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0

This is not a proof of the general statement you ask. However, the answer is no if you're in an inner-product space and the sequence is orthonormal: If $u,v$ are orthonormal, then $$ |u-v|^2 = \langle u-v, u-v \rangle = |u|^2 + |v|^2 = 2 $$ as the cross terms cancel. Therefore, if $\{v_n\}_{n=1}^{\infty}$ is an orthonormal sequence, it will not be Cauchy. ...


0

How about this approach? I use @davcha 's idea to convert the matrix to vector m. so we have the following unconstrained optimization problem: $\min_{m} \|b+V\ m\|_2$ where V denotes $(v^{T}*A)$. Now by decomposing matrix $V=\tilde{V}\ T$, where $\tilde{V}^{T}\tilde{V}=I$, leads to $\min_{m} \|b+\tilde{V}\ \tilde{m}\|_2$ where $\tilde{m}=T\ m$. So ...


2

The inequality which you want to show is often called the Minkowski inequality. An unusual proof can be found here Link. Below I will show a standard one which involves Hölder's inequality. We want to show the triangle inequality, that is, that for any $f$, $g \in C([a, b])$ $$\begin{align*}\| f+g\|_p = \left(\int_{a}^b |f(x)+g(x)|^{p} \ ...


4

I think I got it. There is no such counterexample. I welcome and appreciate any comments on the proof below, which hinges on ideas presented by Klee (1952). $\textbf{Claim:}\quad$If $(X,\|\cdot\|)$ is a completely metrizable normed vector space, then it is a Banach space. Proof:$\quad$Suppose that $(X,\|\cdot\|)$ is a normed vector space and $d$ is such a ...


1

No it is not a norm. If $\|x\|\ne 0$, then $$ \varphi(2x)=\frac{\|2x\|}{1+\|2x\|}=\frac{2\|x\|}{1+2\|x\|}\ne 2\frac{\|x\|}{1+\|x\|}=2\varphi(x). $$


8

The answer is yes, any such space must be a Banach space. This result was proved by Victor Klee in 1952 and answered a question first asked by Banach in 1932. V. L. Klee, Invariant metrics in groups (Solution of a problem of Banach), Proc. Amer. Math. Soc., 3 (1952), 484–487.


1

This was an answer to the original question, which specifically asked for help memorizing the identity. You could try to remember it "in parts" $[f,g] = \|f+g\|^2-\|f-g\|^2$ $\langle f,g \rangle = \frac{1}{4}([f,g]+i[f,ig])$ If someone can offer a geometric interpretation of $[f,g],$ that would certainly help.


2

When you are given just a vector space $X$, then you have as many non-equivalent norms on $X$ as many non-isomorphic normed spaces you can find with the same linear dimension. This is the standard `structure transport' argument. For instance, suppose that you are given a vector space of dimension continuum. Then for each infinite-dimensional Banach space $Y$ ...


3

There are plenty of non-equivalent norms. Let $X$ be an infinite-dimensional normed space with norm $\|\cdot\|_X$. Let $Y$ be another normed space with norm $\|\cdot\|_Y$. Let $T\in \mathcal L(X,Y)$ be compact and injective. Then $$ \|x\|_T:=\|Tx\|_Y $$ is a norm on $X$. Moreover, $\|x\|_T \le \|T\|_{\mathcal L(X,Y)} \|x\|_X$. However, both norms cannot ...


1

When the metric is induced from a norm. This kind of metric space $(X,d)$ must satisfy $$ d(x+a,y+a)=d(x,y)$$ $$ d(\alpha x,\alpha y)=|\alpha|d(x,y)$$ for all $x,y,a\in X$,and scalar $\alpha$. And $X$ must be a vector space.


0

(1) is a composition of the next functions: $$\mkern-1em j_p:\Bbb R\longrightarrow \Bbb R^n$$ $$s\longmapsto sp$$ $$\mkern-1em T_z:\Bbb R^n\longrightarrow \Bbb R^n$$ $$v\longmapsto x+v$$ $$\mkern-1em N_q:\Bbb R^n\longrightarrow \Bbb R$$ $$v\longmapsto \|v\|_q$$ $$\mkern-1em S:\Bbb R\longrightarrow \Bbb R$$ $$t\longmapsto t^2$$ Compute the derivatives of ...


4

Assuming that the operator norm is taken with respect to the Euclidean norm on $\mathbb{R}^n$, we have $$\left\lVert \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}\right\rVert = 1 = \left\lVert \begin{pmatrix} 0 & 0\\ 0 & 1\end{pmatrix}\right\rVert,$$ hence $$2\left\lVert\begin{pmatrix}1 & 0\\0&0 \end{pmatrix}\right\rVert^2 + 2 ...


1

Since $X$ is an infinite dimensional normed space, we can use the Riesz lemma to find a sequence $u_n$ such that $\|u_n\|=1$ and $\|u_n-u_m\| > { 1\over 2}$ for all $m\neq n$. Let $x_n = (1+{1 \over n}) u_n$, and let $F = \{ x_n \}_n$. We see that $\|x_n\| = 1+ {1 \over n} $, and so $d(0,F) = 1 < \|0-y\|$, for any $y \in F$. It remains for us to show ...


