Condition for Implicit Function Theorem

Question

Given $f(x,y)$ continuously differentiable in a domain of the plane, $f(a,b) = 0$ and $f_x(a,b)\ne 0$. Can you show $f(x,y) = 0$ in a finite neighborhood of $(a,b)$?

Examples

1. Sphere tangent to the plane.

   $f(a,b )= 0$ at one point but $f_x(a,b) = 0$ and $f_y(a,b) = 0$ and there is no neighborhood of $(a,b)$ in which $f(x,y)=0$.

2. Sphere intersecting the plane.

   There are points $(a,b)$ (on the circle of intersection) where $f(a,b) =0 $ , $f_x(a,b)\ne 0$ , and $f(x,y)=0$ in a neighborhood of $(a,b)$, (circular arc through $(a,b)$ )

But how do you show this in general without geometry?

Application: Implicit Function Theorem.

Answer 1

Utterly false: $$f(x,y) = x,\qquad f(0,0) = 0,\qquad f_x(0,0) = 1\ne 0.$$

Answer 2

Apostol, Mathematical Analysis, 13.4, The Implicit Function Theorem. Toward the end he states: "Let To = {t;teRk, (0,t)eY}," (0,t) is our (f(x,y)=0,y). You cannot simply "Let" this be true. You have to show To, (f(x,y)=0), exists for the given conditions of the theorem. Examples confirm but do not prove this.

So at this point it appears the Implicit Function Theorem, and hence linearization about a point of f(x,y)=0 and implicit differentiation, is not proven. Implicit Function Postulate?

Apostol Math Analysis is online. You don't have to wade through the whole proof to appreciate this point, which comes at end. One is reminded of Dieudonne's proof which begins "Write the relation f(x,y)=0."- I don't pretend to follow the rest of it, too much work.

Answer 3

Inverse Function Theorem

Given E a subset of Rn and f:E->Rn

If y=f(x), if Difj=dfj/dxi i,j=1,2,3, exist and are continuous on E, and if det (Difj) =’0, at x=a, and if b=f(a) Then y=f(x) is invertible in a finite neighborhood of y=b.

Illustration of terminology:

y1=f1(x1,x2,x3), b1=f1(a1,a2,a3)
y2=f2(x1,x2,x3), b2=f2(a1,a2,a3)
y3=f3(x1,x2,x3), b3=f3(a1,a2,a3)

x1=g1(y1,y2,y3)
x2=g2(y1,y2,y3)
x3=g3(y1,y2,y3)

Finite Neighborhood: not just in the limit as (y1,y2,y3)->(b1,b2,b3)

Implicit Function Theorem

Let z=f(x,y)
z1=f1(x1,x2,y)
z2=f2(x1,x2,y)

Let Difj (D3f1=df1/dy, D3f2=df2/dy) i=1,2,3 j=1,2 exist and be continuous. Let det (D1f1,D1f2; D2f1,D2f2)=’0 at x,y=a,b where f(a,b)=0.

Consider the map F(z,w) where z=f(x,y), w=y
F1=f1(x1,x2)
F2=f2(x1,x2)
F3=y

Then Difj i,j=1,2,3 exists and is continuous. Furthermore det (DiFj)=’0. By the Inverse Function Theorem it follows that F is invertible in a neighborhood of (f(a,b),b)=(0,b):

x1=g1(F1,F2,F3)
x2=g2(F1,F2,F3)
y=y

So far so good. If F1=0 and F2=0 in a subset of the neighborhood of (f(a,b)=0,b), then

x1=g1(0,0,y)
x2=g2(0,0,y)

in a neighborhood of y=b, ie, x=g(y). But you can’t assume this. You have to prove F1=0 and F2=0 in a subset of a neighborhood of (0,b) under given conditions, ie, f(x,y)=0 in a subset of a neighborhood of x,y=a,b. I haven’t seen this proof anywhere.

Conclusion: The Implicit Function theorem is unproven, ie, you can’t start off a discussion of implicit functions by writing;
f1(x1,x2,x3,y1,y2)=0
f2(x1,x2,x3,y1,y2)=0
f3(x1,x2,x3,y1,y2)=0
because you don’t know that condition exists, unless you solve it in which case the Implicit Function Theorem is academic. Assuming it exists is tantamount to assuming it is invertible (under usual conditions).

Ref: Baby Rudin, Inverse and Implicit function theorem. Also 3rd Ed.

Answer 4

Switching to literal notation of Courant, Kaplan, Taylor, etc:

For F(x,y) satisfying conditions of IFT, Courant (D&I Calc II, 119) shows F(x,y) =0 has a solution y=f(x) in a neighborhood of (x0,y0). He says the extension to F(x,y...,u,v) is straight-forward: F(x,y...,u,v)=0 has a solution v=f(x,y,..,u) in a neighborhood of P0.

Under conditions of IFT, Taylor (Adv Calc) shows F(x,y,z,u,v)=0 and G(x,y,z,u,v)=0 are satisfied in a neighborhood of P0 with u=f(x,y,z) and v=g(x,y,z). He does this by first showing v=f(x,y,z,u) exists for F(x,y,z,u,v)=0 per above and then showing u=g(x,y,z) exists for G(x,y,z,f(x,y,z,u))= H(x,y,z,u)=0.

Finally, Taylor indicates the general case can be solved by induction.

Apostol and Rudin typify a different approach, presented abstractly, which led me to believe IFT was unproven because buried in the abstract proof was a false assumption, noted previously.

Answer 5

Under conditions of the Implicit Function Theorem (IFT), Courant's Differential and Integral Calculus, vol. 2 shows a neighborhood $N$ of $P_0=(x_0,y_0)$ where $F(P_0)=0$ exists such that $F(x,y)=0$ has a solution $y=f(x)$ for $x \in N$.

The proof extends to $F(x,y,z,..,u,v)$: A neighborhood $N$ of $P_0$ exists such that $F(x,y,z,..,u,v)=0$ has a solution $v=f(x,y,z,..,u)$ for $(x,y,z,..,u) \in N$.

Given $F_1(x,y,z)$ and $F_2(x,y,z)$ and IFT conditions, a neighborhood $N_1$ of $P_0$ exists such that $F_1(x,y,z)=0$ has a solution $z=f_1(x,y)$, and a neighborhood $N_2$ of $P_0$ exists such that $F_2(x,y,z)=0$ has a solution $z=f_2(x,y)$.

Let $N=N_1 \cap N_2$. Then $z=f_1(x,y)$ for $(x,y) \in N$, and $F_2(x,y,f_1(x,y))$ has a solution $y=g(x)$ for $(x,y) \in N$:

• $Y=g(x)$ for x,y in N,
• $Z=f_1(x,g(x))$ for x,y in N.
• $F_1,x$ or $F_1,y$ & $F_2,x$ or $F_2,y$ =0$.

That sets the stage for proof by induction, I assume.

As a specific example from everyday calculus, if $F_1(x,y,z)=0$ and $F_2(x,y,z)=0$ are equations of surfaces, then $F_1(P_0)$ and $F_2(P_0) = 0$ is the condition that they intersect at a point and $J=0$ is the condition that the normal to both surfaces are unequal at $P_0$ (i.e., not tangent). Then they intersect in a curve in a neighborhood of $P_0$. The difference between this and general case is that you know beforehand that $F_1(x,y,z)$ and $F_2(x,y,z) = 0$ in $S$. In the general case, you show it in a neighborhood of $P_0$.

I haven’t gone into the details because my concern is that the structure make sense, which it doesn’t (for me) for Apostol's or Rudin's proof of IFT, as pointed out earlier. But at least now (for me) the IFT seems sound in principle.