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## New answers tagged normed-spaces

1 vote

$A^\times$ is non-empty means that $A$ is unital. If $1-z\in A^\times$ then $$\forall n,\qquad (1-z)^{-1}=(1-z^{n+1}+z^{n+1})(1-z)^{-1}= \sum_{k=0}^n z^k + z^{n+1}(1-z)^{-1}$$ from which for $\|z\|\... • 70.4k 0 votes ### Prove that$\sum_{n=1}^\infty (-1)^n \frac1n$is convergent but not unconditionally convergent$\lim_{N\rightarrow\infty}\sum_{n=1}^{N}(-1)^n\frac{1}{n}$does exists because $$(-1)^n\frac{1}{n}+(-1)^{n+1}\frac{1}{n+1} \\ = (-1)^n\left[\frac{1}{n}-\frac{1}{n+1}\right] \\ = (-1)^n\... 0 votes ### Prove that \sum_{n=1}^\infty (-1)^n \frac1n is convergent but not unconditionally convergent For the convergence, you can use the Leibnitz Test. We consider a_n=\frac{1}{n} . The (a_n) is a decreasing sequence, and \lim_{n\to \infty}a_n=0. Thus, we have by Leibnitz Test that the series ... 3 votes Accepted ### Why does \langle (U^*U - I)v, v\rangle = 0 \ \forall v imply that U^*U = I, from the polarization identity? Let M = U^*U - I. We are given that \langle Mv,v \rangle = 0 holds for all v \in \mathcal H. On the other hand, the polarization identity allows us to express \langle Mx,y \rangle in terms of ... • 208k -1 votes ### I don't understand which norm exactly should be calculated, the norm of T or T(f)(x)? please give an example First I give the definition of a norm. \left\|T \right\|=\sup\left\|T(f) \right\| over all f such that \left\|f \right\|=1 The a) is not correctly defined. Please make a correction, because T ... • 3,165 -1 votes ### I don't understand which norm exactly should be calculated, the norm of T or T(f)(x)? please give an example C[0,1] is a Banach space under the sup norm. so i think you are supposed to compute:$$||T||=\sup_f\left\{\frac{|Tf(x)|}{|f(x)|}\right\}$$since f\in C[0,1], f is bounded. • 176 1 vote Accepted ### Use of the p-norms in physical world (soft quest.) In a finite dimension vector space, this is the only norm that is invariant under a rotation. Other norms will not preserve lengths. As an example, if you have for example a unitary vector (1,0,0,0,..... • 1,602 0 votes ### Showing that l_2 norm is smaller than l_1 Hint for Squirtle's generalization: calculate derivative of f(x)=(a_1^{x}+a_2^x+...+a_n^x)^{1/x} and prove f'(x)<0,\forall x\geq 1 2 votes ### Construct an Injective and onto unbounded operator. This answer assumes A: The axiom of choice, so all vector spaces have a Hamel basis and B: You are working in a field of Characteristic 0. EDIT: As per @SeverinSchraven 's comment, if working ... • 15.3k 0 votes ### Correct cases for proximal operator of L_2 / Euclidean norm when solving without Moreau's decomposition You may be interested in the website http://proximity-operator.net that lists many different proximity operators (+ code). For instance, the one of \|.\|_2 is the first on this page http://proximity-... 0 votes ### Continuity and norm of operator on l^2 Here is another piece for a complete solution: If the operator A maps all of \ell^2 to \ell^2, then \alpha has to be bounded. Indeed, if \alpha would not be bounded, there would be a ... • 28.3k 0 votes ### Showing a space of functions is a Hilbert space You have to assume that R is strictly positive definite since you don't even get a norm in general. (e.g., when R=0). But in this case there exist positive constants c and C such that c \|x\|^... • 8,896 0 votes Accepted ### Euclidean Norm of a function Every symmetric matrix is diagonalizable i.e. \Phi^{T}\,\Phi which is clearly symmetric can be written as : \Phi^{T}\,\Phi=S^{-1}MS where M is \begin{bmatrix} m_{1} &0 &0... & 0 \\... • 3,165 2 votes Accepted ### Let i :X \rightarrow X^{**} and j : X^* \rightarrow X^{***} be natural embeddings. Then Q := j \circ i^* is a projection with Q(X^{∗∗∗}) = X^∗ For your first question, Q\circ j=j seems useless. Just say$$Q(X^{***})=j\circ i^*(X^{***})=j(X^*),$$the last equality being due to the fact that i^* is onto since i^*\circ j=Id_{X^*} (or: since ... • 2,929 1 vote Accepted ### What is the Banach space for multivariate sequences analogous to the \ell_p-spaces for univariate sequences? What you wrote down is indeed the canonical choice for the norm. The concept you are looking for (albeit a bit overkill in your setting) is Bochner spaces (for the Banach space \mathbb{R}^n and on ... • 5,176 0 votes ### Signum function in the multivariable case The multivariable signum function is often defined as (3) and is commonly invoked in dynamical systems and control literature (for example, pp. 2 in this paper) where it’s common to have x \in \... 1 vote ### Show that T in l^{\infty} to l^p is well defined and compute the operator norm ∥T∥. The inequality you used is true, but there is no reason for \sum_j|\xi_j|^p to be finite; so the inequality is very coarse. Instead, using that x\in\ell^\infty,$$ \|Tx\|^p=\sum_j|\xi_jb_j|^p\leq\... • 184k 1 vote Accepted ### Understanding a step in Wikipedia's proof of the Uniform Boundedness Principle Each$T\in F$is continuous. Also note that, $$X_n = \bigcap_{T \in F} \underbrace{\{x \in X: \|Tx\| \leq n\}}_{:= A_n}.$$ Now let$T\in F$be arbitrary and let$(x_k)_{k \in \mathbb{N}}$be a ... • 432 2 votes ### Non-equivalent norms on$\mathbf{Q}[\sqrt2]$Your intuition is correct.$N_2(u_n)$goes to$0$since$|1-\sqrt 2| < 1$Claim: if$|x|<1$, then$|x^n|\to_{n\to\infty} 0$. Proof: Obvious. Or, let$|x|=1-\epsilon$, where$0\lt\epsilon\le1$. ... • 3,328 0 votes ### Conway( A Course in Functional Analysis) Proposition 3.3 For$r>0$let $$M_r=\sup \{|L(h)|\,:\,\|h\|=r\}$$ By linearity we have$M_r=rM_1.$Furthermore $$\sup \{|L(h)|\,:\,\|h\|<1\} =\sup_{0<r<1} M_r= \sup_{0<r<1} rM_1=M_1= \sup \{|L(h)|\,:... • 9,016 0 votes ### Conway( A Course in Functional Analysis) Proposition 3.3 Let A=\{|L(h)|:\|h\|<1\}. Let B=\{|L(h')|:\|h'\|=1\}. Obviously we have$$\sup A\le \sup B$$because L is linear. And 0\le\sup B<\infty because L is bounded and because B\ne\emptyset.... 0 votes ### Continuity and norm of operator on l^2 If (\alpha_n) is bounded you have already shown that A is continous. So suppose that (\alpha_n) is not bounded, i.e. \forall C > 0 : \exists n \in \mathbb{N} : \vert\alpha_n\vert > C. By ... • 669 1 vote Accepted ### Prove ||S*T||_{op}=||S||_{op}||T||_{op} if T: R^n -> R^n is an isometry In ||S*T||_{op}\le||S||_{op} replace T by T^{-1} (permissible since T^{-1} is also an isometry) and replace S by S*T. You get \|S\|_{op}=\|(S*T)*T^{-1})\|_{op} \leq \|S*T\|_{op} • 8,896 3 votes Accepted ### Best lower bound for sum of vectors "close together" Consider X=\ell^\infty with \sup norm. For 0<\varepsilon <{1\over n-1} and 1\le k\le n let$$x_k=(1-\varepsilon){\mathbf 1}+\varepsilon\delta_k,$$where \delta_k denotes the sequence ... • 9,016 0 votes ### For every polynomial P of degree not greater than 2012 \underset{3\leq t \leq 4}{\max}|P(t)|\leq C \underset{0\leq t \leq 1}{\max}|P(t)|. \textbf{Edit:} As Ryszard Szwarc pointed out there was a mistake so now the second try. I think and explicit constant is easy to access threw watching shifted polynoms$$\underset{t \in [3,4]}{\max}|... 0 votes Accepted ### Completion of the space of bounded linear operators by completing the image. The conclusion is true in the case of separable inner product spaces. Assume$V$is a dense proper subspace of a separable Hilbert space$\mathcal{H}_1$and$W$a dense proper subspace of a separable ... • 9,016 1 vote ### How to understand the norm on the sum of two spaces? This is not a norm on$P_h$, since any nonzero piecewise constant function on the triangulation has norm equal to 0 (and so the homogeneity property required for a norm is not satisfied). You would ... • 11.3k 5 votes ### For every polynomial$P$of degree not greater than 2012$\underset{3\leq t \leq 