Tag Info

New answers tagged

1

The spectral radius is the biggest eigenvalue of a matrix. There is a linked norm called the spectral norm, which is infact the square root of the biggest eigenvalue of the matrix $A^*A$. (So not linked to its own spectral radius, but to the spectral radius of $A^*A$) Norms are always convex. Due to triangle inequality and to the homogeneity. Edit: ...


0

This is true for any linear normed spaces. Let $T:X \to Y$ be a linear operator and $X,Y$ be two normed spaces, then 1- T is continuous iff T is bounded. 2- If T is continuous at a single point $x_0$, then it is continuous. the proof of second ostatement follows from the first statement, and here the details if you want: Assume $T$ is continuous at an ...


0

$||T|| := \inf\limits_{C \geq 0} \{ C: ||Tv|| \leq C||v||, \ \forall v \in V\} = \inf\limits_{C\geq 0}\{C: ||Tv|| \leq C, \forall ||v|| = 1\}$. We established the last equality in comments made above. So if $||T|| \gt \sup\limits_{||v|| = 1} ||Tv||$, then there exists $||Tv|| \gt \sup\limits_{||v|| = 1} ||Tv||$ with $||v|| = 1$, impossible!


1

For the other direction, if $v \neq 0$, then $||Tv|| = ||v|| \cdot||T( \frac{v}{||v||})|| \le ||v|| \cdot \sup_{||v|| = 1} ||Tv||$. So, $||T|| \le \sup_{||v|| = 1} ||Tv||$


0

The Heine-Borel theorem plays a crucial role in the development of both Riemann and Lebesgue integration on $\mathbb{R}^d$. Perhaps the most important result in Riemann integration is that continuous functions on closed intervals $[a,b]$ are integrable. The reason is that $[a,b]$ is compact by Heine-Borel, so continuous functions on it are uniformly ...


0

Let $\{v_1,...,v_n\}$ be a basis of $V$. Define $\Vert v \Vert_1:=\sum |a_i|$ provided that $v:=\sum a_i v_i$. Given an arbitrary norm $\Vert\cdot\Vert$, it holds that for $z\in V$, it holds that $\Vert z\Vert \leq \sum |a_i|\Vert v_i\Vert$. $\Vert z \Vert\leq \Vert z\Vert_1 \max \Vert v_i\Vert $. Take $M:=\max \Vert v_i\Vert$. Conversely, we need some ...


0

Let $x=(x_1,\dots,x_n)\in\mathbb{R}^n$ and $||x||_3^3=\sum_{i=1}^n|x_i|^3$. Using the fact that $\forall 1\leq i\leq n$, $|x_i|\leq||x||_{\infty}$, we can chose $m=n^{-\frac{1}{3}}$. On the other hand: $$||x||_{\infty}^3=\left(\max_{1\leq i\leq n}|x_i|\right)^3=\max_{1\leq i\leq n}|x_i|^3\leq\sum_{i=1}^n|x_i|^3=||x||_3^3$$ So you can chose $M=1$.


0

We have the following: Suppose $z:=(x,y)\in R^2$. First prove that $| x |^3 + | y |^3\leq(| x | + | y |)^3$. Therefore, $2 \Vert z\Vert_\infty \geq \Vert z\Vert_1 \geq \Vert z\Vert_3$. On the other hand, $|x|^3+|y|^3\geq \Vert z \Vert_\infty^3$. Thus, $\Vert z \Vert_3 \geq \Vert z \Vert_\infty$. Summing up: $2 \Vert z\Vert_\infty \geq \Vert z\Vert_1 ...


2

It is enough to show that when one of the norms is 1, the other is bounded between two positive constants, $m$ and $M$. It is easiest to look at the case when $||(x,y)||_\infty=1$. What does that equation tell you about $x$ and $y$? You can use this to bound $||(x,y)||_3$. In terms of the shape of $||.||_3$, you really just want to sketch the curve ...


3

To elaborate on my comment on Michael's answer: The symbol $\left\Vert\mathbf{u}\right\Vert$ for a vetor $\mathbf{u}$ usually stands for the norm of that vector. A norm is "a function that assigns a strictly positive length or size to each vector in a vector space" (quoted from wikipedia). Having a normed vector space enables you to talk about e.g. the ...


1

The norm of a vector $(1,3,4,11,13)$ is $\sqrt{1^2+3^2+4^2+11^2+13^2}$. It is an extension of Pythagoras' Theorem.


