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1 vote

For every normed algebra $\mathbb K$ with $\dim(A)\lt\infty$, prove that $x\mapsto x^{-1}$ continuous on the set of all inverse of $A$

$A^\times$ 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\|\...
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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 ...
-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 $T$ ...
-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.
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1 vote
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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,.....
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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 ...
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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-...
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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 ...
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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\|^...
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0 votes
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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 \\...
2 votes
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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 ...
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1 vote
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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 ...
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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\...
1 vote
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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 ...
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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 $0\lt\epsilon\le1$. ...
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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)|\,:...
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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 $B\ne\emptyset.$...
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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 ...
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1 vote
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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 $\|S\|_{op}=\|(S*T)*T^{-1})\|_{op} \leq \|S*T\|_{op}$
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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 ...
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
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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 ...
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 ...
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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}|...
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)=...
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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{...
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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)^...
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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\|...
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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|+|...
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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 ...
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 ...
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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
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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 $\...
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 $W\times W$...
1 vote
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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
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2 votes
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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\|=...
1 vote
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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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