4}{\max}|P(t)|\leq C \underset{0\leq t \leq 1}{\max}|P(t)|$. The space of polynomials of degree less or equal$2012$is finite dimensional (dimension is equal$2013).All norms on this space are equivalent. In particular the norms \|P\|_1=\max_{0\le x\le 1}|... • 9,016 1 vote ### Show that there is no distance preserving map between Mahattan norm and sup norm In (\mathbb{R}^3,\|\ \|_1) for x=(x_i) with \|x\|_1, x is a canonical basis iff there is only one shortest path from the origin to x : For instance, x=(x_1,x_2,0),\ x_i\neq 0, then c(t)=... • 18.7k 1 vote Accepted ### Lower bound of norm squared This answer is to the clarified version of the question in your comment. We can show that no, we cannot write C_1 as a function of \|x_1\|, \ldots, \|x_n\|. Take, for example, the space X = \Bbb{... • 45.1k 2 votes ### Prove that \sqrt{1-\|u\|_2^2} \cdot \sqrt{1-\|v\|_2^2} \le 1 - |\langle u,v \rangle| The following applies to your V=\mathbb R^n, but more generally to any prehilbert space. Let a=\|u\| and b=\|v\|. Then, (1-|\langle u,v \rangle|)^2\ge(1-ab)^2 (by Cauchy-Schwarz) and (1-ab)^... • 2,929 2 votes ### Prove that \sqrt{1-\|u\|_2^2} \cdot \sqrt{1-\|v\|_2^2} \le 1 - |\langle u,v \rangle| It is even true for any inner product space. \begin{align*} (1-\|u\|^{2})(1-\|v\|^{2})&=1-(\|u\|^{2}+\|v\|^{2})+\|u\|^{2}\|v\|^{2}\\ &\leq 1-2\|u\|\|v\|+\|u\|^{2}\|v\|^{2}\\ &=(1-\|u\|\|v\|... • 53.9k 1 vote ### Prove that \sqrt{1-\|u\|_2^2} \cdot \sqrt{1-\|v\|_2^2} \le 1 - |\langle u,v \rangle| Let u = (u_1,u_2), v = (v_1,v_2). Observe the followings are true: 1). |u\cdot v|^2=|u_1v_1 + u_2v_2|^2\le ||u||_2^2||v||_2^2 . ( CS inequality) 2). |u\cdot v|=|u_1v_1+u_2v_2|\le |u_1v_1|+|... • 4,129 2 votes Accepted ### Prove L^p (1\leqslant p\leq +\infty) Norms on C[0,1] are Not Equivalent Hint:- Do you know that C[0,1] is dense in L^{p} ? . Now do you see the problem?. L^{p} is not equal to L^{q} if p\neq q . But if norms are equivalent then what can you conclude using the ... • 8,357 2 votes Accepted ### Get an estimate on L^{2}(0,1) No. Think of a sequence of f_n such that \| f_n\|_2=1 while \|g f_n\|_2\to 0. And you might as well assume g(x)= x^{\alpha} : the question is at the origin, whatever happens elsewhere could ... • 178 1 vote ### Strict/Strong convexity of non-Euclidean norm I hope this answer can be useful, I also gave it as an answer to that question: Strong convexity of squared \ell_p norm in Bregman divergence I think from that paper (not the original reference but ... 1 vote Accepted ### Another question on norms of products of Banach spaces No, certainly not. If you set V = 0 you are asking: if you take a Banach space W, with norm \| w \|, and then pick a second unrelated norm \| w \|', does there exist a constant C such that \... • 377k 0 votes ### Two questions on norms on Banach space products Let W be a Banach space. Assume the space W\times W is complete with respect to a norm \|(\cdot,\cdot)\|. Assume also that\|(x,0)\|\le c\|x\|,\quad \|(0,y)\|\le c\|y\|$$The space W\times W... • 9,016 1 vote Accepted ### How can I prove the following characterization : \langle u-\pi_ku,w-\pi_ku\rangle\leq 0? See the proof of (1)\Leftrightarrow(2) in the first section of https://fr.wikipedia.org/wiki/Th%C3%A9or%C3%A8me_de_projection_sur_un_convexe_ferm%C3%A9 • 2,929 2 votes Accepted ### Ruston's Theorem The key to this question is to recognize that it is about (strict) convexity. A simple observation is that if \|x\|\leq1, \|y\|\leq1, and \|tx+(1-t)y\|=1 for some t\in(0,1), then \|x\|=\|y\|=... • 184k 1 vote Accepted ### The Directional derivate is maximum when v points in the direction of the gradient of f. The gradient$$ \nabla f= \left( {\partial f\over\partial x}, {\partial f\over\partial y}, {\partial f\over\partial z}, ... \right).$At a particular point$a=(x,y,z,...)\$ the gradient is a vector ...
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