# All norms defined on a finite dimensional normed linear space are equivalent

Given that $$E$$ is a finite dimensional normed linear space. Let $$\dim E=n\geq 1$$ and $$\{e_i\}^{n}_{i=1}$$ be a basis for $$E.$$ Then, there exists unique scalars $$\{\alpha_i\}^{n}_{i=1}$$ such that \begin{align}x=\sum_{i=1}^{n}\alpha_i e_i.\end{align} PROOF

I proved here Prove that $$\| \cdot \|_0$$ defined by $$\| x \|_0=\max\limits_{1\leq i\leq n}|\alpha_i|$$ is a norm on $$E$$. that $$\| \cdot \|_0$$ defined by $$\| x \|_0=\max\limits_{1\leq i\leq n}|\alpha_i|$$ is a norm on $$E$$. So, the next thing to do, is to prove that any norm $$\| \cdot \|$$ defined on $$E,$$ is equivalent to $$\| \cdot \|_0$$.

So, for any $$x\in E,$$ \begin{align}\|x\|=\|\sum_{i=1}^{n}\alpha_i e_i\|\leq \max\limits_{1\leq i\leq n}|\alpha_i|\big\|\sum_{i=1}^{n}e_i\big\| \leq \max\limits_{1\leq i\leq n}|\alpha_i|\sum_{i=1}^{n}\big\|e_i\big\|=\beta\| x \|_0,\end{align} where $$\beta:=\sum_{i=1}^{n}\big\|e_i\big\|.$$ Now, define $$S=\{x\in E: \| x \|_0=1\}.$$ Clearly, $$S$$ is compact. Let \begin{align}\psi:(E&,\| \cdot \|_0)\longrightarrow \Bbb{R},\\& x\mapsto \psi(x)=\|x\|\end{align}

Let $$\epsilon>0$$ and $$x,y\in E$$ be arbitrary such that $$\Big\Vert x-y\Big\Vert_0<\delta,$$ then \begin{align}\left|\psi\left(x \right)-\psi\left(y \right)\right|&=\left|\Vert x\Vert-\Big\Vert y \Big\Vert\right| \\&\leq \Big\Vert x-y\Big\Vert \\&\leq \beta\,\Big\Vert x-y\Big\Vert_0 \\&<\beta \delta. \end{align} So, given any $$\epsilon>0$$, choose $$\delta=\dfrac{\epsilon}{\beta+1}>0,$$ then \begin{align}\left|\psi\left(x \right)-\psi\left(y \right)\right|&<\beta \delta=\beta\left(\frac{\epsilon}{\beta+1}\right)<\epsilon. \end{align} Thus, $$\psi$$ is uniformly continuous on $$E$$ and is automatically continuous on $$E$$. Since $$S\subseteq E$$, then $$\psi$$ is continuous on $$S$$, and the minimum is attained in the set, i.e. there exists $$t_0\in S$$ such that $$\psi(t_0)=\min\limits_{t\in S} \psi(t)$$ and \begin{align}0<\psi(t_0)\leq \psi(t)=\|t\|,\;\;t\in S.\end{align} Let $$u=\frac{x}{\| x\|_0}$$, then $$u\in S$$ and \begin{align}\gamma\leq \psi(u)=\Big\|\frac{x}{\| x\|_0}\Big\|\implies \gamma \| x\|_0\leq \| x\|,\;\;\gamma:=\psi(t_0).\end{align} Finally, we have \begin{align}\gamma\leq \psi(t)=\Big\|\frac{x}{\| x\|_0}\Big\|\implies \gamma \| x\|_0\leq \| x\|\leq \beta\| x \|_0, \;\;\text{for some} \;\;\gamma,\beta>0.\end{align} Therefore, any norm $$\| \cdot \|$$ defined on $$E,$$ is equivalent to $$\| \cdot \|_0$$ and we are done!

Kindly help check if the proof is correct.

