0
$\begingroup$

I want an answer not only in terms of normed vector spaces over the fields $\Bbb R$ or $\Bbb C$, if not a general answer over any kind of normed vector spaces.


Let the linear function

$$A:E\to F$$

Im stuck with this problem. My strategy was:

  1. Trying to prove that $A$ is continuous (I tried composition of continuous functions but I dont found an easy strategy with this, or just tried direct proof for the $\epsilon-\delta$ definition of continuity but again I failed.)

  2. Trying to prove that the set $S=\{x\in E:\|x\|_E=1\}$ is compact (I cant conclude that is complete or something useful to prove compactness.)

If I prove these two things then I can conclude that $A$ is Lipschitz. But I failed to conclude anything about this.


Another basic work around was directly trying to prove that $A$ is Lipschitz:

$$\frac{\|Ax-Ay\|_F}{\|x-y\|_E}=\frac{\|Az\|_F}{\|z\|_E}=\frac{\|\sum_{k=1}^n z_k Ae_k\|_F}{\|\sum_{k=1}^n z_k e_k\|_E}\le \frac{|z_{\max}|C}{{\|\sum_{k=1}^n z_k e_k\|_E}}$$

but for the last equality I failed to conclude a bound. I used above a basis for $E$ of finite dimension $n$.


Some other work around was trying to relate the arbitrary norms to some known norm as the maximum norm and the taxicab norm, then it is easy to check that

$$\|x\|=\left\|\sum_{k=1}^n x_k e_k\right\|\le \sum_{k=1}^n |z_k| \|e_k\|\le \begin{cases}C\sum|z_k|=C\|x\|_1\\Cn|x_{\max}|=Cn\|x\|_{\infty}\end{cases}$$

Above I wasnt sure that I can consider bases with $\|e_k\|=1$ over a non necessarily inner product space, this is the reason for $C$.

But these relations dont help me to conclude anything useful. Some help will be appreciated, thank you.

$\endgroup$
  • $\begingroup$ You need more information to conclude that $A$ is continuous. It in in finite dimensional spaces, but not necessarily otherwise. $\endgroup$ – copper.hat Nov 17 '16 at 19:29
  • $\begingroup$ @copper.hat what you mean? I dont understand. The problem is stated for finite dimensional normed spaces. $\endgroup$ – Masacroso Nov 17 '16 at 19:31
  • $\begingroup$ See math.stackexchange.com/questions/1648275/… $\endgroup$ – Hans Hüttel Nov 17 '16 at 19:37
  • $\begingroup$ Sorry, I can't read :-). $\endgroup$ – copper.hat Nov 17 '16 at 19:38
  • 2
    $\begingroup$ Are you familiar with the fact that all norms on a finite dimensional space are equivalent? $\endgroup$ – copper.hat Nov 17 '16 at 19:40
1
$\begingroup$

Note that I am using two norms below, $\|\cdot\|$ and a new norm $\|\cdot\|_1$.

Pick a basis $e_1,...,e_n$. Without loss of generality, assume that $\|e_k\| = 1$ and define $\|x\|_1 = \sum_k |x_k|$ where $x = \sum_k x_k e_k$. It is easy to see that $\|x\| \le \sum_k |x_k| = \|x\|_1$. In particular, $B_1(0,1) \subset B(0,1)$.

Now I claim that there is some $r>0$ such that $B(0,r) \subset B_1(0,1)$. If not, there there are $x_n$ such that $\|x_n\|_1 = 1$ but $\|x_n\| \to 0$. Since $D=\{ x | \|x\|_1 = 1 \}$ is compact, we can presume (by renumbering the sequence) that there is some $p \in D$ such that $\|x_n -p\|_1 \to 0$. Hence $\|x_n -p\| \to 0$, which means that $p = 0$, a contradiction. Hence there is some $r$ such that $B(0,r) \subset B_1(0,1)$. In particular, we have $\|x\|_1 \le {1 \over r} \|x\|$.

You have $\|A x\| \le \sum_k \|A e_k\| |x_k| \le B \sum_k |x_k| = B \|x\|_1 \le {B \over r} \|x\|$, where $B = \max_k \|A e_k\| $. Hence $A$ is bounded using the norm $\|\cdot\|$.

$\endgroup$
  • $\begingroup$ why $D$ is compact? $\endgroup$ – Masacroso Nov 17 '16 at 20:16
  • $\begingroup$ The map $\phi:\mathbb{K}^n \to E$ given by $\phi(x) = \sum_k x_k e_k$ satisfies $\|\phi(x)\|_1 = \|x\|_1$ (the first norm is on $E$, the second on $\mathbb{K}^n$. Since $\Delta = \{ y \in \mathbb{K}^n | \sum_k |y_k| = 1 \}$ is compact, we see that $D= \phi(\Delta)$ is compact. $\endgroup$ – copper.hat Nov 17 '16 at 20:48
1
$\begingroup$

Hints:

  • Let $\{v_1,\ldots,v_n\}$ be a basis of $E$ and prove that $(E,\|\cdot\|_1) \to (\Bbb R^n, |\cdot|_1)$, $x_1v_1 + \cdots + x_n v_n \mapsto x_1 e_1 + \cdots + x_n e_n$ is an isometry, where $\|\cdot\|_1$ is defined for $x =x_1v_1 + \cdots + x_nv_n$ by $\|x\|_1 = \sum_{i=1}^n |x_i|$. Use Heine-Borel to prove that the unit sphere in $\Bbb R^n$ is compact. Conclude that the unit sphere in $E$ for $\|\cdot\|_1$ is compact.

  • Show that any norm on $E$ is equivalent to the norm $\|\cdot\|_1$ (one direction is easy; for the other one you'll need the compactness of the unit sphere for $\|\cdot\|_1$, the continuity of a norm and the fact that a continuous function on a compact set admits a minimum).

  • Prove that any linear map $L: E \to F$ satisfies: $\exists M > 0$ such that $\|Lx\|_F \le M \|x\|_1$ for all $x \in E$.

$\endgroup$
  • $\begingroup$ The first point is not granted in the book Im following. I will check some proof or try by myself. $\endgroup$ – Masacroso Nov 17 '16 at 20:20
  • $\begingroup$ @Masacroso see the edit. This should make it easier for you. $\endgroup$ – user384138 Nov 17 '16 at 20:35
  • $\begingroup$ I see... Thank you, I was trying these three points (the first not exactly by an isometry). Just a little question: the isometry work for any arbitrary field for both $E$ and $\Bbb R^n$, right? Not necessarily the field must be $\Bbb R$. $\endgroup$ – Masacroso Nov 17 '16 at 20:44
  • 1
    $\begingroup$ @Masacroso the field has to be $\Bbb R$ or $\Bbb C$, otherwise you can't use Heine-Borel. $\endgroup$ – user384138 Nov 17 '16 at 20:48
  • $\begingroup$ Then we can generalize this result for arbitrary normed vector space with arbitrary field? $\endgroup$ – Masacroso Nov 17 '16 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.