2

Since it seems that you want to solve it yourself, I'll just give you a hint. If you want a more accurate answer, just leave a comment. HINT: In a normed space the unit open balls is the set of all elements which have a norm strictly less than 1. Now, what does it mean that $\left\Vert f \right\Vert_1 < 1$? What does it mean that $\left\Vert f ...


2

No, and a counter-example can come from the simple cases. Take $X=\mathbb{R}^2$, $v_0=\hat{i}$ and $v=\hat{i}+\hat{j}$. In this case, $v=v-v_0+v_0=(1,1)$ and so $\|v-v_0+v_0\|=\sqrt{1+1}=\sqrt{2}$ while $\|v-v_0\|=\|\hat{j}\|=1$ and $\|v_0\|=1$. In fact I think the equal sign holds only when $v-v_0$ is a non-negative multiple of $v_0$. For the Pythagorean ...


3

Take $\mathbb R$ over $\mathbb R$ endowed with the absolute value norm, $v = 1$, and $v_0 = 2$. Then $$1 = |v| = |v - v_0 + v_0| \neq |v - v_0| + |v_0| = 1 + 2 = 3.$$ For inner product spaces in particular, $||x||^2 = \langle x, x \rangle.$ We have \begin{align*} ||x + y||^2 &= \langle x+y, x+y \rangle \\ &= \langle x,x \rangle + 2\langle x,y ...


1

I think an easy example of an infinite dimensional vector space would be to consider a sequence space. Concretely consider the space $\mathcal{l}^2(\mathbb{R})$, which consists of all infinite sequences $$ \mathbf{x}=(x_1, x_2, \ldots )$$ with elements $x_i \in \mathbb{R}$. You could view this as vectors, just that you have infinitely many coordinates. ...


1

One standard example the vector space of all real valued continuos functions defined on $[0,1]$, let's call it $V$. Convince yourself that $V$ indeed is a vector space if you define the sum $f+g$ for $f$ and $g$ in $V$ to be the function $x\mapsto f(x)+g(x)$ and $cf$ is defined by $x\mapsto cx$ for any real $c$. One may define an inner product of $V$ by ...


0

This depends very much on convention. If the basis is indexed by lower indices, the tensor product should be lower indices too. One could write the coefficients with upper indices and omit the summation sign like so: $$K = k^{ij}e_i\otimes e_j.$$ That would be the Einstein summation convention which is quite popular in mathematical physics. You can also ...


2

This is a matter of convention that differs between literatures and depending on what you're doing. If you're going to use raised indices in a meaningful way, then I'd expect to see the basis written as $$ B = \{ u^i \}_{i=1}^{\infty} $$ with the index up on vectors. Then your second option for $K$ would be correct. Since you wrote the vectors in the ...


3

Almost: you got a sign wrong. Consider expanding the Leibniz formula for the determinant $\det(\lambda I - A)$ (which is either $f(\lambda)$ or $(-1)^n f(\lambda)$ depending on your convention). The coefficient of $\lambda^j$ comes from cases where you take $j$ of the $\lambda$'s and $n-j$ factors that are entries of $A$. In particular the coefficient of ...


4

Hint: choose any constant function $g$ on the interval $[a,b]$. Then $g \in B$ but $g$ is not the zero function.


4

(a) Consider $g(t)=1$. (b) Consider $g(t)=t^2(1-t)^2$.


0

Balls in general have to be convex. I dont think its possible.


3

A necessary and sufficient condition for a subset $B$ of a vector space $V$ to the unit ball of a norm on $V$ is that $B$ is non-empty, convex and symmetric about the origin ($-B = B$). If $B = \{(x, y) : x^2-1 \le y \le 1-x^2\}\subseteq\mathbb{R}^2$, then $B$ is non-empty, convex and symmetric about the origin and so it is the unit ball of a norm on ...


1

Note that $\operatorname{arg max}(|x|)=\operatorname{arg max}(|y|)$ is a necessary condition to $\max|x_i+y_i|=\max|x_i|+\max|y_i|$. Here, $\operatorname{arg max}(|x|)= k$ such that $\max(|x|)= |x_k|$ Now,$|x_k+y_k|=|x_k|+|y_k|$ . Then either $x_k$ and $y_k$ have the same sign or at least one of then is 0. In the last case, $x=[0,\dots,0]$ or ...


