If $f:U\to\mathbb{R}^3$, with dim $3$ in $U\subset \mathbb{R}^4$, then $|f(x)|$ doesn' t have maximum value for $x\in U$ If $f:U\to\mathbb{R}^3$, where $f\in C^1$ and has dimension $3$ in all points of the open $U\subset \mathbb{R}^4$, then $|f(x)|$ doesn' t have maximum value for $x\in U$
I'm studying a chapter that talks about submersions (functions for which the derivative is surjective) and the theorem of inverse functions, and this is one of the questions it asks. I think that the answer has something to do with the fact that the dimension is $3$ but we' re picking a subset of $\mathbb{R}^4$. I don't even know how to take the maximum value in consideration. Maybe it has something to do with the derivative being not $0$ at this open, but I don' t see the connection
 A: By the local form of submersions, $f$ is an open mapping. Therefore, for any $f(x)$, there exists $t>0$ such that $(1+t)f(x)$ is still on the image. This contradicts existing a maximum value for $|f(x)|$.
A: I assume our OP Guerlando OCs means, by his assertion that $f$ is of dimension $3$ at every point $x \in U \subset \Bbb R ^4$, is that the rank of the linear map $Df(x): \Bbb R^4 \to \Bbb R^3$ is $3$ for every $x \in U$.  If this is indeed the case, we may procede as follows:
First of all, we observe that a point $p \in U$ at which $\vert f \vert$ takes on even a local maximum, we cannot have $\vert f(p) \vert = 0$;  for if that were indeed the case, there would exist a neighborhood $V \subset U$ of $p$ such that
$\vert f(q) \vert \le \vert f(p) \vert = 0 \tag{1}$
for $q \in V$; but then $f(q) = 0$ for every $q \in V$,
and then rank $Df(q)$ would be $0$ on $V$, in contradiction to our hypothesis that $Df$ be rank $3$ everywhere.  So $\vert f \vert \ne 0$ at a local maximum of $f$, and since a global maximum is in fact a local one, $\vert f \vert \ne 0$ at global maxima as well.
We have
$\vert f \vert = \langle f, f \rangle^{\frac{1}{2}}, \tag{2}$
this function is differentable wherever $f(x) \ne 0$, indeed, we compute, for any vector $X_x$ tangent to $U$ at $x \in U$, assuming $f(x)$, and hence $\vert f(x) \vert \ne 0$:
$D_{X_x} \vert f \vert= D_{X_x} \langle f, f \rangle^{\frac{1}{2}} = \dfrac{1}{2}\langle f, f \rangle^{-\frac{1}{2}} D_{X_x}\langle f, f \rangle, \tag{3}$
where we take
$f(x) = (f_1(x)), f_2(x), f_3(x)) \in \Bbb R^3 \tag{4}$
and
$\langle f(x), f(x) \rangle = f_1^2(x) + f_2^2(x) + f_3^2(x); \tag{5}$.
furthermore,
$D_{X_x}\langle f, f \rangle = \langle D_{X_x}f, f \rangle + \langle f, D_{X_x}f \rangle = 2\langle D_{X_x} f, f \rangle; \tag{6}$
combining (3) and (6) we obtain
$D_{X_x} \vert f \vert \ = \vert f \vert^{-1}\langle D_{X_x}f, f \rangle, \tag{7}$
where we have used (2) to simplify the expression (7).  
Now if some point $x \in U$were even a local maximum of $\vert f \vert$, then from (7) we would have
$D_{X_x} \vert f \vert \ = \vert f \vert^{-\frac{1}{2}}\langle D_{X_x}f, f \rangle = 0 \tag{8}$
for all $X_x$; but the hypothesis that $Df$ is of rank $3$ everywhere on $U$, and at $x$ in particular, implies that the four rows of the matrix 
$Df = \begin{bmatrix} f_{1, 1} & f_{2, 1} & f_{3, 1} \\ f_{2, 1} & f_{2, 2} & f_{3, 2} \\ f_{3, 1} & f_{3, 2} & f_{3, 3} \\ f_{4, 1} & f_{4, 2} & f_{4, 3} \end{bmatrix} \tag{9}$
span the tangent space to $\Bbb R^3$, which itself may be identified with $\Bbb R^3$,  at $f(x)$.  Thus we can choose some tangent vector $Y_x$ to $U$ at $x$ such that
$D_{Y_x}f(x) = Y_x^T Df(x) = f^T(x) = \begin{pmatrix} f_1(x) \\ f_2(x) \\ f_3(x) \end{pmatrix}, \tag{10}$
and from this we have, by (8),
$D_{Y_x} \vert f \vert \ = \vert f \vert^{-\frac{1}{2}}\langle D_{Y_x}f, f \rangle = \langle f(x), f(x) \rangle > 0; \tag{11}$
(11) stands in ooposition to (8); thus we conclude that $\vert f \vert$ has no maxima, either global or local, on $U$.  
In fact we have shown that $\vert f \vert$ can have no critical points $x$, either minima, maxima, or saddles, within $U$ unless $f(x) = 0$; such $x$ are in fact global minima for $\vert f \vert$.  The proof is essentially the same as that given above, with only perhaps some minor modifications.
A: If $a\in U$ and $f(a)= 0,$ then $f$ cannot have a maximum at $a.$ If it did, then $f=0$ in a neighborhood of $a,$ hence $Df(a) = 0,$ contradiction.
Suppose $a\in U$ and $|f(a)|>0.$ Let $\partial f\partial x_j(a) = v_j, j = 1,2,3,4.$ Then
$$Df(a)(x_1,x_2,x_3,x_4) = x_1v_1+x_2v_2+x_3v_3+x_4v_4.$$
Because $Df(a)$ is surjective, not all of the vectors $v_j$ can be perpendicular to $f(a).$ Suppose WLOG $v_1$ is not perpendicular to $f(a).$ Then as $t\to 0\in \mathbb R,$
$$f(a +(t,0,0,0)) = f(a) + tv_1 + o(t).$$
Now geometrically we're done. We can think of $v_1$ as pointing out of $f(a)^\perp$ at $f(a).$ Either it points "away" from the origin, in which case $|f(a) + tv_1|>|f(a)|$ for small positive $t,$ or it points "towards" the origin, in which case $|f(a) + tv_1|>|f(a)|$ for small negative $t.$ The $o(t)$ term will not affect this if $|t|$ is small.
Analytically,
$$|f(a +(t,0,0,0))|^2 = |f(a)|^2 + 2\langle f(a),tv_1 + o(t)\rangle +|o(tv_1 + o(t))|^2$$ $$\tag 1  = |f(a)|^2 + 2t\langle f(a),v_1\rangle +o(t).$$
We know $\langle f(a),v_1\rangle \ne 0.$ If this inner product is positive, then $(1)$ is greater than $|f(a)|^2$ for small positive $t.$ If this inner product is negative, then $(1)$ is greater than $|f(a)|^2$ for small negative $t.$ It follows that $|f(a)|$ cannot be even a local maximum value for $|f|$ in $U,$ and we're done.
