PROOF OF IMPLICIT FUNCTION THEOREM
Let f1,· · · , fm be real valued functions on an open subset E of Rn×Rm = Rn+m and p = (x0,u0) be a point of E. Let us write fi = fi(x,u) for 1 ≤ i ≤ m where x = (x1,· · ·, xn) and u= (u1,· · · , um).We would like to study the solvablility of the system of equations
f1(x1,· · ·, xn, u1,· · ·, um) =c1
f2(x1,· · ·, xn, u1,· · ·, um) =c2 ...
fm(x1,· · ·, xn, u1,· · ·, um) =cm
Herec= (c1,· · ·, cm)∈Rm.Letf :E →Rmbe the mapf = (f1,· · · , fm) onE.The above system of equations is equivalent to the solvability of the equation
f(x,u) =c.
Assume thatf(p) =cand that
∆ =
∂f1
∂u1(p) · · · ∂f1
∂um(p) ... . .. ...
∂fm
∂u1(p) · · · ∂fm
∂um(p)
6= 0.
Then we can find an open neighborhoodU ofpand an open subsetV ofRn with the properties that (1) for eachx∈V,there exists a uniqueu∈Rmwith (x,u)∈U andf(x,u) =c
(2) for eachx∈V,we denote the above uniqueubyg(x).Then we obtain a functiong:V →Rm such thatg isC1 andu0=g(x0);
(3) If we denote T = Df(p) and set T1(x) = T(x,0) for x ∈Rn and set T2(u) = T(0,u) for u∈Rm,thenDg(u0) =−T2−1◦T1.
Let us define a new functionF :E→Rn+m by
F(x,u) = (x, f(x,u)).
ThenF is aC1-function on the open setE.Furthermore,
DF(x,u) =
Im Dxf(x,u) 0 Duf(x,u)
. Here
Dxf(x,u) =
∂f1
∂x1
(x,u) · · · ∂f1
∂xn
(x,u)
∂f2
∂x1
(x,u) · · · ∂f2
∂xn
(x,u) ... . .. ...
∂fm
∂x1(x,u) · · · ∂fm
∂xn(x,u)
, Duf(x,u) =
∂f1
∂u1
(x,u) · · · ∂f1
∂um
(x,u) ... . .. ...
∂fm
∂u1
(x,u) · · · ∂fm
∂um
(x,u)
The Jacobian ofF atpis given by
JF(p) = detDF(p) = ∆6= 0.
By assumption, there exists an open neighborhoodU of pand open neighborhood U0 ofF(p) such thatF :U →U0 is bijective whose inverse is alsoC1.LetG:U0→U be the inverse toF :U →U0 and write G(y,v) = (G1(y,v), G2(y,v)) whereG1 :U0→Rn andG2:U0→Rm areC1-functions.
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2 PROOF OF IMPLICIT FUNCTION THEOREM
Let us choose an open rectangleI inRnand an open rectangleJ inRmsuch thatI×J⊆U0.Since F :U →U0 is a continuous, F−1(I×J) is open.
For (y,v)∈I×J,we have
(y,v) =F(G(y,v)) =F(G1(y,v), G2(y,v)) = (G1(y,v), f(G1(y,v), G2(y,v))).
In other words,G1(y,v) =yand hencef(y, G2(y,v)) =vfor any (y,v)∈I×J.ChooseV =I⊆Rn and defineg:V →Rm byg(x) =G2(x,c).SinceG2 isC1,so is g.Furthermore,
f(x, g(x)) =f(x, G2(x,c)) =cfor any x∈V .
Uniqueness of solution tof(x,u) =cis guaranteed by the fact thatF :U →U0 is bijective. (This should be verified by the readers.)
Sincef(p) =c, f(x0,u0) =c,we see thatu0=g(x0).This can be proved by (x,u) =G(F(x,u)) = (x, G2(x, f(x,u)))
andG2(x, f(x,u))) =ufor any for any (x,u)∈F−1(I×J).In fact, u0=G2(x0, f(x0,u0)) =G2(x0,c) =g(x0).
Sincef(x, g(x)) =c,by chain rule,
Dxf(x, g(x)) +Duf(x, g(x))Dg(x) = 0.
Pluggingp= (x0,v0) into the above equation, we obtain that Dxf(p) +Duf(p)Dg(x0) = 0.
SinceT1=Dxf(p) andT2=Duf(p),we obtain that Dg(x0) =−T2−1◦T1.
