5

Given $a,b,c\in \mathbb{R}$ such that $|x|\leq 1$ and $|ax^2+bx+c|\leq 1$, find the maximum value of

$$ |2ax+b| $$

My Attempt:

Set $x=1$ in $|ax^2+bc+c|\leq 1$ to get

$$ \tag{$\star$}|a+b+c|\leq 1 $$

Similarly, set $x=-1$ in $|ax^2+bc+c|\leq 1$ to get

$$ \tag{$\star\star$}|a-b+c|\leq 1 $$

Similarly, set $x=0$ in $|ax^2+bx+c|\leq 1$ to get

$$ \tag{$\star\star\star$}|c|\leq 1 $$

Now, adding $(\star)$ and $(\star\star)$ gives

$$ |a+c|\leq 1 $$

Subtracting $(\star)$ from $(\star\star)$ gives

$$ |b|\leq 1 $$

Now, we have

$$ -1\leq (a+c)\leq 1\\ -1-c\leq a \leq 1-c $$

Since $-1\leq c\leq 1$, we get

$$ -2 \leq a \leq 2,\; -1 \leq b\leq 1 $$

How can I complete the solution from this point?

Fabrosi
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juantheron
  • 53,015

2 Answers2

3

By the Markov brothers' inequality, for a polynomial $p$ of degree $n$ we have $$ \max_{-1\le x\le 1} |p'(x)| \le n^2\max_{-1\le x\le 1} |p(x)| $$ with equality for Chebyshev polynomials. Taking $p(x)=ax^2+bx+c$ we get $|2ax+b|\le 4$.

1

Without loss of generality we can assume that $b = 0$, otherwise graph of the function $x \mapsto ax^2+bx+c$ could be shifted ($x \mapsto x-b/(2a)$).

So we know that for $x$ such that $|x| \le 1$ we have $|ax^2+c| \le 1$ and want to maximalize $|2ax|$. Assume that $a > 0$ (if $a<0$, just multiply everything by $-1$). So $|2ax| = 2a |x|$. Due to the symmetry ($b = 0$) we can assume even that $x > 0$, so we are maximalizing $2ax$ for $x \in (0,1)$. Thus we are interested in how big $2a$ can get. Setting $x \in \{0, 1\}$ we notice that $|c+a| \le 1$ and $|c| \le 1$. It means that $c$ lies in the interval $[-1,1]$ and $a$ lies in $[-2,2]$. Can $a$ be equal to $2$? Yes, if $c=-1$.

Conclusion: for $a = 2$, $b = 0$, $c = -1$, $x = 1$ we have $|2ax + b| = 4$ and $4$ can't be replaced by any larger number.