# Tag Info

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 ...

4

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 ...

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

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 $\phi(\chi_I) = \chi_J,$ ...

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 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 $C\lt2N$.

2

Let $M=\sup_{k\ge 1} |\lambda_k|$. By assumption, the partial sums $s_n=\sum_{k=1}^n \lambda_k x_k$ satisfy $\|s_n\|_\infty\le MK$ for all $n$. For bounded sequences, weak* convergence in $\ell_\infty$ is precisely the coordinate-wise convergence. Thus, we only need to check that for each fixed index $j$ the numeric series $\sum_{k=1}^\infty \lambda_k ... 2 As the OP corrected guessed the right subset to study is the set$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 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 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 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 The solution is correct. Just to beef up this post, I'll sketch a slightly different proof: the complement of$l_0$is open. If$x\notin l_0$, let$r=\frac12\limsup_{k\to\infty} |x(k)|$. If$\|x-y\|\le r$, then $$\limsup|y(k)| \ge \limsup_{k\to\infty} |x(k)|-r =r$$ hence$y\notin l_0$. By the way, this is the first time I see notation$l_0$used for ... 1$D((0,T)\times \Omega)\subset L^2((0,T);H^1(\Omega))\subset L^1_{loc}((0,T);H^1(\Omega))$. Define $$f(v) = \int_0^Tdt \langle f(t,\cdot),v (t,\cdot)\rangle_{H^{-1};H^1},$$ then $$|f(v)|\le \int_0^Tdt \| f(t,\cdot)\|_{H^{-1}}\|v (t,\cdot)\|_{H^1} \le | K |\int_0^Tdt \| f(t,\cdot)\|_{H^{-1}}\sqrt{\|v (t,\cdot)\|_{L^\infty}^2+\|\nabla_x v ... 1 This is a very interesting issue. In fact, this isomorphism is not as "natural" as one might have thought. As an exercise, one should see that, given a map of Hilbert spaces V\to W it is rarely the case that the square of maps involving W^*\to V^* and the "Riesz-Fisher" dualities ... commutes. This is fairly crazy, yes, given the standard curriculum. A ... 1 If you're talking linear algebra (finite or infinite dimensional), any element in the subspace$$\;\sum_{i\in I}S_i\;,\;\;S_i\;\;\text{a vector subspace}\;\;\forall\,i\in I$$is a finite expression of the form$$s_{i_1}+…+s_{i_k}\;,\;\;s_{i_m}∈S_{i_m}$$If you allow infinite sums then it is because you have some kind of analytic structure in your linear ... 1 Let \pi,H_1 be a Strinespring dilation of \phi on A. Now let K be some Hilbert space with \dim K>\dim H_1 and T\in B(H,K) some operator. Let H_2=H_1\oplus K, V_2:H\to H_2 be given by V_2:\xi\mapsto V\xi\oplus T\xi, \pi_2:A\to B(H_2) by \pi_2(a)=\pi(a)\oplus 0. Then$$ V_2^*\pi_2(a)V_2\xi=(V^*\oplus T^*)(\pi(a)\oplus 0)(V\xi\oplus ... 1 I'll assume that you meant for$\{ x_{n}\}$to be a sequence of unit vectors. Otherwise, you could let$x_{n}=0$for all$n$, and there would exist such a sequence regardless of whether or not$i\lambda \in\sigma(T)$. Note that there is nothing gained by introducing$i$into the discussion, especially because you made no assumption about$\lambda$being ... 1 Your reference is talking about matrices. Which spectral mapping theorem are you interested in? For example, here's one version (see e.g. Rudin, "Functional Analysis", theorems 10.28 and 10.33): Suppose$T$is a bounded linear operator on the complex Banach space$X$,$\Omega \subseteq \mathbb C$open with$\sigma(T) \subset \Omega$, and$f$analytic in ... 1 $$f(a+h)-f(a)=f'(a)h+\frac{1}{2}f''(a)h^2+o(h^2)$$ and $$f'(a+h\theta) = f'(a)+f''(a)h\theta + o(h\theta).$$ In the last equation I just used the definition of$f''(a)$. Hence $$f'(a)h+\frac{1}{2}f''(a)h^2 + o(h^2)=f'(a)h+f''(a)\theta h^2 + o(h^2\theta).$$ Dividing by$h^2$we deduce $$\frac{1}{2}f''(a) = f''(a) \theta +o(1).$$ If$f''(a) \neq 0$, ... 1 You want to define a function$f:Y^*\rightarrow M^\perp$. That way, for$\varphi\in Y^*$you must have an associated$f(\varphi)$instead of$f\circ\varphi$. Also, in (i), you have$[x],[z]\in Y$, so$x,\in X$and$\varphi\in Y^*$, so you cannot compute$\varphi(x)$nor$\varphi(z)$(actually, this step is not necessary). Surjectivity is not good at all. You ... 1 If the space$X$is banach it is an easy consequences of the open mappig theorem. Anyway with the norm induced topology over$ X $you in fact are resizing a ball so it is a ball again and it is open by definition of the topology. So the map sends open ball in open ball therefore it is open. This reasoning heavily rely on the "absolute omogeneity" of the ... 1 It appears that my (first) edition of the book does not contain this statement, but I think I understand it. The elements of$H^{-1} $are bounded linear functionals on$H^1_0$: $$H^{-1}=\{f:H_0^1\to\mathbb R \ ; \ |f(u)|\le C\|u\|_{H^1}\}$$ Then what do we mean by saying that$L^2\subset H^{-1}$? It means that some functionals on$H^1_0\$ admit a bound by ...

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