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## Hot answers tagged functional-analysis

4

Under the assumption that $\epsilon$ is constant across $a$, then no additional assumptions are needed. Suppose that $x_a$ is the fixed point of $f_a$, and choose $e>0$. Then there is a $\delta$ such that for all $b$ within $\delta$ of $b$, you have $e\epsilon\geq \vert f_b(x_a)-f_a(x_a)\vert=\vert f_b(x_a)-x_a\vert$. Now, repeatedly apply $f_b$ to ...

3

A solution, by my professor: Consider the closed unit ball in $\ell^{\infty}(I).$ The extreme points of this are functions $f: I \to \{-1, 1\}.$ Notice that if $\phi: \ell^{\infty}(I) \to \ell^{\infty}(J)$ is an isometric isomorphism, then $\phi$ takes extreme points to extreme points. We may assume without loss of generality that $\phi(\chi_I) = \chi_J,$ ...

3

You ask what is the definition of a positive element in $E$; the element $a \in E$ is positive if $a$ is positive when we think of it as an element in $A$. But an operator system is more than that. It also allows us to define an order structure on $M_n(E)$ for each $n$ and to say whether each matrix $[a_{ij}]\in M_n(E)$ is positive or not. The statement "an ...

3

I do not think that $\mathscr D(\Omega)$ is sequential. On the other hand this is probably not used: By definitionn of the locally convex inductive limit topology of $\mathscr D(\Omega)= \lim X_n$ (where $X_n$ are the Frechet spaces of smooth functions with support in $K_n$ for a compact exhaustion) a linear map with values in any locally convex space is ...

2

If two different norms induce the same topology, then the identity map is a continuous linear map when $X$ is considered with one of the norms in the domain and the other one in the codomain. So it is bounded, which leads to the norms being comparable, and so equivalent as we can reverse roles.

2

If you do not need any control over the norm of the vector $\alpha$, then, yes, such a vector exists. Take any vector with norm less than $\min (\frac{\varepsilon}{2}, \frac{ \varepsilon}{ 2 \Vert T \Vert})$. If you add the assumption that $\Vert \alpha \Vert=1$, such a vector $\alpha$ need not exist. Consider the operator $T(x)=-x$. Its restriction to any ...

2

$T$ doesn't enter into this problem. You have $y\in Y$ ($Tx_0$ in your notation), $V$ a neighborhood of $0$, and you want to know if $y+V$ is a neighborhood of $y$. Yes it is. $V$ contains an open ball $B$ of radius $r$ centered at $0$, so $y+V$ contains $y+B$. Suppose $x\in Y$ and $|x-y|<r$. Then $x-y\in B$ and so $x=y+(x - y)\in y+B$, so $x\in V$.