QUESTION:

What gives the assurance that $$\psi(t_0)>0?$$

• You say, "Clearly $S$ is compact." Compact with respect to which topology? You have two norms whose topologies you may not assume coincide. Also, I'm not certain compactness is that self-evident! – Theo Bendit Dec 26 '18 at 21:48
• @Theo Bendit: I made such a claim since it is closed and bounded. – Omojola Micheal Dec 26 '18 at 21:49

The key of the argument is the fact that $$S$$ is compact, and you are glossing over that. It's not very hard, but it is not the right part of the proof to say "clearly": it's precisely the part of the proof where you have to use that $$E$$ is finite-dimensional.

Once you know that $$S$$ is compact, you are done. You have already proven that $$\psi$$ is uniformly continuous, as your argument started with $$\psi(x)\leq\beta\|x\|_0$$. And you don't even need uniform continuity, which is automatic for a continuous function on a compact set. Once you know that $$S$$ is compact and $$\psi$$ is continuous, it is standard that it attains a max and a min on $$S$$, and you are done.

Finally, your (unneeded) argument to show that $$\psi$$ is uniformly continuous starts by saying that $$x/\|x-y\|_0\in S$$, which is not the case. And the argument cannot be right because it doesn't use that $$S$$ is compact, so it doesn't use that $$E$$ is finite-dimensional; nor it uses what $$\|\cdot\|_0$$ is (which you did use in your original proof of $$\psi(x)\leq\beta\|x\|$$): it is a "proof" that any two norms on a vector space are comparable, something that is not true.

Finally, you have $$\psi(t_0)>0$$ because $$\psi$$ is a norm; since $$t_0\in S$$, you have $$\|t_0\|_0=1$$, so $$t_0\ne0$$ and then $$\psi(t_0)>0$$.

• +1) Thank you for your critical examination of the proof. Can you please, tell me where the proof is faulty? Or can you provide one? – Omojola Micheal Dec 27 '18 at 16:21
• The argument is faulty because you cannot say that $x/\|x-y\|_0\in S$; these elements can be unbounded in general and your argument does not apply. And I don't need to provide a proof, because you already did.You need to erase from "We claim" all the way until "Thus", as none of that is needed. As I mentioned, $\psi$ is continuous by your first estimate, and that together with the compactness of $S$ gives you the minimum you need. – Martin Argerami Dec 27 '18 at 16:29
• Thanks and let me do that right away! – Omojola Micheal Dec 27 '18 at 16:34
• Can you please, check now? – Omojola Micheal Dec 27 '18 at 16:45
• Looks ok now. I also added the explanation why $\psi(t_0)>0$. – Martin Argerami Dec 27 '18 at 17:50

Since the minimum is attained in the set $$S$$, it's pretty straightforward. Recall that $$S$$ is defined as :

$$S = \{x \in E : \|x\|_0 = 1\}$$

You defined $$\psi(t)$$ as :

\begin{align}\psi:(E&,\| \cdot \|_0)\longrightarrow \Bbb{R},\\& x\mapsto \psi(x)=\|x\|\end{align}

But, note that for any norm, it is :

$$\|x\| = 0 \Leftrightarrow x = 0$$

The importance of $$\Leftrightarrow$$ as an if and only if operator is noted here.

Specifically, it is : $$\psi(t_0) = \|t_0\|$$ with $$t_0 \in S$$. But $$\|t_0\|_0 = 1 > 0 \Leftrightarrow t_0 > 0 \Leftrightarrow \|t_0\| \equiv \psi(t_0) >0$$.

• Thanks for that clarification! Kindly check the proof if correct. – Omojola Micheal Dec 26 '18 at 21:41
• @Mike Seems okay. – Rebellos Dec 26 '18 at 21:44
• Thanks a lot! I am grateful! – Omojola Micheal Dec 26 '18 at 21:45
• @Mike Always happy to help. Merry Christmas ! – Rebellos Dec 26 '18 at 21:46
• Same, same here! Goodbye! – Omojola Micheal Dec 26 '18 at 21:47

We know that if and only if $$||x||=0$$ then $$x=0$$. And we know the same for $$||x||_0$$. Since $$||t_0||_0=1$$ we get that $$||t_0||\neq 0$$. And since $$||x||\geq0$$ we get positivity of $$||t_0||=\psi(t_0)$$

• Thanks a lot! I really appreciate you! – Omojola Micheal Dec 26 '18 at 21:46