Tag Info

Hot answers tagged

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 ...


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 ...


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.


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 $g(t)=t^2(1-t)^2$.


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 ...


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 ...


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 ...


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 ...


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

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


2

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


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

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 ...


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 ...


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 ...


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

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 ...


2

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


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

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 ...


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, $||T|| \le \sup_{||v|| = 1} ||Tv||$


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: ...


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 ...


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.


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 ...



Only top voted, non community-wiki answers of a minimum length are eligible