0

If you sum $n$ pairs of numbers, the maximum of the $n$ results is the sum of the maximums if you had to add precisely these two maximums, and if they had the same sign.


0

Can't you just say that since $A$ is weakly closed then $A$ is strongly closed, then $A$ is sequentially closed? Cause I think every weakly closed set is strongly closed and in metric spaces (we are in a normed space which means in a metric space) (strongly) being closed set is equivalent to being sequentially closed.


2

In general there is no such surjection (the C*-case is quite special in that respect). Indeed, let $V=c_{00}$ be the vector space of finitely supported vectors. For $(\xi_n)$ in $V$ define $\|(\xi_n)\|_1 = \sup_n |\xi_n|$ and $\|(\xi_n)\|_2^p = \sum_{k=1}^\infty |\xi_k|^p$ for some fixed $p\in (1,\infty)$. Conspicuously, the respective completions are ...


2

Given the Banach spaces $\mathcal{c}_0$ and $\ell^\infty$. Consider the identity: $$T_0:\mathcal{c}_0\to\mathcal{c}_0:\quad T_0:=\mathbb{1}$$ It has No continuous extension: $$T:\ell^\infty\to\mathcal{c}_0:\quad T\restriction_{\mathcal{c}_0}=T_0$$ For the details see: Werner


1

No. Consider $F=\mathbb R^2$. And $T(e_1)=e_1$, $T(e_2)=2 e_2$. The subspace generated by $e_1$ could be your $E$. The norm of $T_0$ then is 1 and the norm of $T$ is 2. Even better: Let $F$ be some Banach space and $T:F\to F$ a bounded non-zero operator. Take $E=\{0\}$. Then we have norm 0 for $T_0$ and a positive norm for $T$ because $T$ is non-zero. ...


2

Setup and a hint: Let $x \in F$. Let $y \in E \setminus F$, $\| y \|=1$. Then $\| x - (x+\varepsilon y/2) \|=\varepsilon/2<\varepsilon$. So if $x+\varepsilon y/2 \not \in F$, then $B_\varepsilon(x) \not \subset F$. Now show that for every $\varepsilon > 0$, $x+\varepsilon y/2 \not \in F$. My hint: suppose it is in $F$, and conclude that $y \in F$, ...


1

Since $(X, ||\cdot || ) $ and $ (X, ||\cdot ||' )$ are homeomorphic the topologies generated by these norms are identical. Hence the identity map $\mbox{Id} :(X, ||\cdot || ) \to (X, ||\cdot ||' )$ is continiuous so the norms are equivalent. And from the equivalence of the norms it easily follows your assertion.


1

Take a Cauchy sequence in $(X,||\cdot||)$ and show that is also a Cauchy sequence in $(X,||\cdot||')$. This proves the assertion since convergence is a topological property which is preserved by homeomorphisms.


1

As a sketch$\ldots$ Well if you've a homeomorphism $f$, then if $\{x_n\}_{n \in \Bbb N}$ is cauchy wrt $\|\cdot \|$, and converges to $x \in X$, then $\|x_n-x\|<\delta$ for $n$ sufficiently large, and $\|x_n-x_m\|<\delta$, for $n,m$ sufficiently large. Since $f$ is continuous we can make $\|f(x_n)-f(x_m)\|'<\epsilon$ for $\|x_n-x_m\|<\delta$, so ...


0

For the case $p\neq 2$ the famous parallelogram law dos not hold so the space $l^p$ is not Hilbert. It is a theorem in many functional analysis text book such as Erwin Kreszig (Intoductory functional analysis with application)


3

What it means is not that hard: We have a sequence of real numbers $\langle x_n,y_n\rangle$ which converges to a real number $\langle x,y \rangle$. It doesn't matter what your interpretation of what an inner product "means" is. So it means that the absolute value of $\left|\langle x_n,y_n \rangle - \langle x,y \rangle \right|$ gets as small as we like by ...


-2

I am not a mathematician but sometimes getting an answer from non-professional can help with intuition so here goes. The way I view it, spaces are houses made of a set of building blocks that can change shape. So take the most generic rectangular brick and build a house. Then change the shape of the brick to something else, lets say a rhombus. Then the ...


4

The relaxed condition also implies $$ \left\|\frac1\alpha \alpha x\right\|\le\left|\frac1\alpha\right|\|\alpha x\|$$ and hence $$ \|\alpha x\|\le |\alpha|\|x\|\le |\alpha|\left|\frac1\alpha\right|\|\alpha x\|=\|\alpha x\|,$$ which implies equality throughout.


