### 29th Putnam 1968

Problem A5

Find the smallest possible α such that if p(x) ≡ ax2 + bx + c satisfies |p(x)| ≤ 1 on [0, 1], then |p'(0)| ≤ α.

Solution

Answer: 8. Extreme case is ±(8x2 - 8x + 1).

Note that |p'(0)| = |b|. So the question is how large we can make |b| and still be able to find a and c such that p(x) lies between -1 and 1 on [0 ,1].

wlog a > 0. Write p(x) = a(x + b/2a)2 + c. If -b/2a ≤ 0, then p(x) has its minimum on [0, 1] at 0 and its maximum at 1, so we require p(1) - p(0) ≤ 2 (we can then adjust c to get |p(x)| ≤ 1 on the entire interval). But p(1) - p(0) = a + b, so b ≤ 2 (and the extreme case is p(x) = 2x - 1).

If -b/2a ≥ 1, then p(x) has its maximum at 0 and its minimum at 1 (on the interval). So we require p(0) - p(1) < 2 and hence a + b ≥ -2. But -b ≥ 2a, so a ≤ 2 and b ≥ -4 (and the extreme case is 2x2 - 4x + 1).

If 0 < -b/2a ≤ 1/2, then p(x) has its minimum at -b/2a and its maximum at 1. So we require a + b + b2/4a ≤ 2 or (b + 2a)2/4a ≤ 2. But 0 < -b ≤ a, so (b + 2a)2 ≥ a2 and hence a2/4a ≤ 2 or a ≤ 8 and 0 < -b ≤ 8.

Finally, if 1/2 ≤ -b/2a < 1, the p(x) has its minimum at -b/2a and its maximum at 0. So we require b2 /4a ≤ 2. So again a2 ≤ b2/4a ≤ 2. Hence a ≤ 8. But b2 ≤ 8a, so -b < 8. Hence -8 ≤ b < 0.