1
$\begingroup$

The proposition I have been trying to prove is that the set $A=\{x\in E:N(x)=1\}$ is a compact subset of the (real) finite-dimensional vector space $E$ for any norm $N:E\to \mathbb{R}$.

I am reading the proof of this proposition from a textbook of topology. Briefly, the main reasoning of the proof is as follows. There exists a norm $N^{'}(x)$ such that the metric space $(E,\rho^{'}),~\rho^{'}(x,y)=N^{'}(x-y)$, is isometric with the Euclidean metric space $(\mathbb{R}^n,|\cdot|)$. Furthermore, the set $B=\{x\in\mathbb{R}^n:|x|=1\}$ is a closed and bounded subset of $\mathbb{R}^n$, therefore it is compact. Given that compactness is a topological property and that $B$ is compact, the set $C=\{x\in E:N^{'}(x)=1\}$ is also compact.

It is further known that all the norms of $E$ are equivalent, given that $E$ is finite-dimensional. It is finally claimed in the proof that since i) $N$ and $N^{'}$ are equivalent norms, ii) compactness is a topological property and iii) $C$ is compact, $A$ is compact too.

This is all understood apart from one point; I have to prove that there exists a homomorphism $f:A\to C$. I don't see how this follows from the equivalence of norms $N$ and $N^{'}$.

I am aware that equivalence of norms implies equivalence of metrics, which means that the metric spaces $(E,\rho),~\rho(x,y)=N(x-y)$, and $(E,\rho^{'})$ have the same topology, but this hasn't helped me construct the homomorphism of interest.

I thought of defining the function $f:A\to C$ as $f(x)=\frac{x}{N^{'}(x)}$, which can be shown to be bijective, but I can't prove that it is continuous, so as to prove that it is homomorphic (in fact I am not even sure if it is continuous).

I am trying to prove this proposition without constraining the problem to inner product spaces (aka Hilbert spaces).

$\endgroup$
  • $\begingroup$ Just define $f \colon (A, N) \to (C, N^\prime)$ as $f(x) \colon= x$ for all $x \in A$. To prove continuity, use the equivalence of the norms $N$ and $N^\prime$. $\endgroup$ – Saaqib Mahmood Aug 12 '14 at 14:15
  • $\begingroup$ You mean to define $f$ to be the identity function? The $f$ you proposed is certainly continuous, as this follows from the fact that $(E,N)$ and $(E,N^{'})$ have the same topology. However, the identity function does not map a vector $\mathbb{x}\in A$ to a vector in $C$, since $N^{'}(x)\ne 1$ in general. $\endgroup$ – scidom Aug 12 '14 at 14:39
0
$\begingroup$

If $A$ has the norm $N$ and $N'$ is an equivalent norm, $N'$ is continuous. Since $N'(x)\neq 0$ for all $x\in A, f$ is continuous as well. A similar argument can be used to show the inverse is continuous.

There is an easier way however. Since $E$ is finite-dimensional, there is an isomorphism $L:E\rightarrow \mathbb{R}^n$ for some $n$. If $N$ is the norm on $E$, we can define a norm $N'$ on $\mathbb{R}^n$ by $N'(x):=N(L^{-1}(x))$ (you have to check this is a norm). Then $B=\{x\in \mathbb{R}^n:N'(x)=1\}$ is closed and bounded, since all norms on $\mathbb{R}^n$ are equivalent, hence B is compact. But then so is $A$, since $A=L^{-1}(B)$ and $L$ is an isomorphism.

$\endgroup$
  • $\begingroup$ Thanks Vincent, your answer completed the proof. You are right that $f(x)=\frac{x}{N^{'}(x)}$ is continuous (it suffices to show that $x$ and $\frac{1}{N^{'}(x)}$ are continuous). It is not needed to show that the inverse is continuous, because it holds in general that if a function $f:E_1\to E_2$ is bijective and continuous and $E_1$ is compact, then $f^{-1}$ is also continuous. $\endgroup$ – scidom Aug 13 '14 at 13:14

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.