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I am studying the rank theorem on the book of Zorich, Mathematical analysis I, sec. 8.6, p. 505. The proof is clear, until this sentence appears:

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I don't post screenshots of the whole proof of the theorem for copyright reasons, but I kindly ask anyone who has the book, if he can give me a hand in understanding how the above last statement can be demonstrated.

Thanks in advance.

Nameless
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  • The first statement is clear from the Jacobi matrix obtained just above it. The second statement is because the points are discrete, and thus the shrinking to a (convex) ball is possible. The last statement is due to the fact that the partial derivative is a measure of the 'rate of change' of one function with the other. – David Nov 16 '19 at 20:00
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    It is basically an one-dimensional fact that if $f'(x)=0$ for all $x$ in an interval then $f$ is constant there (i.e. does not depend on $x$). Apply this fact to every $u^i$ at a time. Convexity ensures that the domain restriction (when only $u^i$ varies) is an interval. – A.Γ. Nov 16 '19 at 20:06
  • as far I can see this is just the implicit function theorem. Im trying to see how it is proved in this text but it is a lot messy work with matrices –  Nov 16 '19 at 20:38
  • So, using Lagrange mean value theorem for real functions of several variables in a ball, we can easily conclude that if the partial derivative of f w.r.t. $x_i$ is zero in the ball, than f is independent from $x_i$ in that ball. Right? – Nameless Nov 16 '19 at 23:36

1 Answers1

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That actually is a very good question as it is the kind of details one overlooks.

$\bullet\enspace$ The fact that the partial derivatives $\displaystyle \frac{\partial g^j}{\partial u^i}(u),\ i,j > k$ vanish comes from the assumption that $f$ has rank $k$ in the neigbhorhood $\tilde{O}(u_0)$ and as he explains, $g$ also has rank $k$. A possible characterization of this is that there is no $p\times p$ invertible submatrix ($p\geq k+1$) of the Jacobi matrix

$$J:=\begin{pmatrix} 1 & & 0 & & & \\ & \ddots & & & 0 & \\ 0 & & 1 & & & \\ \frac{\partial g^{k+1}}{\partial u^1}(u) & \cdots & \frac{\partial g^{k+1}}{\partial u^k}(u) & \frac{\partial g^{k+1}}{\partial u^{k+1}} (u) & \cdots & \frac{\partial g^{k+1}}{\partial u^{m}} (u)\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ \frac{\partial g^{n}}{\partial u^1}(u) & \cdots & \frac{\partial g^{n}}{\partial u^k}(u) & \frac{\partial g^{n}}{\partial u^{k+1}} (u) & \cdots & \frac{\partial g^{n}}{\partial u^{m}} (u) \end{pmatrix}$$ In particular if one takes the upper left corner $k\times k$ identity matrix and adds just one extra row $j_0$ and column $i_0$ (unusual convention), one must take $i_0,j_0 \geq k+1$ (otherwise, one would just add a row without column or an extra column but no extra row).

Explicitly, our $p\times p $ submatrix $S$ with $p:= k+1$ has coefficients $S_{rc}:= J_{rc}$ ($r$ row indice, $c$ column indice) with $r\in \{1,2, \cdots, k\}\cup \{j_0\}$ and $c\in \{1,2, \cdots, k\}\cup \{i_0\}$ and looks like $$S:=\begin{pmatrix} 1 & & 0 & 0\\ & \ddots & & \vdots \\ 0 & & 1 & 0 \\ \frac{\partial g^{j_0}}{\partial u^1}(u) & \cdots & \frac{\partial g^{j_0}}{\partial u^k}(u) & \frac{\partial g^{j_0}}{\partial u^{i_0}} (u) \end{pmatrix}$$ By assumption it is not invertible (otherwise $g$ would be of rank $\geq k+1$), its determinant must vanish, i.e. $\displaystyle \frac{\partial g^{j_0}}{\partial u^{i_0}} (u) = 0$. This should hold for any possible $i_0, j_0$.

$\bullet\enspace$ Contrary to the first comment, there is no restriction upon some "points are discrete" criterion. The fact that one can choose a smalller convex open neighborhood has to do with the fact that the space (here $\mathbb{R}^m$ with any norm) is a locally convex topological space. There is no need of such generality, what one should understand is just that the topology is generated by balls which are convex. (Any open can be written as union (possibly uncountable) of balls with different centers and radii. Or from a different viewpoint, any open contains a ball.)

$\bullet\enspace$ Indeed the point of restricting to a convex set $C\subseteq \mathbb{R}^m$ is to get back to the one variable situationand apply what the author calls "Lagrange mean value" thm: any two points in a convex subset is connected by a straight segment which can be parametrized by some $\gamma: [0,1] \to C\ \in \mathcal{C}^1$. One can then consider the function of one variable $g\circ\gamma$.

In fact one usually restricts oneself to convex subset (in the several variable case) to obtain some more precise inequality, e.g. "Calcul différentiel" (1997), André Avez, Théorème des accroissements finis 1.6. and Corollaire 1.7. p.19)

Otherwise, one has the general statement that if the differential $df$ (of a differentiable map $f:E\to F$ between normed vector spaces) vanishes on an connected open subset then $f$ is constant on that subset. (cf. e.g. "Calcul différentiel" (1997), André Avez, Thm § 2. p.20, or this question or that one, or available from a google search, "Calculus on Normed Vector Spaces" (2012), Rodney Coleman, Thm 3.3. p.63). (The idea is really just to replace segments by piecewise $\mathcal{C}^1$ and the fact that for normed spaces path-connectedness is equivalent to connectedness).

In our case, the map we show is constant is $\pi\circ g\circ \iota$ where $\pi: \mathbb{R}^n \to \mathbb{R}^{n-k}$ is the projection on the last $n-k$ components and for any $(u^1,\cdots, u^k)$ given, $\iota : (u^{k+1}, \cdots, u^m)\longmapsto (u^1,\cdots, u^k, u^{k+1}, \cdots, u^m) $. "For any fixed $(u^1,\cdots, u^k)$, $g$ does not depend on $(u^{k+1}, \cdots, u^m)$".

Noix07
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