0

Assume that $\{e_1,\ldots,e_k\}$ is a basis of $Y$. Let $$ d=\inf_{\lvert c_1\rvert+\cdots+\lvert c_k\rvert=1}\|c_1e_1+\cdots+c_ne_k\|. \tag{1} $$ Clearly, $d>0$, since the set $K=\{(c_1,\ldots,c_k):\lvert c_1\rvert+\cdots+\lvert c_k\rvert=1\}$ is compact, and $f(c_1,\ldots,c_k)= \|c_1e_1+\cdots+c_ne_k\|$ is continuous and non-vanishing on $K$. $(1)$ ...


2

Based on your statement of the problem, you don't assume that $X$ is closed. So I don't think it is clear on the front end that the limit of a sequence is an element of $X$. Instead here is a hint. Assume $(y_k)_{k=1}^\infty$ is a cauchy sequence. Express each $y_k = a_{1,k}e_1 + \cdots a_{n,k}e_n$ using your basis. Use the definition of cauchy to prove ...


0

Hint: If $x_n\rightarrow x$, then you reduce to the finite dimensional case by adding $x$ to the basis and using the inherited norm.


0

let $V=\{(t_1 ,t_2 , ... ,t_n )\in\mathbb{R}^{n+1} : t_1 , t_2 , ....,t_{n+1} \geq 0 \wedge t_1 + t_2 +...+t_{n+1} =1 \}.$ By Caratheodory theorem the function $T: V\times S^{n+1}\to \mbox{conv} (S) ,$ $$T(t_1 ,t_2 , ... t_{n+1} , x_1 , x_2 , ..., x_{n+1} ) =\sum_{j=1}^{n+1} t_j x_j $$ is suriection. And since $V\times S^{n+1}$ is compact then so is ...


0

Hint: consider the continuous map $f : \Delta_{n-1} \times S^{n} \rightarrow \mathrm{co}(S)$ defined by $f(\lambda_1, \ldots, \lambda_n, x_1, \ldots, x_n) = \sum_{i=1}^n \lambda_i \cdot x_i$, where $n = \dim X + 1$, and $\Delta_{n-1}$ is the simplex $\{ (\lambda_1, \ldots, \lambda_n) \in \mathbb{R}_+ : \sum_i \lambda_i = 1 \}$.


0

You can only if $X$ is bounded itself. This is because of the continuity of the norm. Consider a bounded dense part $B$, 2 sequences $x_n,y_n \in B$ such as $x_n\to x, y_n\to y$ and such as, $D$ being the diameter of $X$, $\|x-y\| > D- \epsilon$. Let $D_B$ be the diameter of $B$. $$ D_B\ge \|x_n-y_n\| \to \|x-y\| > D-\epsilon $$ using the continuity ...


1

Observe that $$ \{1\}\times [0,\infty)\subset \overline{ch(S)}, $$ while $$ \{1\}\times [0,\infty)\cap {ch(S)}=\varnothing. $$


0

It is enough to show that: $$\left\|x-y\right\|-\left\|w-z\right\|\leq \|x-w\| + \|y-z\| $$ $$\left\|w-z\right\|-\left\|x-y\right\|\leq \|x-w\| + \|y-z\| $$ $$\left\|x-y\right\|-\left\|w-z\right\|\leq \|x-z\| + \|y-w\| $$ $$\left\|w-z\right\|-\left\|x-y\right\|\leq \|x-z\| + \|y-w\|$$ and so on. But this follows directly from triangle inequality. For ...


1

i'm a bit out of my depth here, so apologies if this is nonsense, but i think your idea works. suppose $\|x\|=N$ and define $f_n=\sum_{j=0}^n \frac{x^j}{j!}$. this sequence converges absolutely for all $N$, but is not a polynomial.


5

Take $$ f_n(x)=\left\{\begin{array}{ccc} \frac{1}{x} & \text{if} & \frac{1}{n}\le x\le 1,\\ n^2x & \text{if} & 0\le x\le \frac{1}{n}. \end{array}\right. $$ Then $\{f_n\}_{n\in\mathbb N}\subset C[0,1]$, and for $m\ge n$, $$ \|f_m-f_n\|=\sup_{x\in [0,1]}x^2\lvert\,f_m(x)-f_n(x)\lvert \overset{\ast}=\sup_{x\in ...


0

To show $\lim_{n\to \infty} ||x_n||=||x_0||$, we need to show $\forall \epsilon >0,\exists N\in \mathbb{N}, s.t. \forall n\geq N$,$|(||x_n||-||x_0||)| <\epsilon$. Thus we try to find an upper bound for $|(||x_n||-||x_0||)|$. How can we do that? We try to use the assumption, which is $||x_n-x_0||\to 0$.Then how can we connect these two things? We try to ...


