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366 Answers to Exercises and Solutions to Problems

5.6. We deal first with the quadratic situation. Let

f(x) = (x - v)(x - u) = (v - x)(u - x) = (x - (v + 4/2)2 - ((v - u)/2)2

where -1 5 u 5 v 5 1. When u < t 5 v, then -1 5 f(z) < 0 with
equality on the left iff v = -u = 1, x = 0. Also

f(l)f(-1) = (1 - u2)(1 - v2) 5 1

with equality iff u = v = 0. Hence, if f(z) # z2, at least one of f(1) and

f(-1) is less than 1. If u + v < 0, then 0 < f(-1) < 1, and if u + v > 0,

then 0 < f(1) < 1.

There are certain cases in which the result is clear:

(1) All of the Zi have the same sign.

(2) The degree of p(z) is 1.

(3) The degree of p(z) is 2, from the above analysis.

Assume the result holds for polynomials of degree less than n 1 3. Let
-1 < a < 0 < b < 1 and p(z) = (2 - ci)(z - 22). .. (x - I,,) where
xi < 0 < x, and -1 5 xi 5 x2 5 ... < x, 5 1. With no loss of
generality, we can take --a < b. Suppose that xi 5 a < 0 < b 5 I,. Then,
if q(x) = (x - xs)...(x - x,-i), then

Ip( = I(2 -x1)(3: - xn>l lq(x)l I ldz)l for 21 5 x I zn.

By the induction hypothesis applied to q(x), Ip( 2 1 and Ip( 2 1
cannot both occur.
Now
p(a)p(b) = fig(xi) where g(t) = (U - t)(b - x).
i=l

Since [g(x)1 5 1 f or a^5 x^5 1, if also g(-1) <^1 or a^5 xi, it follows that
Ig(zi)l 5 1 for each i and that Ip( > 1 cannot occur.
This leaves the situation that -1 5 xi < Q < 0 < z, < b < 1 and
g(-1) = (1+ a)(1 + b) > 1. Now 2(1 + e) > g(-1) > 1, so that a > -l/2.
Suppose if possible that Ip( > 1 and Ip(^2 1. Then we must have that
xn - a > 1, so that x, > l/2. Then

and

IP( Ip(a) I

Ma)l = (a - x1)(x, - a) 1 (a + l)(l - a) ’ Ma)1 ’ l

IP(b) I IP(b) I

Iqcb)l = (b - xl)(b - x,) ’ (b + l)(b - l/2)

Mb) I

’ 2(V)

= IP( > 1