# Tag Info

## Hot answers tagged normed-spaces

13

The answer to the question exactly as you asked it is yes; your space is isomorphic as a vector space, with no topology, to various Banach spaces. (See various comments for details.) Edit: The assertion that the answer is yes has met with vigorous disbelief. Also there's a technical point that I realized after some thought I simply didn't know how to do. ...

5

As a complement to the earlier (good) answer and comments: the space of all sequences (whether real or complex) arises in at least one fairly natural way, namely, as the continuous dual to the LF-space (strict inductive limit of Frechet spaces) $\mathbb R^\infty=\bigcup_n \mathbb R^n$, where $\mathbb R^n$ has its usual topology and is included in $\mathbb ... 5 By definition of bilinearity,$\langle x,y\rangle - \langle x,z\rangle = \langle x,y-z\rangle$. On the other hand in general there is no formula for$\langle a,b\rangle \pm \langle c,d\rangle$. 4 I think that if you want to work with norms on vector spaces over fields in general, then you have to use the concept of valuation. Valued field: Let$K$be a field with valuation$|\cdot|:K\to\mathbb{R}$. This is, for all$x,y\in K$,$|\cdot|$satisfies:$|x|\geq0$,$|x|=0$iff$x=0$,$|x+y|\leq|x|+|y|$,$|xy|=|x||y|$. The set$|K|:=\{|x|:x\in ...

3

The answer for infinitely many functionals is no; there's a counterexample here. For finitely many functionals it must be yes... Right. First, there is a norm on any real vector space $X$, for example if $B$ is a (Hamel) basis define $$\left\vert\left\vert\sum_{b\in B}c_b b\right\vert\right\vert=\sum_{b\in B}|c_b|.$$Now if $||\cdot||$ is a norm on $X$ ...

3

Your answer is incorrect. $\langle f,g \rangle$ is allowed to take any value, but $\langle f,f \rangle$ must be non-negative. Try to come up with an example of two vectors whose dot-product is negative, noting that the dot-product is the prototypical inner product. The property of inner products that fails here is that $$\|f\|^2 = \langle f,f \rangle = 0 ... 3 There are two cases: 0\in C and 0\notin C. In the first case the minimum norm is =0, and x_0=0 is the unique element with this property. If x_0 with minimum norm has already been found (and your question indicates that this is already known to you) assume x_1\neq x_0 has the same norm and it is an element of C. It is easy to see that x_1 and ... 3 y=(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...) \notin c_{00}, y_1=(1,0,0,...), y_2=(1,\frac{1}{2},0,0,...) and so on. y_n \in c_{00} and \|y_n -y \|_{\infty}=\frac{1}{n+1} which tends to 0 when n tends to \infty. 3 Note, that \|\cdot\| is somehow the composition of the 2-norm and the 3-norm on \mathbb R^2. We have$$\|x\| = \|(\|(x_1,x_2)\|_2,x_3)\|_3$$and with this at hand the proof is easy (but a bit ugly). We have by the triangle inequality for the 2-norm$$a:= \|(x_1+y_1,x_2+y_2)\|_2 \leq \|(x_1,x_2)\|_2 + \|(y_1,y_2)\|_2 =:band since for all ... 3 This is just an elaboration on what others have already said in the comments. Let V be a finite-dimensional normed linear space (over a subfield \mathbb F of \mathbb C). As you say, all norms on a finite-dimensional vector space are equivalent, so we may assume that we have the usual 2-norm on \mathbb C^n. The definition of weak convergence is ... 3 The \ell^p spaces are a special case of the L^p spaces obtained by using the counting measure on the set of natural numbers. If you squint closely at the integral it looks like a sum or indeed as Forever Mozart points out: summation is just integration with the trivial measure on \mathbb{N}. 2 It is true, it is a consequence of the uniform boundness principle. https://en.wikipedia.org/wiki/Uniform_boundedness_principle Apply the contraposition. Consider the family T^n(x) =(Tx)_n and suppose that \sup_{T^n,\|x\|=1}\|T^n(x)\|=\infty. Then there exists x such that \sup_n\|(Tx)_n\|=\infty. This is impossible since T must be defined at x. ... 2 Yes, the inequalities hold. Let B be the closed unit ball for the norm \|\cdot \|. The assumptions imply that \pm e_j, the standard basis vectors, are in B. Hence, their convex hull is contained in B. This convex hull is the unit ball for the \ell^1-norm, which implies \|\cdot \|\le \|\cdot \|_1. Similarly, we need to prove that B is ... 2 Yes, it is. In fact, we can say more. If K has the trivial valuation, X is a finite-dimensional K-vector space, and p is any norm on X, then there exist positive constants c_1 and c_2 such that c_1\leq p(x)\leq c_2 for all nonzero x\in X. To get c_2, note that if \{e_1,\dots,e_d\} is a basis for X, then writing x=\sum a_i e_i, we ... 2 You need to ask yourself, is V is a closed subspace of X? Namely, given a sequence v_n\in V such that for some v\in X, \|v_n-v\|_\infty\to0, does that imply that v\in V as well? (You could try and show that X\setminus V is open, but that's harder.) 1 It seems that the uniqueness part has already been settled. In your argument regarding the existence part, you have proved that if the sequence (x_n)_{n=1}^\infty you have constructed converges, then its limit belongs to C. But so far you have only that \|x_n\| \rightarrow s. It remains to be shown that (x_n)_{n=1}^\infty does converge. To this end, ... 1 I think the answer is yes. A sign- and permutation- invariant norm defined on \mathbb C^n is called a symmetric gauge function. It is known that every unitarily invariant norm on M_n(\mathbb C) is induced by a symmetric gauge function. See, e.g. theorem 7.4.24 on pp.438-440 of Horn and Johnson (1985), Matrix Analysis, 1/e, Cambridge University Press. To ... 1 Let x\in X and r>0, and consider the open ball B(x;r)=\{y\in X:\|x-y\|<r\}. For any y,z\in B(x;r) and t\in(0,1), we have \begin{align} \|ty + (1-t)z - x\| &= \|t(y-x) + (1-t)(z-x)\|\\ &\leqslant t\|y-x\| (1-t)\|z-x\|\\ &<r, \end{align} so B(x;r) is convex. Write f(t) = t(a-b) + t, then it is clear that f is an affine ... 1 Using the Axiom of Choice, one can construct in a given infinite dimensional normed linear space X, a discontinuous linear functional (see this, e.g.). The kernel of such a functional is a proper and dense subset of X (see this). This would provide a subspace Y of X that is proper and such that for every x\in X, \text{dist}\,(x,Y)=0. From ... 1 You have M=\mathcal{N}(f). Then g\in M^{\perp} iff \mathcal{N}(f)\subseteq\mathcal{N}(g). Suppose f(u)\ne 0 for some u\in X.Then f\left(y-\frac{f(y)}{f(u)}u\right)=0 holds for all y\in X, which gives 0= g\left(y-\frac{f(y)}{f(u)}u\right)=g(y)-\frac{f(y)}{f(u)}g(u),\;\;\;y\in X \\ \implies g = ...

