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I am struggling with an inductive proof of the implicit function theorem, concretely with the final part of construction of a function, up to this final point everything is perfectly clear to me. First the following is known to be true:

Theorem 1: Let $F : G \subseteq \mathbb R^{n+1} \to \mathbb R$, $G \ne \emptyset$, $F \in C^1(G)$, $G$ open. For $x^0 \in \mathbb R^n, y^0 \in \mathbb R$ let $$ F(x^0, y^0) = 0 \quad \mbox{ and } \quad F_y(x^0, y^0) \ne 0. $$ Then there exists some neighborhood $U(x^0) \subseteq \mathbb R^n$ and a function $f : U(x^0) \to \mathbb R$ such that $$ y = f(x_1, \ldots, x_n), \qquad y_0 = f(x_1^0, \ldots, x_n^0) $$ and $$ F(x, f(x)) = 0 $$ and $f \in C^1(U(x^0))$.

Now the Implicit Function Theorem reads as

Theorem 2 (Implicit Function Theorem): Let $F : G \subseteq \mathbb R^{n+m} \to \mathbb R^m$, $G\ne \emptyset$, $F \in C^1(G)$, $G$ open. Also $(x^0, y^0) \in G$ with $x^0 \in \mathbb R^n, y^0 \in \mathbb R^m, F(x^0,y^0) = 0$ and $$ \det\left( \frac{\partial F}{\partial y} \right)_{y=y_0} \ne 0. $$ Then there exists neighborhoods $U(x^0) \subseteq \mathbb R^n, V(y^0) \subseteq \mathbb R^m$ and a function $g : U(x^0) \to V(x^0)$ such that $F(x, g(x)) = 0$ on $x \in U(x^0)$.

Proof: The proof proceeds by induction on $m$, if $m = 1$ it is Theorem 1, so we assume $m > 1$ and suppose (by induction hypothesis) the statement holds for $m-1$. Let $F : G \subseteq \mathbb R^{n+m} \to \mathbb R^n$ be given, then because of $$ \det\left( \frac{\partial F}{\partial y} \right)_{y=y_0} \ne 0 \quad \mbox{ or } \quad \det \begin{pmatrix} \frac{\partial F_1}{\partial y_1} & \frac{\partial F_1}{\partial y_2} & \cdots & \frac{\partial F_1}{\partial y_m} \\ \vdots \\ \frac{\partial F_m}{\partial y_1} & \frac{\partial F_m}{\partial y_2} & \cdots & \frac{\partial F_m}{\partial y_m} \end{pmatrix} \ne 0. $$ So we can suppose that every row contains an entry $\ne 0$, suppose w.l.o.g. that $\frac{\partial F_m}{\partial y_m} \ne 0$. By Theorem 1 we can solve (locally) for $y_m$, that means there is some neighborhood $U$ of $(x, y_1, \ldots, y_{m-1})$ and a function $$ y_m = \varphi(x, y_1, \ldots, y_{m-1}) $$ and $\varphi$ is continously differentiable with ($x \in \mathbb R^n$) $$ F_m(x, y_1, \ldots, y_{m-1}, \varphi(x, y_1, \ldots, y_{m-1})) = 0 $$ for all $(x,y_1,\ldots,y_{m-1}) \in U$. Now set $$ \Phi_i(x, y_1, \ldots, y_{m-1}) := F_i(x, y_1, \ldots, y_{m-1}, \varphi(x, y_1, \ldots, y_{m-1})) $$ for $i = 1, \ldots, m-1$. Then $\Phi : U \subseteq \mathbb R^{n+m-1} \to \mathbb R^{m-1}$ and $\Phi$ fulfills als prerequisites to apply the induction hypothesis (here the proof shows this, but I omit it because it is quite long and does not apply to my question). Then there exists neighborhoods $W \subseteq \mathbb R^n, V \subseteq \mathbb R^{m-1}$ and a function $g : W \to V$ with $$ \Phi(x, g(x)) = 0 $$ for all $x \in W$. Now set $h(x) = (g(x), \varphi(x,g(x))$, then we have $$ F(x, h(x)) = 0 $$ and the proof is finished. $\square$

My question is on the last part. Namely the construction of the function $h(x)$,

1) Because $h(x) = (g(x), \varphi(x, g(x))$ it must be the case that $(x, g(x) \in U$, because $\varphi : U \to \mathbb R$, but I do not see that this must be the case?

2) The same issue if I want to show that $F(x, h(x)) = 0$, if $i \ne m$ it is $$ F_i(x, h(x)) = \Phi_i(x, g(x)) $$ by definition, but for $$ F_m(x, h(x)) = F_m(x, g(x), \varphi(x,g(x)) = F_m(x, g_1(x), \ldots, g_{m-1}(x), \varphi(x, g_1(x), \ldots, g_{m-1}(x)) $$ and to conclude that this equals $0$ it also must be that $(x, g(x)) \in U$, but I do not see that this must be the case?