2

Unless you are asking if $V + Tx_0$ is a neighbouhood of $Tx_0$, your question makes no sense. Otherwise, the assertion is clearly true, and we have the following much more general fact : If $E$ is a normed space, $x\in E$ and $V$ a neighbourhood of $0$ in $E$, then $x + V$ is a neighbourhood of $x$. This follows from the observation that $x \in ... 2 I'd like to offer another approach to building$\tilde h$. First step:$h$is$C^1$and non-zero on$K$, hence there exists$\epsilon>0$such that$h(x)\ne0$whenever$dist(x,K)\le\epsilon$. Second step: let's take a function $$\phi(x) = \begin{cases}c\exp\left(-\frac{1}{1-|x|^2}\right), &|x|<1,\\0,&\text{otherwise.}\end{cases}$$ where$c$is ... 2 We can view$\bar{x}_a$as a minimizer of the continuous function$x\mapsto \|f_a(x)-f_a(f_a(x))\|$. If$f$is jointly continuous as a function of$\mathbb{R}^n\times\mathbb{R}$, then the argmin correspondence that maps$a$to the set of minimizers of this functions is upper hemi-continuous by Berge's maximum theorem (one has to show that locally all ... 2 You can use the contraction mapping estimates directly. You have the estimate$\|\bar{x}_a - f_a^{(k)}(x_0)\| \le {(1-\epsilon)^k \over \epsilon} \|f_a(x_0) - x_0\|$, so we can see that if we let$B = \sup_{a \in B(\hat{a},1)} \|f_a(x_0) - x_0\|$, then$\|\bar{x}_a - f_a^{(k)}(x_0)\| \le {(1-\epsilon)^k \over \epsilon} B$for all$a \in B(\hat{a},1)$. So, ... 2 Suppose that$\lambda\not\in\{0,1\}$. Then $$a(a-\lambda)=a^2-\lambda a=a-\lambda a=(1-\lambda)a.$$ Similarly, $$(1-a)(a-\lambda)=(1-a)a-(1-a)\lambda=-(1-a)\lambda.$$ Let $$b=\frac1{1-\lambda}\,a-\frac1\lambda\,(1-a).$$ Then $$b(a-\lambda)=ba-\lambda b=\frac1{1-\lambda}\,a-\frac\lambda{1-\lambda}\,a+1-a=1,$$ and one can also check that ... 2 You have $$\lvert f(x)\rvert = \sup_{\vartheta\in [0,2\pi]} \lvert\Re f(e^{i\vartheta}x)\rvert,$$ since for an appropriate choice of$\vartheta$,$f(e^{i\vartheta}x)is real and non-negative. Thus $$\lVert f\rVert = \sup_{\lVert x\rVert = 1} \lvert f(x)\rvert = \sup_{\lVert x\rVert = 1}\sup_\vartheta \lvert\Re f(e^{i\vartheta}x)\rvert = \sup_{\lVert ... 2 As the OP corrected guessed the right subset to study is the set A\subset\ell^\infty consisting of all the sequences with zeros and ones, i.e., \{a_n\}\in A if and only if a_n\in\{0,1\}, for all n. Clearly if \{a_n\},\{b_n\}\in A and \{a_n\}\ne\{b_n\}, then$$ \|\{a_n\}-\{b_n\}\|_\infty=1, $$since for some n |a_n-b_n|=1. Also ... 2 If null(f) is not dense in X, you can find x\in X and r>0 such that B(x,r)\cap null(f)=\varnothing. if y\in X is such that |f(y)|\geq|f(x)|, then for some \alpha with |\alpha|\leq 1 we have f(\alpha y)=f(x), so x-\alpha y\in null(f), hence x-\alpha y\not\in B(x,r), so \Vert y\Vert\geq\Vert\alpha y\Vert\geq r. What this just said ... 2 The ingredients you have are thus S(w_n) is bounded, which guarantees the existence of weakly convergent subsequences, and S is such that the only possible limit of a subsequence is S(w) (by pointwise a.e. convergence of a further subsequence). That means the full sequence S(w_n) converges weakly to S(w), because of the Theorem: Let (x_n) be ... 2 1) \|e^{itB}\|=1 if t\in\mathbb R and B=B^*. This is simply because e^{itB} is a unitary. 2) The expression bU-aV is the real part of \lambda T: that is, 2(bU-aV)=\lambda T+(\lambda T)^*. So it is selfadjoint, and e^{2i(bU-aV)} is a unitary as in part 1. 3) Note that up to here you haven't used that AT=TA. This forces AT^n=T^nA for all ... 2 Fix a basis v_1,\ldots,v_m \in V_k. Then the map \sum \alpha_jv_j\to\left(\sum |\alpha_j|^2\right)^{1/2} defines a norm on V_k, and this norm is induced by the inner product \langle \sum \alpha_jv_j,\sum \beta_jv_j\rangle = \sum \alpha_j\overline{\beta_j}. In a finite-dimensional space all norms are equivalent, so the identity map becomes a ... 