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Solutions to Problems; Chapter 7 365

For case (3), it suffices to consider functions g(z) whose minimum value

is -1 and maximum is +l. Then g(z) = -1 + 2(1- c)-~(z - c)~, so that

g’(u) = 4(1 - c)-“(u - c). We wish to maximize ]g’(u)] over all possible

values of c. By using calculus or the fact that the equation

4(u = 4(u - c>

(1 - t>2 (1 - c)2

has c as a double root, one finds that the maximum value is (1 - u)-‘.

Indeed

4(u - c)

&-qyy2=

(1+ c - 2u)2 > o

(1 - u)(l- c)2 -

with equality iff c = 2u - 1. Hence [g’(u)1 < (1 - u)-’ with equality iff
g(z) = -1 + [(z + 1 - 2u)2]/[2(1- u)“].
For case (4), we can consider functions of the form g(z) = -1 +
2(1 + c)-~(z - c)~ where g’(u) = 4(1 + c)-~(u - c). Now, checking the
extreme case c = 0 and c = 1, we find that

4u 4(u - c> 4c(uc+u+l) > o

--= (l+ c)Z (1+c)2 -

$=$-(u-l)=

(1 - c)[(3u + 1) - (u + l)c] > o

(1+c)2 -

Hence ]g’(u)] 5 max(4u, 1 - u),
Thus, for ]u] 5 l/2,

Is’(u)l L m&l + 14, (1 - ]~])-~,4]u], 1 - 14) = (1 - bI>-‘.

In conclusion, we have

Condition on u K, Function yielding equality

l/2 < 1’111 5 1 44 222 - 1

0 5 u 5 l/2 (1 - u)-’ -1+ (z + 1 - 2u)2/[2(1- u)2]

-l/2 5 u < 0 (1+ u)-’ -1 + (2 - 1 - 2u)2/[2(1+ u)“].

Remark. Similar problems were posed in the Putnam Competition (6
Al and 29 A5). For some perspective on the situation, see A.M. Gleason,
R.E. Greenwood, L.M. Kelly, The William Lowell Putnam Mathematical
Competition Problems and Solutions: 1938-1964 p. 207. The Tchebychef
polynomials occur yet again: if p(z) is a polynomial of degree n for which
-1 5 p(z) 5 1 for -1 5 t 5 1, then ]p’(z)] 5 n2 with equality occurring
for some z when p(z) is a Tchebychef polynomial.