\leq \|(x,y)\|_\infty$ and $|y|\leq \|(x,y)\|_\infty$ then $$|x+y| \leq |x|+|y| \leq 2 \|(x,y)\|_\infty$$ this implies that $$ \|f\|\leq 1 $$ but $f(1,1)=1$ so $\|f\|=1$ so $$ ...


1

Consider $u\in B(H_1\oplus H_2, H_1\oplus H_2)$ and since $u^*u$ is projection, then $u$ is partial isometry, which means that $uu^*u=u.$


0

Let $A=M_2(\mathcal O_n)$. Fix a nontrivial automorphism $\alpha$ of $\mathcal O_n$, and let $$ A_0=\left\{\begin{bmatrix}a&0\\0&\alpha(a)\end{bmatrix}:\ a\in\mathcal O_n\right\}. $$ Then $A_0\simeq\mathcal O_n$. Now let $$ A_t=u_t\,A_0\,u_t^*, $$ where $$ u_t=\begin{bmatrix}\cos t&\sin t\\ -\sin t&\cos t\end{bmatrix}. $$ The continuity of ...


1

I will elaborate on the previous poster's answer by proving the aforementioned corollary, which is often referred to as "$X^*$ norms $X$." In particular: Claim: For any $x\in X$ there exists $T\in X^*$ such that $\vert\vert T\vert\vert=1$ and $\vert T(x)\vert=\vert\vert x\vert\vert$. Pf.: Let $L$ be the one dimensional subspace defined by $x$, that is $L=\{...


3

There's a neat corollary of Hahn-Banach saying that for every $x$ there exists a linear operator $T$ with unitary norm such that $T(x)=||x||$. Using this and the definition of the norm on the dual space, you have your proof.


1

The connection with the spectral measure $P$ is $$ P(E) = \chi_{E}(A). $$ So, for example, $$ P[a,b] = \chi_{[a,b]}(A), \;\; P(a,b) = \chi_{(a,b)}(A) \\ P[a,b] = P(a,b) + P\{a\}+P\{b\} $$ The spectral measure $P$ is regular in the strong topology, which gives you \begin{align} P[a,b]x & = \lim_{\...


0

Using the dual basis, the matrix representation of $T^+$ is given by the transpose, $T^t$.


3

False. The closed span of $e_k$ for $k \ge n$ is invariant.



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