Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$.

Question

Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$.

I made quite a few attempts but could not explicitly derive the required condition. I can understand that it happens, but cannot derive that. Please help.

Answer 1

The proof for 3/2 follows, though intuitively, I think that you can limit it to 5/4: it's a quadratic and if you draw an extreme graph then it should go through (-1, 1), (0, -1), (1, -1) with a minimum at (1/2, -5/4), or some reflection of this. Back to 3/2 ...

If $f$ is monotonic in [-1, 1] then trivially, $|f| \le 1$, so assume not.

Then $f$ is differentiable and (being quadratic) can only have one extreme point at $f'(x_0) = 0$, i.e $2ax_0 + b$ = 0, so that $ax_0 = -b/2$. The extreme value of the function at this point is therefore $f(x_0) = x_0(-b/2) + bx_0 + c = bx_0/2 + c$.

Taking the possible values at -1, 0, and 1 gives

$|f(-1)| = |a - b + c| \le 1$; $|f(0)| = | c| \le 1$; $|f(1)| = |a + b + c| \le 1$

So $|c| \le 1 $ and it follows from these inequalites that $|b| \le 1$

Substituting in the value of $f(x_0)$,

$|f(x_0)| = |bx_0/2 + c| \le |b|.|x_0|/2 + |c| \le 1.1/2 + 1 = 3/2$

Continuation of proof to show that $|f(x)| \le 5/4$

(1) If $|a| \ge 1$

At the extremum, $x_0 = -b/2a$ so by substitution $f(x_0) = ab^2/4a^2 -b^2/2a +c$

$|f(x_0)| = |c - b^2/4a| \le |c| + |b|^2/4|a|$. We already proved $|c|, |b| \le 1$ and by assumption $|a| \ge 1$ so $|f(x_0)| \le 1 + 1/4 = 5/4$

(2) If $|a| \le 1$

Take the Taylor expansion of $f(x)$ about $x_0$: $f(x) = f(x_0) + (x-x_0) f'(x_0) + (x-x_0)^2 f''(z)/2$ for $z$ between $x$ and $x_0$. Note that $f'(x_0) = 0$ at the extremum, and $f''(z) = 2a $ for all $z$, so

$f(x) = f(x_0) + a(x-x_0)^2$, and $|f(x_0)| = |f(x) - a(x - x_0)^2| \le |f(x)| + |a|(x-x_0)^2$. By assumption, $|a| \le 1$, so

$|f(x_0)| \le |f(x) | + (x-x_0)^2$

For $x = -1, 0, or +1$ and $x_0$ in [-1, 1], one of these values makes $|x - x_0| \le 1/2$ (take 0 if $-1/2 \le x_0 \le +1/2 $ ; +1 if $x_0 > 1/2$; -1 if $x_0 \le -1/2$), and in all cases the corresponding function value satisfies $|f(x)| \le 1$, so at this x,

$|f(x_0)| \le 1 + (1/2)^2 = 5/4.$

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Footnote in response to comment 24 April 2022....

A) $|f(-1)| = |a - b + c| \le 1$
B) $|f(0)| = | c| \le 1$
C) $|f(1)| = |a + b + c| \le 1$

From (A), $ -1 \le a - b + c \le 1$
So, $-1 \le - a + b - c \le 1$
And, (C) $-1 \le a + b + c \le 1$
Then $-2 \le 2b \le 2 $
Giving $|b| \le 1$

Answer 2

WLOG, let us suppose $f(x)=a(x-u)^2-v$ for $a>0$ and $0< u\le 1$, $v\ge 0$.

Then the premise says $1\ge au^2-v\ge -1$, $1\ge a(1-u)^2-v\ge -1$, $1 \ge a(1+u)^2-v\ge -1$. We need to show that $v\le \frac{3}{2}$.

The above inequalities implies that $1\ge au$ and $2\ge a(1+2u)$.

If $u\in (0,1/2]$, then

$$ v\le 1+au^2\le 1+2u^2/(1+2u)\le\frac{5}{4}. $$

If $u\in [1/2, 1]$, then $$ v\le 1+a(1-u)^2\le 1+(1-u)^2/u\le \frac{5}{4}. $$

Let $u=\frac{1}{2}$, $a=2$, $v=\frac{5}{4}$, then $f=2(x-\frac{1}{2})^2-\frac{5}{4}$ attains the extreme value.

I suppose there should be more effective proof.