1

If $X$ is a finite-dimensional complex normed space with basis $\{ x_1,x_2,\cdots,x_N \}$, then you can define $L : \mathbb{C}^{N} \rightarrow X$ by $$ L(\alpha_1,\cdots,\alpha_n)=\alpha_1 x_1+\alpha_2 x_2+\cdots +\alpha_N x_N. $$ This map is continuous because $$ \begin{align} \|L(\alpha_1,\cdots,\alpha_n)\|_{X} & \le ...


8

Remember the definition of equivalence is that there exists numbers $c,d$ such that $c\Vert x\Vert_2 \leq \Vert x\Vert_1 \leq d\Vert x\Vert_2$. What the c and d do in the picture is to stretch or shrink the shapes you've drawn. What the inequality represents is one shape fitting inside another. What equivalence means is that I could shrink the circle for ...


2

I'm not sure I understand this question. The distance is defined to be the norm of the difference, and the definition is then the same as it is in metric spaces. You can get geometric intuition from Euclidean space, where $\| x - y \|$ is the length of the line segment connecting $x$ and $y$. If you replace $x$ with another sequence $y_m$, the situation is ...


2

If $V$ is a vector space of finite dimension $n$, any collection of $n + 1$ vectors is linearly dependent. Using this property you can easily show that the vector spaces in the first posting are not finite-dimensional by proving the existence of a linearly independent set of arbitrary size. For example, let's take a look at $C([0, 1])$. For every natural ...


6

Given the axiom of choice, every vector space has a basis (though it will be a very unnatural basis), and you are correct that infinite-dimensional vector spaces are exactly those where the basis is infinite. But this kind of basis (often called a Hamel basis) is rather useless and impossible to visualize. So, a more concrete way of thinking about it might ...


1

So, (1) is wrong as you stated it. The key inequality is simply $$ \|\nabla f(c)\| \leq \sup \|\nabla f\|. $$ (2) is rather trivial: with the choice of $h_0$ you get $$ \langle f(b)-f(a),h_0 \rangle = \frac{\|f(b)-f(a)\|^2}{\|f(b)-f(a)\|} = \|f(b)-f(a)\|. $$ Finally, let me remark that $X$ looks like an inner product space rather than a general normed space. ...


3

Of course that Cauchy Schwarz works ! $$\left<a,i\right>\underset{C.S.}{\leq} \|a\|\underbrace{\|i\|}_{\leq 1}\leq \|a\|$$


0

You can write $$ \|T\|_{\rm op}=\inf\{c>0: \|Tv\|\leq c,\ v\in V,\ \|v\|\leq1\}. $$ If $\|x\|<1$, then $\|\epsilon\,x\|<\epsilon$, and so $\|T(\epsilon x)\|\leq k$, i.e. $$ \|Tx\|\leq\,\frac k\epsilon. $$ Thus, $$ \|T\|_{\rm op}\leq\frac k\epsilon. $$ Technically, you still need to adjust for $\|x\|\leq1$ instead of $\|x\|<1$, but that is not a ...


2

This is true for any function f--the norm isn't allowed to take on infinite values.


1

By contradiction. Assume $f(x_1)\neq f(x_2)$. Define $y=x_2-x_1$, $g(t) = f(x_1+ty)$. You know that $g'(t)=f'(x_1+ty;y)=0$ for all $t\in[0,1]$. Lagrange theorem gives $g(1)-g(0) = g'(s)(1-0)$, with $s\in[0,1]$. But $g'(s)=0$ and $g(1)-g(0)=f(x_2)-f(x_1)\neq 0$. Contradiction.


0

Try writing hf'(x,y)=f(x+hy)-f (x) then as you know f'(x,y) always exists and is 0 make an argument for why f (x+epsilon×y)=f (x) for some epsilon.


2

As pointed out by Clement C in the comments: Equivalent norms induce the same topology. Also the other direction of implication is true: When two norms induce the same topology then they are equivalent. Take two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a vector space and you ask yourself: When are the topologies from those norms the same? This is the case ...


2

Some application of this result and more generally that all norms are equivalent on finite dimensional spaces: The compact subspaces are the closed bounded spaces. All linear maps are continuous. More generally all multilinear maps are continuous. All linear maps are bounded on the unit ball.



Top 50 recent answers are included