# Tag Info

9

Your question is equivalent to the following: does every pre-Hilbert space (= inner product space) $M$ admit an orthonormal basis? It turns out that the answer is "no". A counter-example can be found in N. Bourbaki's Topological Vector Spaces, Exercise V.2.2. I tried to solve the exercise, but I'm not quite sure that I understood it. Let ...

4

Let $F\subset T(X)$ be a closed (in $Y$) subspace. Then $E = T^{-1}(F)$ is a closed subspace of $X$, and $T\lvert_E \colon E \to F$ is a compact surjective operator. Since $F$ is closed in $Y$, the open mapping theorem implies that $F$ is finite-dimensional. So: the image of a compact operator cannot contain an infinite-dimensional Banach space.

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

2

No, in general you cannot expect this. Simply consider $(X,\|\cdot\|)=(\mathbb{R},|\cdot|)$, and set $x=0$. Then the inequality reads $$|y|^3 \leq \frac{1}{2} |y|.$$ Obviously, this is in general not true for $y>1/\sqrt{2}$. Yes. The equality \begin{align*} \|x\|^2 x - \|y\|^2 y &= (x-y)\|x\|^2 + y (\|x\|^2-\|y\|^2) \\ &= (x-y)\|x\|^2 + y ... 2 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 ... 2 Having a dense span is just one of the conditions a family \{e_n\} must satisfy to be a Schauder basis. The much more restrictive condition is that there are (continuous) coordinate linear forms \xi_n such that for every x\in X we have\lim_{n\to\infty} \left\lVert x - \sum_{k=0}^n \xi_k(x)\cdot e_k\right\rVert = 0.$$For the space C([0,1]), the ... 2 The Fourier transform of a tempered distribution is a distribution. Let \hat f be the Fourier transform of f\in\mathcal S'(\Bbb R). Your equation leads to$$(1+4\pi|\xi|^2)\hat f =0$$in the sense of distributions. If we have an equation gT=0, where g\in C^\infty(\Bbb R), T\in D'(\Bbb R), then supp T\subset \{x\in \Bbb R: g(x)=0\}. In our case ... 2 I'd like to offer another approach to building \tilde h. First step: h is C^1 and non-zero on K, hence there exists \epsilon>0 such that h(x)\ne0 whenever dist(x,K)\le\epsilon. Second step: let's take a function$$\phi(x) = \begin{cases}c\exp\left(-\frac{1}{1-|x|^2}\right), &|x|<1,\\0,&\text{otherwise.}\end{cases}$$where c is ... 2 We can view \bar{x}_a as a minimizer of the continuous function x\mapsto \|f_a(x)-f_a(f_a(x))\|. If f is jointly continuous as a function of \mathbb{R}^n\times\mathbb{R}, then the argmin correspondence that maps a to the set of minimizers of this functions is upper hemi-continuous by Berge's maximum theorem (one has to show that locally all ... 2 You can use the contraction mapping estimates directly. You have the estimate \|\bar{x}_a - f_a^{(k)}(x_0)\| \le {(1-\epsilon)^k \over \epsilon} \|f_a(x_0) - x_0\|, so we can see that if we let B = \sup_{a \in B(\hat{a},1)} \|f_a(x_0) - x_0\|, then \|\bar{x}_a - f_a^{(k)}(x_0)\| \le {(1-\epsilon)^k \over \epsilon} B for all a \in B(\hat{a},1). So, ... 2 When you are given a matrix-valued function {\bf x}\to A({\bf x}) and know for sure that it is the Jacobian of some vector-valued function$${\bf f}:\quad{\mathbb R}^n\to{\mathbb R}^m,\qquad {\bf x}\to{\bf f}({\bf x})$$then it is easy to recover {\bf f} from A. Indeed, the columns of A are nothing else but the partial derivatives$${\bf ...

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

I see no indication that $M$ and $M'$ are supposed nontrivial invariant subspaces. But by part $(d)$ if theorem 12.22, If $\omega \subset \Delta$ is open and nonempty, then $E(\omega) \neq 0$ and the monotonicity, whenever the spectrum contains more than one point, there are partitions of $\Delta$ into disjoint Borel sets $\omega$ and $\omega'$ such ...

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

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, $$since for some n |a_n-b_n|=1. Also ... 2 Let w=\dfrac{\partial v}{\partial x}. Since v\in W^{1,\infty}, we have w\in L^\infty. The space L^\infty is dual to L^1. Thus, pairing a fixed L^\infty function with a convergent sequence of L^1 functions produces a convergent sequence of numbers. This is why$$\int b(u_k) w\to \int b(u)w$$For the other integral, note that a(u_k)w ... 2 To see that L\subset\ker\phi, if a\in L then by Kadison's inequality$$ 0\leq|\phi(a)|^2=\phi(a)^*\phi(a)\leq\phi(a^*a)=0, $$so \phi(a)=0. And now for b\in L^*, then b^*\in L, so \phi(b)=\overline{\phi(b^*)}=0 (this last part is just the fact that the kernel of a selfadjoint functional contains adjoints; it is an ideal, in fact). 2 For any f \in L^1 we have$$\lim_{\omega \to \infty} \underbrace{\int_{\mathbb{R}} f(x) \cdot e^{-\imath \, x \omega} \, dx}_{=:\hat{f}(\omega)} = 0, \tag{1}this result is known as Riemann-Lebesgue-Lemma. Since f \in L^1 we can choose a sequence (f_n)_{n \in \mathbb{N}} of simple functions such that f_n \stackrel{L^1}{\to} f. Since ... 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 It suffices to show that \|(T-T^*)x\|=0, for all x\in H. We have \begin{align} \|(T-T^*)x\|^2&= \langle (T-T^*)x,(T-T^*)x\rangle \\& =\langle Tx,(T-T^*)x\rangle-\langle T^*x,(T-T^*)x\rangle\\&= \langle x,T^*(T-T^*)x\rangle-\langle x,T(T-T^*)x\rangle \\&=\langle x,T^*Tx-x\rangle- \langle x,x-TT^*x\rangle=2\|Tx\|^2-2\|x\|^2. \end{align} ... 1 Since A is densely defined, we can characterise D(A^\ast) as\begin{align} D(A^\ast) &= \left\lbrace f \in H : g \mapsto (f, Ag) \text{ is continuous}\right\rbrace\\ &= \left\lbrace f\in H : \bigl(\exists K_f < \infty\bigr)\bigl(\forall g\in D(A)\bigr)\bigl(\lvert (f,Ag)\rvert \leqslant K_f\cdot \lVert g\rVert\bigr)\right\rbrace. ...

1

It's easy to see that $A$ is symmetric on its dense domain, which means that $A^{\star}$ extends $A$. In order to show that $A$ is selfadjoint, it is enough to show that $(A\pm iI)$ are surjective, which is something that is trivial to demonstrate in this case. Indeed, if $f \in L^{2}$, then $(x\pm i)^{-1}f$ are in $L^{2}$ with their respective images under ...

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}$$If you allow infinite sums then it is because you have some kind of analytic structure in your linear ... 1 If f and g are in L^{1}, then$$ \left|\int_{-\infty}^{\infty}fe^{i\omega t}\,dt-\int_{-\infty}^{\infty}ge^{i\omega t}\,dt\right| \le \|f-g\|_{1}. $$So, if you can prove the limit property$$ \lim_{|\omega|\rightarrow\infty}\int_{-\infty}^{\infty}f(t)e^{i\omega t}\,dt=0  for a dense subspace of $L^{1}$, then you'll have it for all ...

1

$T$ 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 $x\in V$.

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