# Tag Info

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


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$Pis regular in the strong topology, which gives you \begin{align} P[a,b]x & = \lim_{\... 0 Using the dual basis, the matrix representation ofT^+$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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