Kenneth Kuttler

190 CHAPTER 7. THE DERIVATIVE

7.5 A Mean Value InequalityThe following theorem will be very useful in much of what follows. It is a version of themean value theorem as is the next lemma. The mean value theorem depends on the functionhaving values in R and in the lemma and theorem, it has values in a normed vector space.

Lemma 7.5.1 Let Y be a normed vector space and suppose h : [0,1]→Y is continuousand differentiable from the right and satisfies

∥∥h′ (t)∥∥≤M, M≥ 0. Then ∥h(1)−h(0)∥≤M.

Proof: Let ε > 0 be given and let

S≡ {t ∈ [0,1] : for all s ∈ [0, t] ,∥h(s)−h(0)∥ ≤ (M+ ε)s}

Then 0 ∈ S. Let t = supS. Then by continuity of h it follows

∥h(t)−h(0)∥= (M+ ε) t (7.12)

Suppose t < 1. Then there exist positive numbers, hk decreasing to 0 such that

∥h(t +hk)−h(0)∥> (M+ ε)(t +hk)

and now it follows from 7.12 and the triangle inequality that

∥h(t +hk)−h(t)∥+∥h(t)−h(0)∥= ∥h(t +hk)−h(t)∥+(M+ ε) t > (M+ ε)(t +hk)

Thus∥h(t +hk)−h(t)∥> (M+ ε)hk

Now dividing by hk and letting k→∞,∥∥h′ (t)∥∥≥M+ε,a contradiction. Thus t = 1. Since

ε is arbitrary, the conclusion of the lemma follows. ■

Theorem 7.5.2 Suppose U is an open subset of X and f : U → Y has the propertythat Df (x) exists for all x in U and that, x+ t (y−x) ∈U for all t ∈ [0,1]. (The linesegment joining the two points lies in U.) Suppose also that for all points on this linesegment, ∥Df (x+t (y−x))∥ ≤M. Then ∥f (y)−f (x)∥ ≤M |y−x| . More generally if∥Dvf (y)∥≤M for all y on the segment joiningx andx+v, then ∥f (x+av)−f (x)∥≤Ma. Also Davf (x) = aDvf (x) if a ̸= 0.

Proof: Let h(t) ≡ f (x+ t (y−x)) .Then by the chain rule applied to h(t), h′ (t) =Df (x+ t (y−x))(y−x) and so∥∥h′ (t)∥∥= ∥Df (x+ t (y−x))(y−x)∥ ≤M ∥y−x∥

by Lemma 7.5.1, ∥h(1)−h(0)∥= ||f (y)−f (x)|| ≤M ||y−x|| . For the second part, leth(t)≡ f (x+ tav). Then if a ̸= 0,

h′ (t) = limh→0

h(t +h)−h(t)h

≡ limh→0

aha

(f (x+ tav+hav)−f (x+ tav))

= Dvf (x+ tav)a.

This shows that Davf (x) = aDvf (x) . Now for the inequality, there is nothing to showif a = 0 so assume a ̸= 0. Then by assumption and Lemma 7.5.1, ∥h(1)−h(0)∥ =∥f (x+av)−f (x)∥ ≤Ma. ■

190 CHAPTER 7. THE DERIVATIVE7.5 A Mean Value InequalityThe following theorem will be very useful in much of what follows. It is a version of themean value theorem as is the next lemma. The mean value theorem depends on the functionhaving values in R and in the lemma and theorem, it has values in a normed vector space.Lemma 7.5.1 Let Y be a normed vector space and suppose h : [0,1] + Y is continuousand differentiable from the right and satisfies ||h' (t)|| <M, M > 0. Then \\h(1) —h(0)|| <M.Proof: Let € > 0 be given and letS= {t € [0,1]: forall s € [0,7], ||h(s) —h(0)|| < (W+e)s}Then 0 € S. Let t = sup S. Then by continuity of h it follows\|h (t) —h(0)|| = (M+e)t (7.12)Suppose ¢t < 1. Then there exist positive numbers, /, decreasing to 0 such that[Fa (t+ hg) — h(0)|| > (M+ €) (t+ he)and now it follows from 7.12 and the triangle inequality that[h(t +h) — h(t) || + ||h (2) — hb (0)|= |Rht+hg)-h@||+(M+e)t > (M+) (t+lx)Thus|b (t+ hy) —h(t)|| > (M+ €) heNow dividing by h; and letting k — 9, ||’ (t) | > M-+€,a contradiction. Thus t = 1. Since€ is arbitrary, the conclusion of the lemma follows.Theorem 7.5.2 Suppose U is an open subset of X and f :U —Y has the propertythat Df (a) exists for all x in U and that, x+t(y—«x) €U for allt € [0,1]. (The linesegment joining the two points lies in U.) Suppose also that for all points on this linesegment, ||Df (w+t(y—a))|| <M. Then || f (y) — f (x)|| <M|y—2|. More generally ifDu f (y)|| <M for all y on the segment joining x and x + v, then || f (x+av) — f (x)|| <Ma. Also Dav f (") = aDvf (x) ifa 40.Proof: Let h(t) = f (w7 +t(y—2)).Then by the chain rule applied to h(t), h’(t) =Df («+t(y—2))(y—a) and so||P’ (¢)|| = |DF (a +t (y—2)) (y—2)|| < M|ly—2|by Lemma 7.5.1, ||/h (1) — h(0)|| = ||f (y) — f (@)|| < M||y — x]|. For the second part, leth(t) = f (w@+tav). Then ifa 40,h(t) = tim PUM BO = lim — (f (w+tav +hav)— f (@+tav))= Dyf (x@+tav)a.This shows that Day f (x) = aD, f (2). Now for the inequality, there is nothing to showif a= 0 so assume a #0. Then by assumption and Lemma 7.5.1, ||h (1) —A(0)|| =\|f (a +av)—f(a)||<Ma.