0

HINT: $| \, ||x|| - ||x_n|| \,| \leq || x - x_n ||$


2

As Daniel Fischer noted, condition (P) is equivalent to the norm being strictly convex. Indeed, if the norm is not strictly convex, then the unit sphere contains a line segment, and removing the midpoint of this line segment from the closed unit ball creates a nonconvex set. Conversely, if the norm is strictly convex, then any $X$ as you described is ...


0

This question has been answered in a comment: A Hilbert space doesn't have to be separable. The definition only requires that it's complete. – Svinepels Jan 14 at 17:31


2

$\psi$ is linear, so we need only check that $\psi$ is bounded. $|p(0)| = |\int_0^1 p(0) dx| = |\int_0^1 (p(0)-p(x) + p(x)) dx| \le \int_0^1 (|p(0)-p(x)| + |p(x)| ) dx$. Note that $p(0)-p(x) = \int_0^x -p'(t) dt$ (this is the key part of this solution), hence $|p(0)-p(x)| \le \int_0^1 |p'(t)| dt$. Then $|p(0)| \le \int_0^1 (\int_0^1 |p'(t)| dt ) + |p(x)| ...


0

Using $2ab\le a^2+ b^2$ and monotonicity of $x\mapsto \sqrt x$ we obtain the first $$ \|x\|_1 = \sqrt{ (|x_1|+|x_2|)^2} = \sqrt{ |x_1|^2+2|x_1|\cdot|x_2|+ |x_2|^2} \le \sqrt{ 2|x_1|^2+2|x_2|^2} = \sqrt2\|x\|_2. $$ Again by monotonicity $$ \|x\|_2 = \sqrt{ |x_1|^2+|x_2|^2} \le \sqrt{ \|x\|_{max}^2+\|x\|_{max}^2} = \sqrt2\|x\|_{max}. $$ Since $|x_i|\le ...


1

If $g\in B_\infty(f,\epsilon)$ then $||f-g||_\infty<\epsilon$ and consequently, $||f-g||_1=\int_0^1|f-g|dx<\epsilon$.So $||f||_1-\epsilon \leq||g||_1\leq||f||_1+\epsilon$. Therefore if you take $\epsilon=\frac{1-||f||_1}{2}$, you obtain $B_\infty(f,\epsilon)\subset B_1(0,1)$.


0

Assuming $A\in\mathbb{C}^{n\times n}$ with columns $a_1,\ldots,a_n$ and that you already know the equivalence constants for vector $p$-norms: For $\|A\|_{\infty}\leq\sqrt{n}\|A\|_2$: We have for all $x\in\mathbb{C}^n$, $\|x\|_{\infty}\leq\|x\|_2\leq\sqrt{n}\|x\|_{\infty}$, so $$ \|A\|_{\infty}=\max_{x\neq 0}\frac{\|Ax\|_{\infty}}{\|x\|_{\infty}} ...


1

First I want make a remark on the fact of not specifying the space on which we work(your matrices are square or not?), it is also important to give us your norms definitions, so your question will be self-understood, especially if you consider the differences found in the notations used in literature. In case you are not restricted to search for a ...


1

You have, for $x$ with $\|x\|_2=1$ and using Cauchy-Schwarz, $$ \|Ax\|_\infty=\max_k\,\left|\sum_{j=1}^nA_{kj}x_j\right|\leq\max_k\left(\sum_{j=1}^n|A_{kj}|^2\right)^{1/2}=\max_k \|R_k(A)\|_2, $$ where $R_k(A)$ is the $k^{\rm th}$ row of $A$. Now suppose that we fix $k_0$ such that $\|R_{k_0}(A)\|_2$ is maximum among the rows. Let ...


2

The gradient of a function is a linear functional that gives an approximation to the function: $$ f(x+h) = f(x) + \color{blue}{\nabla f(x) h} + o(\|h\|), \quad \|h\|\to0 $$ (I use Fréchet gradient definition here.) In scalar case, $\nabla f(x) h$ is just $f'(x)h$, multiplication of two numbers. In $\mathbb R^n$, it is $\nabla f(x) \cdot h$, inner ...


0

Let $z\in \overline Z$, then there exists $z_n\in Z$ such that $z_n\to z$. $z_n=y_n+\alpha_n x$ with $y_n\in Y$ and $\alpha_n\in \Bbb K$. If $\alpha_n$ is bounded, then it admit a convergent subsequence $\alpha_{n_j}$, and denote $\alpha$ its limit. We get $y_{n_j}=z_{n_j}-\alpha_{n_j}x$ which is convergence, as $Y$ is closed , we have $y_{n_j}\to ...


1

The induced norm of the matrix $A$ as a map from $(\mathbb R^n , \Vert \cdot \Vert_1)$ to $(\mathbb R^n, \Vert \cdot \Vert_2)$ is given by $$ \max_{j=1, \dots ,n} \Vert A^j \Vert_2, $$ where $A^j$ is the $j^{th}$ column of $A$. You can find this as equation (2c) in "On the Calculation of the $l_2 \to l_1$ Induced Matrix Norm" Konstantinos Drakakis and ...



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