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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 $\psi(t_0)>0?$

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    $\begingroup$ 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! $\endgroup$ – Theo Bendit Dec 26 '18 at 21:48
  • $\begingroup$ @Theo Bendit: I made such a claim since it is closed and bounded. $\endgroup$ – Omojola Micheal Dec 26 '18 at 21:49
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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 $\psi(t_0)>0$.

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  • $\begingroup$ +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? $\endgroup$ – Omojola Micheal Dec 27 '18 at 16:21
  • $\begingroup$ 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. $\endgroup$ – Martin Argerami Dec 27 '18 at 16:29
  • $\begingroup$ Thanks and let me do that right away! $\endgroup$ – Omojola Micheal Dec 27 '18 at 16:34
  • $\begingroup$ Can you please, check now? $\endgroup$ – Omojola Micheal Dec 27 '18 at 16:45
  • $\begingroup$ Looks ok now. I also added the explanation why $\psi(t_0)>0$. $\endgroup$ – Martin Argerami Dec 27 '18 at 17:50
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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 $\|t_0\|_0 = 1 > 0 \Leftrightarrow t_0 > 0 \Leftrightarrow \|t_0\| \equiv \psi(t_0) >0$.

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  • $\begingroup$ Thanks for that clarification! Kindly check the proof if correct. $\endgroup$ – Omojola Micheal Dec 26 '18 at 21:41
  • $\begingroup$ @Mike Seems okay. $\endgroup$ – Rebellos Dec 26 '18 at 21:44
  • $\begingroup$ Thanks a lot! I am grateful! $\endgroup$ – Omojola Micheal Dec 26 '18 at 21:45
  • $\begingroup$ @Mike Always happy to help. Merry Christmas ! $\endgroup$ – Rebellos Dec 26 '18 at 21:46
  • $\begingroup$ Same, same here! Goodbye! $\endgroup$ – Omojola Micheal Dec 26 '18 at 21:47
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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 $||t_0||=\psi(t_0)$

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

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