2 The trace satisfies the following inequality:$$ |\tau(ya)|\leq\|y\|\,\tau(|a|). $$This implies directly that \|\tau_a\|\leq\tau(|a|). For the reverse inequality, if a=v|a| is the polar decomposition, then |a|=v^*a and so \tau(|a|)=\tau(v^*a)\leq\|\tau_a\|. Suppose that X=\{\tau_a:\ a\in N\} is not norm-dense in N_*. Note that X is a ... 2 Every C^*-algebra has a faithful representation that is a direct sum of cyclic representations. So, yes, if \pi(a)>0 for every cyclic representation \pi, then the element a must be positive. Take a look at the wikipedia page for the Gelfand-Naimark Theorem for further details. 2 You have canonical embeddings A\hookrightarrow A^{**}, B\hookrightarrow B^{**}. The embedding is given by a(f)=f(a) for all a\in A, f\in A^*. Restriction allows us to embed B^{**} into A^{**}. Now, a map E:A\to B like the conditional expectation admits a dual map E^*:B^*\to A^* given by (E^*f)(a)=f(E(a)). And similarly we can obtain a ... 2 Let$$\begin{align} p(t)&=1-t+t^2-t^3+\cdots+t^{2N-2}-t^{2N-1}\\ &=(1-t)(1+t^2+t^4+\cdots+t^{2N-2})\\ &=(1-t){1-t^{2N}\over1-t^2}\\ &={1-t^{2N}\over1+t} \end{align}$$Clearly \sup_{t\in[0,1]}|p(t)|\le1 and |a_0|+|a_1|+\cdots=1+1+\cdots=2N. So whatever C is, take N so that C\lt2N. 1 1) what do you mean by explicitly? A coordinate map is a bijection on it's image, so if v\in \alpha(U) there exists exactly one u\in U such that \alpha(u)= v. But in general there is not way to say more. 2) The tangent space of \mathbb{R}^m at a point u in \mathbb{R}^m is just \mathbb{R}^m. The tangent space of the manifold \alpha(U) in v ... 1 You are right. Lower semicontinuity is equivalent to:$$ \forall x \ \ \ \{ y| \delta_S(y)\le x \} $$is closed. Let us prove this from OP's definition: \delta_S is lsc iff \forall x\ \liminf_x \delta_S = \delta_S(x) if \liminf_x \delta_S = \delta_S(x), then let (y_n)\in \{ y| \delta_S(y)\le x\}^{\Bbb N} a convergent sequence.$$ ... 1 It is true and it is dense. Iff\in C^\infty([0,T]\times \Gamma)$then$\partial^n_t u$clearly is$C^\infty$is$t$and belongs to$H^1(\Gamma)$, for all$t\in[0,T]$, as it is sufficiently smooth with respect to the variables which paprametrize$\Gamma$. Also, as$\Gamma$is compact and smooth,$H^1(\Gamma)$is the completion$C^\infty(\Gamma)$. 1 Let$d$be the metric. Hint 1: If$d(x,y)<\epsilon$, then $$\overline{B}(y,\rho-\epsilon) \subseteq \overline{B}(x,\rho) \subseteq \overline{B}(y,\rho+\epsilon)$$ Hint 2: $$\overline{B}(x,\rho) = \bigcap_{n \in \mathbb N} \overline{B}(x,\rho+1/n)$$ I will let you do open balls. 1 Some words might be missing and sloppy notations do not help but the linearity is the following. Assume that the bivariate distribution$p$of$(X,Y)$is such that$p(x,y)=r(x)q(y\mid x)$for every$(x,y)$in the state space of$(X,Y)$, where$r$is the distribution of$X$. (Then$q$is the conditional distribution of$Y$conditionally on$X$.) Let$s$... 1 So since$f$is non-zero,$S$is not zero That's correct, but and hence we must have$E(w)$is the identity. isn't. All that follows from$S = E(w)S$and$S\neq 0$is that$E(w)$is the identity on the range of$S$, but$\mathcal{R}(S)$need not be the full space (and isn't, generally).$E(w)$is a projection, so$S = E(w)S$is equivalent to ... 1 $$h=f\circ s,$$ where$s(x,y)=x-y$. Thus $$h^{-1}(A)=s^{-1}\big(\,f^{-1}(A)\big)$$ So if$A\in{\mathscr B}(\mathbb R)$, then$f^{-1}(A)\in{\mathscr B}(\mathbb R)$, since$f$is Borel measurable. Also, if$B\in{\mathscr B}(\mathbb R)$, then$s^{-1}(B)\in{\mathscr B}(\mathbb R^2)$, since$s$is continuous, and hence Borel measurable. 1 The Sobolev embedding theorem implies that in two dimensions,$H^1$(which is$W^{1,2}$) embeds into$L^q$for every$1\le q<\infty$. (By the way, it's not enough for$\Omega$to be bounded; we need some assumption on$\partial \Omega$, such as smoothness. This is assumed at the beginning of the paper.) But in (9) the authors use$H^{3/4}\$ (which is ...

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