I am studying analysis and I have had a lot of uncertainties. For instance, I cannot solve this exercise:

If $f:U\rightarrow\mathbb{R}^3$ has class $C^1$ and rank $3$ in all of the points of the open $U\in\mathbb{R^4}$, show that $|f(x)|$ do not assume maximal value for $x\in U$.

(I guess this is the comand, but I'm so sorry if I did mistakes. My language and the language of the comand is Portuguese)

Well. I know that $f$ is a submersion. So, it's an open map. From here can I get the required? If I know that $f$ is an open map, have I that $|f(x)|$ is an open set and so that it has not a maximum?

Edit - September, 25

I was taking another look at this question I decided try a formal proof in here:

Once the rank of $f$ is maximal (the dominium dimension) and the contradominium has a lower dimension, so $f$ is a submmersion. But the submmersions are opened applications. So, $A=\{f(x);x\in U\}\subset \mathbb{R}^4$ is open.

Consider a value $|f(x_0)|,x_0\in U$ and we'll prove that it isn't maximal. Once $A$ is open, there is $\delta>0$ such that $B[f(x_0),\delta]\subset A$.

Taking $f(x_0) $ as a vector, we have $(1+\delta)f(x_0)\in A$. So, there is $x_1\in U$ such that $f(x_1)=(1+\delta)f(x_0)$.

But $|f(x_1)|=(1+\delta)|f(x_0)|>|f(x_0)|$ and this completes the proof.

What do you think? Thanks very much.

  • $\begingroup$ Your second paragraph makes no sense to me. $\endgroup$ – zhw. Jul 16 '18 at 20:08
  • $\begingroup$ Sorry, I did a mistake and I hope now the question is correct. Thanks. $\endgroup$ – Na'omi Jul 16 '18 at 21:52
  • 1
    $\begingroup$ Your proof is correct. The idea is very geometric, once you know how to prove that submersions are open maps. $\endgroup$ – Laz Sep 25 '18 at 22:06
  • $\begingroup$ @Laz, thanks very much for the reply. $\endgroup$ – Na'omi Sep 25 '18 at 22:51

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