Kenneth Kuttler

164 CHAPTER 7. IMPLICIT FUNCTION THEOREM

Definition 7.0.4 Let δ ,η > 0 satisfy: B(x0,δ )×B(y0,η)⊆U where f : U ⊆Rn×Rm→ Rp is given as

f (x,y) =

f1 (x,y)f2 (x,y)

...fp (x,y)

and for

(x1 · · · xn

)∈ B(x0,δ )

pand y ∈ B(y0, η̂) define

J(x1, · · · ,xp,y

)≡

 f1,x1

(x1,y

)· · · f1,xn

(x1,y

)...

...fp,x1 (x

p,y) · · · fp,xn (xp,y)

 . (∗)

Thus, its ith row is D1 fi(xi,y

). Let K,r be constants.

By Theorem 6.9.5 f is differentiable and its derivative is the p× (n+m) matrix,(D1f (x,y) D2f (x,y)

Also, by Lemma 7.0.1, and (x,y) ∈ B(x0,δ )×B(y0,η)⊆U, and h,k sufficiently small,there are xi on the line segment between x and x+h such that

f (x+h,y+k)−f (x,y) = f (x+h,y+k)−f (x,y+k)+f (x,y +k)−f (x,y)

= J(x1, · · · ,xp,y+k

)h+D2f (x,y)k+o(k) (7.1)

Proposition 7.0.5 Suppose g : B(x0,δ )×B(y0,η0)→ [0,∞) is continuous and alsothat g(x0,y0) = 0 and if x ̸= x0,g(x,y0)> 0. Then there exists η < η0 such that if y ∈B(y0,η) , then the function x→ g(x,y) achieves its minimum on the open set B(x0,δ ).

Proof: If not, then there is a sequence yk→ y0 but the minimum of x→ g(x,yk) forx ∈ B(x0,δ ) happens on ∂B(x0,δ ) ≡ ∂B ≡ {x : |x−x0|= δ} at xk. Now ∂B is closedand bounded and so compact. Hence there is a subsequence, still denoted with subscript ksuch that xk→ x ∈ ∂B and yk→ y0. Let 0 < 2ε < min{g(x̂,y0) : x̂ ∈ ∂B} .

Then for k large, |g(xk,yk)−g(x,y0)|< ε, and |g(xk,yk)−g(xk,y0)|< ε , the sec-ond inequality from uniform continuity. Then from these inequalities, for k large,

g(x0,yk) ≥ g(xk,yk)> g(xk,y0)− ε

> min{g(x̂,y0) : x̂ ∈ ∂B}− ε > 2ε− ε = ε

Now let k→ ∞ to conclude that g(x0,y0)≥ ε , a contradiction. ■Here is the implicit function theorem. It is based on the mean value theorem from one

variable calculus, the extreme value theorem from calculus, and the formula for the inverseof a matrix in terms of the transpose of the cofactor matrix divided by the determinant.

Theorem 7.0.6 (implicit function theorem) Suppose U is an open set in Rn×Rm.Let f : U → Rn be in C1 (U) and suppose

f (x0,y0) = 0, D1f (x0,y0)−1 exists. (7.2)

164 CHAPTER 7. IMPLICIT FUNCTION THEOREMDefinition 7.0.4 Ler 5,1 > 0 satisfy: B(ao,5) x B(yo,n) CU where f :U CR" xR” — R? is given asfi oe)fr(f,yf (x,y) = : :fp (@5y)andfor( av! +» a" Je B(ao,6)’ and y € B(yo,f}) defineSis (x! ,y) ue Sin (x',y)J(x!,---,a?,y) = : : . (x)Sox (a?,y) ue Soxn (x? y)Thus, its i" row is Dy fi (a, y). Let K,r be constants.By Theorem 6.9.5 f is differentiable and its derivative is the p x (n-+m) matrix,( Dif (x,y) Dof (x,y) ).Also, by Lemma 7.0.1, and (x,y) € B(x0,5) x B(yo,n) CU, and h, k sufficiently small,there are x' on the line segment between x and x + h such thatf(xth,yt+k)—f (x,y) =f(a@th,yt+k)—f(aytk)+f (x,y +k)—f (x,y)=J(a',---,2?,y+k)h+Dof (x,y)k+o(k) (7.1)Proposition 7.0.5 Suppose g : B(xo,5) x B(yo,;No) — [0,°°) is continuous and alsothat g(x0,Yo) =0 and if x # Lo, g (x, Yo) > 0. Then there exists N <p such that if y €B(Yo,1), then the function x + g(x,y) achieves its minimum on the open set B (a0, 5).Proof: If not, then there is a sequence y,; — Yo but the minimum of x > g(x, y;) forx € B(ao,6) happens on 0B (ao, 6) = OB = {x : |x — x9| = 5} at xy. Now OB is closedand bounded and so compact. Hence there is a subsequence, still denoted with subscript ksuch that x, > x € OB and y, — yo. Let 0 < 2€ < min{g (&, yo) : & € OB}.Then for k large, |g (@x,Yx) — 8 (5 Yo)| < €, and |g (xx, yx) — 8 (x, Yo)| < €, the sec-ond inequality from uniform continuity. Then from these inequalities, for k large,8(@0,Y~) > (Le YE) > S(#e,Yo) —E> min{g(é,yo):@¢ 0B}—e>2e—-e=€Now let k — © to conclude that g (ao, yo) > €, a contradiction. liHere is the implicit function theorem. It is based on the mean value theorem from onevariable calculus, the extreme value theorem from calculus, and the formula for the inverseof a matrix in terms of the transpose of the cofactor matrix divided by the determinant.Theorem 7.0.6 (implicit function theorem) Suppose U is an open set in IR" x R”™.Let f :U — R" be in C! (U) and supposef (0, Yo) = 9, Dif (x0,Yo) exists. (7.2)