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$f(x)=ax^2+bx+c$ where $a, b, c \in R $ and $|f(x)|\leq 1$ on the interval $|x|\leq1$. Prove that $|f'(x)\leq4|$ on the same interval.

I've tried a few approaches - I put $x = 0, 1, -1$ and figured out that $a\leq 2$ and $b,c \leq1$. I tried multiplying the polynomial by powers of $x$ and substituting as well. So far I have managed to prove that $|f'(x)\leq5|$ but I can't prove it for 4.

user1299784
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1 Answers1

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A more strong statement for $f(x)=ax^2+bx+c$:

$$|f(x)|\le 1 \space for \space x=-1,0,1 \implies |f'(x)|\le 4 \space for \space |x|\le 1$$

So, we have $A=a+b+c,B=a-b+c,C=c\in [-1,1]$.

Then, $\frac{3}{2}A+\frac{1}{2}B-2C=2a+b=f'(1)\in [-4,4]$.

Also, $\frac{-1}{2}A+\frac{-3}{2}B+2C=-2a+b=f'(-1)\in [-4,4]$.

And since $f'(x)$ is linear, $f'(x)\in [-4,4]$ for all $x\in [-1,1]$.

Vadim
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    Nice +1, and beat me by $\approx 20$s. A stronger statement - if $p_n(x)$ is a polynomial of degree $n$ s.t. $\lvert p_n(x) \rvert \le 1$ on $\lvert x \rvert \le 1$, then $\lvert {p_n}'(x) \rvert \le n^2$ in the same interval! – Macavity Feb 12 '14 at 04:59
  • Oops, you did not have to remove your answer, I did not mean that in my comment. Sorry, if I was misunderstood. – Vadim Feb 12 '14 at 05:03
  • Thats fine, it essentially was the same as the logic you used. – Macavity Feb 12 '14 at 05:11
  • I posted your other statement about a polynomial of arbitrary degree as a question here: http://math.stackexchange.com/questions/673367. You are welcome to prove it. :) – Vadim Feb 12 '14 at 05:14
  • OK. Will wait a while, if no one else gets it in a few hours will post an answer. – Macavity Feb 12 '14 at 05:25