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As a minor point, you forgot to write that the implicit function $g$ on Theorem 2 is of class $C^1$.

More important, the theorems 1 and 2 that you enunciated do not claim the (local) uniqueness of the implicit solution. This feature is important.

By reading your question I guess that you are trying to understand the inductive proof of the implicit function theorem given in ``The Implicit Function Theorem - History, Theory, and Applications'', pp.39-41, by S. G. Krantz and H. R. Parks.

Regarding your specific question, notice that $$0=\Phi_1(x,g(x))=F_1\Big(x,g_1(x),\ldots,g_{m-1}(x),\varphi\big(x,g_1(x),\ldots,g_{m-1}(x)\big)\Big).$$ Analogously for $F_2,\ldots,F_{m-1}$ and for $F_m$. Thus, $$h(x)=\Big(g_1(x),\ldots,g_{m-1}(x),\varphi\big(x,g_1(x),\ldots,g_{m-1}(x)\big)\Big).$$

One approach to understand an inductive proof is to reproduce it for n=1, n=2, and n=3. Since you already understood the case n=1, I suggest that you develop a complete proof for the case n=2 (this case should be easy). Then, try to develop a complete proof for the case n=3 (if you overcome this case then you are almost done). Finally, try the general case.

I also suggest that you take a look at the article ``The Implicit and the Inverse Function Theorems: Easy Proofs'', Real Analysis Exchange, Vol. 39(1), pp. 207-218 (2013/2014), by O. de Oliveira. You can find a preprint on the internet.

Best wishes,

Oswaldo

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The proof has to proceed by induction* on n because then you are always working with neighborhoods of P0.

F1(x1,x2,y1,y2)=0
F2(x1,x2,y1,y2)=0
Have a solution in a neighborhood of P0. (F1->F2)

F1(x1,x2,y1,y2)=0
F2(x1,x2,y1,y2,y3)=0
P01 is entirely different than P02. (F1->’F2)

See appendix of “Rutherford Aris; Vectors, Tensors, and the Basic Equations of Fluid Mechanics” for proof by induction on n.

EDIT: Elaboration and Examples of Aris Induction proof:

Fundamental Assumption (FA):
If F(x1,x2,..xn) satisfies conditions of Implicit Function Theorem (IFT), F(x1,x2,..,xn) has a solution xn=f(x1,x2,…xn-1) in a neighborhood of P0=x10,x20,..,xn0.

Given, under conditions of IFT
F1(x1,y1,y2), F1(P0)=0
F2(x1,y1,y2), F2(P0)=0

Then
F1(x1,y1,y2)=0 has a solution y2=f1(x1,y1) in a neighborhood of P0 by FA. Then
F2(x1,y1,f1(x1,y1))= 0 has a solution by FA in a neighborhood of P0, and
y1=y1(x1)
y2=y2(x1)
In a neighborhood of P0.

Geometrically, F1(x1,y1,z1)=0 and F2(x1,y1,z1)-0 are surfaces through P0 in box neighborhoods of P0 and their common solution is the line of intersection through P0 in a box neighborhood of P0.

Another Example
Given, under conditions of IFT:
F1(x1,x2,y1,y2,y3)=0
F2(x1,x2,y1,y2,y3)=0
F3(x1,x2,y1,y2.y3)=0..

Assuming you have solved F1 and F2,
y1=f1(x1,x2,y3)
y2=f2(x1,x2,y3)

Substituting into F3 gives an Equation in y3 which can be solved by FA. Then
y1=y1(x1,x2)
y2=y2(x1,x2)
y3=y3(x1,x2)
in a neighborhood of P0.

Note: A neighborhood of P0 arises from the requirement that dF/dxn =’ 0 is the same sign throughout a neighborhood of P0, which is true if dF/dxn is continuous.

At this point the general procedure should be obvious from above.

The Inverse Function Theorem follows as a special case. Ex:
x1=f1(y1,y2) -> F1(x1,x2,y1,y2)=0
x2=f2(y1,y2) -> F2(x1,x2,y1,y2)=0
Then:
y1=y1(x1,x2)
y2=y2(x1,x2)

  • The terminology "induction on n" is Aris'. Though true in a sense, it's a little confusing, which is the reason for this post. The crucial notion is a common neighborhood P0 as you increase the number of variables you are solving for by 1.
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  • $\begingroup$ The question of continuous differentiability of solutions, and neighborhoods, of induction proof, is addressed here: mymathforum.com/math/… $\endgroup$ – hartlw Aug 4 '15 at 0:18

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