5
$\begingroup$

Let $X$ be a finite-dimensional vector space over $\mathbb{F}$. ($\mathbb{R}$ or $\mathbb{C}$)

Theorem: All norms on $X$ are equivalent.

Proof: $a_k$s and $c_k$s will refer to elements of $\mathbb{F}$. Let $(e_k)_{k=0}^{n-1}$ be a basis of $X$. Define $\lVert \sum_{k=0}^{n-1} c_k e_k \rVert_1 = \sum_{k=0}^{n-1} |c_k|$. Evidently $\lVert \cdot \rVert_1$ is a norm. Let $\lVert \cdot \rVert$ be another norm on $X$. Then if we define $c_0 = 1/\max(\lVert e_k \rVert)_{k=0}^{n-1}$, for $x = \sum_{k=0}^{n-1} a_k e_k$, $$ c_0\lVert x \rVert = c_0\left\lVert \sum_{k=0}^{n-1} a_k e_k \right\rVert \leq c_0 \sum_{k=0}^{n-1} |a_k| \lVert e_k \rVert \leq \sum_{k=0}^{n-1} |a_k| = \lVert x \rVert_1 $$ Conversely, observe that if $c_1 > 0$ so that $\lVert x \rVert_1 \leq c_1 \lVert x \rVert$, (We may assume that $\lVert x \rVert_1 \neq 0$ by the identity of indiscernibles) $$ 1 = \frac{\lVert x \rVert_1}{\lVert x \rVert_1} \leq \frac{c}{\lVert x \rVert_1} \lVert x \rVert = c_1\left\lVert \frac{x}{\lVert x \rVert_1} \right\rVert $$ so it suffices to show that $\lVert\cdot\rVert$ is bounded below on the unit sphere $S =\{x \in X\mid \lVert x \rVert_1 = 1\}$.

However, this is the part I'm having trouble with. Obviously if $\mathbb{F} = \mathbb{R}$, I can use the fact that every bounded set in $\mathbb{R}^n$ is totally bounded, and thus show the compactness of $S$.(which implies every sequence of norms on $S$ has a convergent subsequence) But what if $\mathbb{F} = \mathbb{C}$? How does the argument work here? Also, the sequence of norms mentioned earlier converges iff $\lVert\cdot\rVert$ is continuous, which is what I'm trying to show.

$\endgroup$
  • $\begingroup$ You have shown that the natural map $(X,\|\cdot\|_1)\rightarrow(X,\|\cdot\|)$ is continuous. The unit sphere $S_1 = \{ x\in X : \|x\|_1 = 1 \}$ is compact. Therefore $S_1$ is compact in $(X,\|\cdot\|)$. Because $0 \notin S_1$, this gives the existence of $\alpha$ such that $0 < \alpha \le \inf_{x\in S}\|x\|$. Hence, $\alpha \le \|\frac{1}{\|x\|_1}x\|$ for $x \ne 0$, or $\|x\|_1 \le \frac{1}{\alpha}\|x\|$. $\endgroup$ – DisintegratingByParts Feb 24 '16 at 5:19
  • $\begingroup$ I realised that the $1$-norm is not the usual Euclidean norm on $X$, so how is $S_1$ compact? $\endgroup$ – Henricus V. Mar 1 '16 at 17:21
  • $\begingroup$ Every coordinate is bounded, and there you can find a convergent subsequence. Take the new subsequence and do the same in the next coordinate. Etc. until you have a convergent subsequence. Show that the limit is on unit sphere in the $1$ norm. $\endgroup$ – DisintegratingByParts Mar 1 '16 at 18:01
0
$\begingroup$

The same argument work, because compact set in $\mathbb{C}^n$ are exactly closed and bounded set (think about $\mathbb{C}^n$ as $\mathbb{R}^{2n}$), but you can also avoid all this argument by using Open mapping theorem as above :

the identity maps : $$ \begin{array}{c} I : (E,\|.\|_1) \rightarrow (E, \|.\|) \end{array} $$
Is surjective and continuous (that what you have proved) and the space $E$ is a Banach space for $\|.\|_1$ so it inverse will be continuous.

$\endgroup$
  • $\begingroup$ I think the continuity of identity map only implies that the norms induce the same topology, but not equivalent ($\exists c_0,c_1 > 0\mid \forall x \in X, c_0 \lVert x \rVert \leq \lVert x \rVert_1 \leq c_1 \lVert x \rVert$). I'm using the "strong equivalence" definition of wikipedia. $\endgroup$ – Henricus V. Feb 24 '16 at 2:14
  • $\begingroup$ The are equivalent in the case of normed space because saying that the identity is an homeomorphism mean that (using the fact that $I$ is linear) the norms will be equivalent $\endgroup$ – Hamza Feb 24 '16 at 2:17
  • $\begingroup$ see remarks (1) page 28 math.washington.edu/~greenbau/Math_554/Course_Notes/ch1.2.pdf $\endgroup$ – Hamza Feb 24 '16 at 2:20

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.