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3

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 ...


0

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 ...


1

Its not defined a norm on $C^1[0,1]$ since, for example $\|1\|=0.$


1

We have that $$ (fg)^{(k)}(x)=\sum_{j=0}^k\binom{k}{j}f^{(j)}(x)g^{(k-j)}(x), $$ hence $$ \lvert (\,fg)^{(k)}(x)\rvert\le\sum_{j=0}^k\binom{k}{j }\lvert\, f^{(j)}(x)\rvert\lvert g^{(k-j)}(x)\rvert\le \sum_{j=0}^k\binom{k}{j }\max_{x\in[a,b]}\lvert\, f^{(j)}(x)\rvert \cdot \max_{x\in[a,b]}\lvert\, g^{(k-j)}(x)\rvert, $$ and thus $$ ...


1

$C^n([a,b])$ is the space of the functions over $[a,b]$ such that $f^{(n)}$ is continuous (otherwise, we would not be able the take a $\max$ in the definition of the norm, but just a $\sup$). Anyway, $$\frac{1}{k!}\frac{d^k}{dx^k}(f\cdot g)=\sum_{j=0}^{k}\frac{f^{(j)}}{j!}\cdot\frac{g^{(k-j)}}{(k-j)!}$$ while: $$\|f\|\cdot\|g\| = \sum_{0\leq a,b\leq ...


2

The right naming here would not be "Heine-Borel" but "Banach-Alaoglu".


1

The metric on X is induced by the norm, so that if $x = \{x_i\}$ and $y = \{y_i\}$ are elements of X, then $d(x,y) = \sup_{i \in \mathbb{N}} |x_i - y_i|$. Now I claim that $X_{++}$ is not an open set. To show a set in a metric space is open, I have to show that for any element in that set, I can find an open ball around that element entirely contained in ...


5

In a unital Banach algebra the set of invertible elements is open and nonvoid, hence the set of noninvertible elements is closed ( and also nonvoid, since it contains zero). Any proper ideal is contained in the set of noninvertible elements. Hence, no dense proper ideals in a unital Banach algebra.


1

Two of the three properties of a metric are obviously satisfied by $D_1$, namely that $D_1(x,y)\ge0$ and $D_1(x,y)=0$ if and only if $x=y$ $D_1(x,y)=D_1(y,x)$ because they hold for $D$. Also, from $$ \frac{a}{1+a}\le 1,\qquad \frac{a}{1+a}\le a $$ which are easily verified for $a\ge0$, you can derive $$ \frac{a}{1+a}\le \min\{1,a\} $$ Now set $a=D(x,y)$ ...


2

A plan for show this kind of non-inequality is to show a sequence of functions where the left side is bounded, but the right diverges. Therefore there can be no such constant. Can you think of a simple sequence of functions that are bounded in their extreme values, but have unbounded slopes?


0

Take any convergent sequence $(x_n)$of points on this line. Since each $x_n$ is on this line there is a sequence $(t_n)$ such that $x_n=a+t_nd$ where $0 \le t_n \le 1.$ Show that limit of $x_n$ is also on this line. Then you will have the closedness.


0

To show a set is closed, you just have to take an arbitrary point not in that set and find an open neighborhood of that point which does not intersect your set. So, is there a point not on the line but such that every open neighborhood of the point meets the line? If so, what point?


17

A compact set must be bounded. Otherwise we can take $\{ x_n \}_{n=1}^\infty$ such that $\| x_n \| \geq n$. This will have no convergent subsequence, which we can prove by showing that it has no Cauchy subsequence. A compact set must be closed. Otherwise we can pick a sequence which converges to a point in the closure which is not in the set. This will ...


4

The following result is available: Let $(X,\lVert\cdot\rVert)$ be a Banach space and $S$ a subset of $X$. Then $S$ is compact if and only if the following two conditions hold: $S$ is bounded; for each positive $\varepsilon$, there exists a finite dimensional space $F=F(\varepsilon)$ such that for all $x\in S$, we have $d(x,F)=\inf\{\lVert ...


5

Hint: Show that $$\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{1-\sqrt{5}}{2}\right)^n$$ is an integer. The second term has limit $0$. Remark: There are many ways to show that our expression $\alpha^n+\beta^n$ is an integer. One way is by induction, using the fact that ...



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