Page 45 - Vector Analysis

§2.4 Conditions for Differentiability 41

Then B f (0, 0) = lim f (h, 0) ´ f (0, 0) = lim h = 1. Similarly, B f (0, 0) = 1; thus if f is
Bx hÑ0 h h[Ñ0 h ] By

differentiable at (0, 0), then (Df )(0, 0) = 1 1 . However,

ˇ[ 1 ] [] ˇ ˇˇf (x, + y)ˇˇ
ˇf (x, y) ´ f (0, 0) ´ 1 x ˇ = y) ´ (x ;
ˇ y ˇ

thus if xy ‰ 0,

ˇˇf (x, y) ´ (x + y)ˇˇ = |1 ´ x ´ y| Û 0 as (x, y) Ñ (0, 0), xy ‰ 0.

Therefore, f is not differentiable at (0, 0).

2.4 Conditions for Differentiability

Proposition 2.29. Let U Ď Rn be open, a P U , and f = (f1, ¨ ¨ ¨ , fm) : U Ñ Rm. Then f is
differentiable at a if and only if fi is differentiable at a for all i = 1, ¨ ¨ ¨ , m. In other words,
for vector-valued functions defined on an open subset of Rn,

Componentwise differentiable ô Differentiable.

Proof. “ñ” Let (Df )(a) be the Jacobian matrix of f at a. Then

@ ε ą 0, D δ ą 0 Q ››f (x) ´ f (a) ´ (Df )(a)(x ´ a)››Rm ď ε}x ´ a}Rn if }x ´ a}Rn ă δ .

Let tejumj=1 be the standard basis of Rm, and Li P L (Rn, R) be given by Li(h) =

eiT[(Df )(a)]h. Then Li P B(Rn, R) by Remark 1.79, and if }x ´ a}Rn ă δ,

ˇˇfi(x) ´ fi(a) ´ Li(x ´ a)ˇˇ = ˇˇei ¨ ( (x) ´ f (a) ´ (Df )(a)(x ´ a))ˇˇ
f

ď ››f (x) ´ f (a) ´ (Df )(a)(x ´ a)››Rm ď ε}x ´ a}Rn ;

thus fi is differentiable at a with derivatives Li.

“ð” Suppose that fi : U Ñ R is differentiable at a for each i = 1, ¨ ¨ ¨ , m. Then there exists

Li P B(Rn, R) such that

@ε ą 0, D δi ą 0 Q ˇˇfi(x) ´ fi(a) ´ Li(x ´ a)ˇˇ ď ε ´ a}Rn if }x ´ a}Rn ă δi .
m }x

Let L P L (Rn, Rm) be given by Lx = (L1x, L2x, ¨ ¨ ¨ , Lmx) P Rm if x P Rn. Then

L P B(Rn, Rm) by Remark 1.79, and

m

››f (x) ´ f (a) ´ L(x ´ a)››Rm ď ÿ ˇˇfi(x) ´ fi(a) ´ Li(x ´ a)ˇˇ ď ε}x ´ a}Rn

i=1

if }x ´ a}Rn ă δ = min ␣δ1, ¨ ¨ ¨ , δm(. ˝