The real polynomial p(x) = ax3 + bx2 + cx + d is such that |p(x)| ≤ 1 for all x such that |x| ≤ 1. Show that |a| + |b| + |c| + |d| ≤ 7.
Solution
Putting x = 1, -1, 1/2, -1/2 we get:
|a + b + c + d| ≤ 1 (1)
|-a + b - c + d| ≤ 1 (2)
|a + 2b + 4c + 8d| ≤ 8 (3)
|-a + 2b - 4c + 8d| ≤ 8 (4)
(1) and (2) give |a + c| + |b + d| ≤ 1. The easiest way to see this is to consider separately the four cases: a + c ≥ 0, b + d ≥ 0; a + c ≥ 0, b + d < 0; a + c < 0, b + d ≥ 0; a + c < 0, b + d < 0.
But |a| - |c| ≤ |a + c|, so |a| + |b + d| - |c| ≤ 1 (5). Similarly, from (3) and (4) we get
-|a| + |2b + 8d| + |4c| ≤ 8 (6).
Now 5(5) + 2(6) gives: 5 |b + d| + 4 |b + 4d| + 3 |a| + 3 |c| ≤ 21. So it is sufficient to show that 3 |b| + 3 |d| ≤ 5 |b + d| + 4 |b + 4d| (*).
However, (*) is true for any b, d. It is obvious for d = 0. If d is not zero, we can divide by d. Putting y = b/d we have to show that 4 |y + 4| + 5 |y + 1| - 3 |y| - 3 ≥ 0. The easiest way to see this is to consider separately the various ranges for y. Put f(y) = 4 |y + 4| + 5 |y + 1| - 3 |y| - 3. For example if y ≤ -4, we have f(y) = -4(y + 4) - 5(y + 1) + 3y - 3 = -6y - 24, which is ≥ 0 (with equality iff y = -4). For -4 < y ≤ -1, we have f(y) = 4(y + 4) - 5(y + 1) + 3y - 3 = 2y + 8 > 0. For -1 < y ≤ 0, we have f(y) = 4(y + 4) + 5(y + 1) + 3y - 3 = 12y + 18 > 6. For y > 0, we have f(y) = 4(y + 4) + 5(y + 1) - 3y - 3 = 6y + 18 > 18.
Note that equality is achieved by a = 4, b = d = 0, c = -3.
© John Scholes
jscholes@kalva.demon.co.uk
16 Dec 2002