1

No, not even if $X=Y=H$, a Hilbert space. Say $H$ is an infinite-dimensional Hilbert space and let $B$ be a Hamel basis for $H$. Say $m:B\to(0,\infty)$ is unbounded and define $T:H\to H$ by $$T\left(\sum_{b\in B}c_b b\right)=\sum_{b\in B}m(b)c_b b$$(where $c_b=0$ except for finitely many $b\in B$.) Then the kernel of $T$ is $\{0\}$, but $T$ is unbounded ...

1

Since $f$ is continuous (at $0$), there is a neighbourhood $U$ of $0$ such that $f(U)\subset(-1,1)$. Choose $\delta>0$ such that $\{x\in X|\|x\|\leq\delta\}\subseteq U$. Then, if $x\in X$ is such that $\|x\|\leq \delta$, we have $x\in U$, and hence, $|f(x)|\leq 1$. Since $\|\frac{\delta x}{\|x\|}\|=\delta$, it follows that for all $x\in X$ we have $$... 1 Suppose F is such an extension. Then F(x) = x for x \in Y. But since Y is dense, that implies F(x) = x for all x \in X. So we must have Y = X. 1 An infinite-dimensional vector x = (x_1, x_2, x_3, \ldots) (or I guess we can just say sequence) is in \ell^2 iff \displaystyle \sum_{k=1}^{+\infty} |x_k|^2 < +\infty. Well, no matter what the value of n is in x = (1, 0, 0, \dots, 0, -n^2, 0, 0, 0, \dots), we'll always have \displaystyle \sum_{k=1}^{+\infty} |x_k|^2 = 1 + n^4, and this is ... 1 Suppose w is in the image of T then w=(b_1/1,b_2/2,\dots) for some bounded sequence (b_n), but then$$\frac{b_n}{n}=\frac{1}{\sqrt n}$$hence b_n=\sqrt n which is not bounded. An accumulation point is the same as a limit point it just means that for any neighbourhood U of w there is a point x\in \mathrm{Im}(T) such that x\in U. 1 Yes. First f is surjective. Indeed, let y \in Y. \frac{y}{||y||} \in B_Y = f(B_X) hence there is x \in B_X s.t f(x)=\frac{y}{||y||} and so f(||y||x)=y by linearity. We conclude that f is surjective and hence bijective (we already know it is injective by assumption) Because of the facts that B_Y = f(B_X) and that f is bijective, we have ... 1 If a=\lim a_n and b=\lim b_n , then you have to show that a+b=\lim (a_n+b_n) . 1 Notice that we can write$$c_1^2 + c_2^2 = |\nabla v|^2 = (\nabla v \cdot \tau)^2 + (\nabla v \cdot \nu)^2,$$where \tau is the direction of the tangent line. By assumption, we know that the second term on the RHS is 0 on \Gamma_0. What is left to show is that also the tangential part of the gradient is 0. Do you see why this is true? EDIT: take ... 1 (i) is correct. (ii) Show that \{ f \in X : f(0)=0\} and \{ f \in X : f(1)=0\} are closed subspaces of X: from this, you will have that V is an intersection of closed sets, hence closed. This can be done using sequences, and (i). Pick a sequence f_n \subset \{ f \in X : f(0)=0\} converging to some f in X. But (i) shows that$$f(0) = \lim_n ...

1

This is perhaps best understood from a topological perspective. A base for the weak topology is formed by finite intersections of sets of the form, $$\{u:a < \phi(u) < b\},$$ for any continuous linear functionals $\phi$. Geoometrically, one of these sets looks like the infinite slab between two parallel hyperplanes. In finite dimensions, you can ...

Only top voted, non community-wiki answers of a minimum length are eligible