Kenneth Kuttler

100 CHAPTER 4. THE DERIVATIVE

where εk→ 0 and det(Dg+ εkI)≡ det(Dgk) ̸= 0. Then

∑j(cof(Dg))i j, j = lim

k→∞∑

j(cof(Dgk))i j, j = 0 ■

4.8 Implicit Function TheoremThe implicit function theorem is one of the greatest theorems in mathematics. There aremany versions of this theorem which are of far greater generality than the one given here.The proof given here is like one found in one of Caratheodory’s books on the calculusof variations. It is not as elegant as some of the others which are based on a contractionmapping principle but it may be shorter and is based on more elementary ideas. For a moreelegant proof which generalizes better see my book Real and Abstract Analysis. The proofgiven here is based on a mean value theorem in the following lemma.

Lemma 4.8.1 Let U be an open set in Rp which contains the line segment t → y+t (z−y) for t ∈ [0,1] and let f : U → R be differentiable at y+ t (z−y) for t ∈ (0,1) andcontinuous for t ∈ [0,1]. Then there exists x on this line segment such that f (z)− f (y) =D f (x)(z−y) .

Proof: Let h(t)≡ f (y+ t (z−y)) for t ∈ [0,1] . Then h is continuous on [0,1] and has aderivative, h′ (t) = D f (y+ t (z−y))(z−y), this by the chain rule. Then by the mean valuetheorem of one variable calculus, there exists t ∈ (0,1) such that

f (z)− f (y) = h(1)−h(0) = h′ (t) = D f (y+ t (z−y))(z−y)

and we let x = y+ t (z−y) for this t. ■Also of use is the following lemma.

Lemma 4.8.2 Let A be an m×n matrix and suppose that for all i, j,∣∣Ai j∣∣≤C. Then the

operator norm satisfies ∥A∥ ≤Cmn.

Proof: Note that if z is a vector, |z|= sup|y|≤1 (z,y) . Indeed, for |y| ≤ 1, the right sideis no more than |z| thanks to the Cauchy Schwarz inequality and this can be achieved byletting y = z/ |z|.

∥A∥ ≡ sup|x|≤1|Ax|= sup

|x|≤1sup|y|≤1|(Ax,y)|= sup

|x|≤1sup|y|≤1

∣∣∣∣∣∑i∑

jAi jx jyi

∣∣∣∣∣≤ sup

|x|≤1sup|y|≤1

∑i

∑j

C∣∣x j∣∣ |yi| ≤C∑

i∑

j|x| |y|=Cmn. ■

Definition 4.8.3 Suppose U is an open set in Rn ×Rm and (x,y) will denote atypical point of Rn×Rm with x ∈ Rn and y ∈ Rm. Let f : U → Rp be in C1 (U) meaningthat all partial derivatives exist and are continuous. Then define

D1f(x,y) ≡

 f1,x1 (x,y) · · · f1,xn (x,y)...

...fp,x1 (x,y) · · · fp,xn (x,y)

 ,

D2f(x,y) ≡

 f1,y1 (x,y) · · · f1,ym (x,y)...

...fp,y1 (x,y) · · · fp,ym (x,y)

 .

100 CHAPTER 4. THE DERIVATIVEwhere €, — 0 and det (Dg + €;/) = det (Dg,) #0. Then¥ (cof (Dg));;,; = lim Y} (cof (Dgk));;,; = 0j J4.8 Implicit Function TheoremThe implicit function theorem is one of the greatest theorems in mathematics. There aremany versions of this theorem which are of far greater generality than the one given here.The proof given here is like one found in one of Caratheodory’s books on the calculusof variations. It is not as elegant as some of the others which are based on a contractionmapping principle but it may be shorter and is based on more elementary ideas. For a moreelegant proof which generalizes better see my book Real and Abstract Analysis. The proofgiven here is based on a mean value theorem in the following lemma.Lemma 4.8.1 Let U be an open set in R? which contains the line segment t > y+t(z—y) fort € [0,1] and let f : U > R be differentiable at y+t(z—y) fort € (0,1) andcontinuous for t € [0,1]. Then there exists x on this line segment such that f (z) — f (y) =Df (x) (z—y).Proof: Let h(t) = f(y +t(z—y)) fort € [0,1]. Then / is continuous on (0, 1] and has aderivative, h’ (t) = Df (y+t(z—y)) (z—y), this by the chain rule. Then by the mean valuetheorem of one variable calculus, there exists ¢ € (0,1) such thatf (2) — f(y) =h(1) —h(0) =A! (1) = Df (y +t (z—y)) (zy)and we let x = y+¢(z—y) for this ¢.Also of use is the following lemma.Lemma 4.8.2 Let A be an mx n matrix and suppose that for alli, j, |Aij| <C. Then theoperator norm satisfies ||A|| <Cmn.Proof: Note that if z is a vector, |z| = supjy\<, (z,y). Indeed, for |y| < 1, the right sideis no more than |z| thanks to the Cauchy Schwarz inequality and this can be achieved byletting y = z/ |z|.[Asup |Ax| = sup sup |(Ax,y)| = sup sup|x|<1 [x|<1 |y|<1 [x|<1 |y|<1Aisiijsup sup YY C|x;| lvil < chin ly| =Cmn.Ix[<tly|<li jIADefinition 4.8.3 suppose U is an open set in R" x R™ and (x,y) will denote atypical point of R" x R” with x € R" andy € R". Let f: U — R? be in C! (U) meaningthat all partial derivatives exist and are continuous. Then definefix, %Y) + fim (%Y)Dif (x,y) : :fox (%Y) + Spain &Y)fiy:Q@Y) + fiym &Y)Dof (x,y) : :San (x,y) ue Sym (x,y)