Page 57 - Vector Analysis

§2.5 Properties of Differentiable Functions 53

2.5.4 The Mean Value Theorem

Theorem 2.55. Let U Ď Rn be open, and f : U Ñ Rm with f = (f1, ¨ ¨ ¨ , fm). Suppose that
f is differentiable on U and the line segment joining x and y lies in U. Then there exist
points c1, ¨ ¨ ¨ , cm on that segment such that

fi(y) ´ fi(x) = (Dfi)(ci)(y ´ x) @ i = 1, ¨ ¨ ¨ , m.

Moreover, if U is convex and sup }(Df )(x)}B(Rn,Rm) ď M , then
xPU
}f (x) ´ f (y)}Rm ď M }x ´ y}Rn @ x, y P U .

Proof. Let γ : [0, 1] Ñ Rn be given by γ(t) = (1 ´ t)x + ty. Then by Theorem 2.49, for each
i = 1, ¨ ¨ ¨ , m, (fi ˝ γ) : [0, 1] Ñ R is differentiable on (0, 1); thus the mean value theorem
(Theorem A.9) implies that there exists ti P (0, 1) such that

()
fi(y) ´ fi(x) = (fi ˝ γ)(1) ´ (fi ˝ γ)(0) = (fi ˝ γ)1(ti) = (Dfi)(ci) γ 1(ti) ,

where ci = γ(ti). On the other hand, γ 1(ti) = y ´ x.
Let g(t) = (f ˝ γ)(t). Then the chain rule implies that g 1(t) = (Df )(γ(t))(y ´ x); thus

}g 1(t)}Rm ď }(Df )(γ(t))}B(Rn,Rm)}y ´ x}Rm ď M }x ´ y}Rn .
()
Define h(t) = g(1) ´ g(0) ¨ g(t). Then h : [0, 1] Ñ R is differentiable; thus by the mean
value theorem (Theorem A.9) we find that there exists ξ P (0, 1) such that

()
h(1) ´ h(0) = h1(ξ) = g(1) ´ g(0) ¨ g 1(ξ) ;

thus by the fact that g(0) = f (x) and g(1) = f (y),

}f (x) ´ f (y)}R2 m = h(1) ´ h(0) ď }g(1) ´ g(0)}Rm}g 1(ξ)}Rm
ď M }f (x) ´ f (y)}Rm}x ´ y}Rn

which concludes the theorem. ˝

Example 2.56. Let f : [0, 1] Ñ R2 be given by f (t) = (t2, t3). Then there is no s P (0, 1)
such that

(1, 1) = f (1) ´ f (0) = f 1(s)(1 ´ 0) = f 1(s)

since f 1(s) = (2s, 3s2) ‰ (1, 1) for all s P (0, 1).