Minimum and Maximum of Quadratic on an Interval

Evaluate at x = 0, ½, 1:
• f(0) = c
• f(½) = a/4 + b/2 + c
• f(1) = a + b + c

Compute the linear combination:
f(0) − 2f(½) + f(1) = c − 2(a/4 + b/2 + c) + (a + b + c)
= c − a/2 − b − 2c + a + b + c = a/2

Bound via triangle inequality

So a = 2[f(0) − 2f(½) + f(1)].
By the triangle inequality and the constraint |f(x)| ≤ 1 on [0,1]:
a = |2f(0) − 4f(½) + 2f(1)|
≤ 2|f(0)| + 4|f(½)| + 2|f(1)|
≤ 2(1) + 4(1) + 2(1) = 8

Equality holds when each |f(0)| = |f(½)| = |f(1)| = 1 and the signs satisfy:
f(0) = 1, f(½) = −1, f(1) = 1

Solving the system: c = 1, a/4 + b/2 + c = −1, a + b + c = 1
gives a = 8, b = −8, c = 1.
The maximum value of a is 8. a ≤ 8

8 positive values of a satisfy the equation