# Maximum value of absolute sum in a polynomial

Suppose that for any $$-1\leq x\leq 1$$, $$|ax^2+bx+c|\leq 1$$. Find the maximum possible value of $$|a|+|b|+|c|$$.

My attempts:

It is suffices to find the maximum value of $$|a|+|b|+|c|$$ if for any $$-1\leq x\leq 1$$, $$(ax^2+bx+c)^2\leq 1$$. I inserted $$x=-1, x=0, x=1$$, and I got that $$c\leq 1, a+c\leq 1, a+b+c\leq 1$$. How to continue? Or maybe it is not ppssoble to solve the problem with my attempts? Any solution?

• Almost identical question: math.stackexchange.com/q/314686/42969 – the same method can be used to solve this question. Mar 3, 2019 at 13:06
• @Phillips Hubert In which year it was posted in Qvant? Mar 3, 2019 at 13:10

Hint: We have $$-1\le a+b+c\le 1$$ and $$|c|\le 1$$ and $$-1\le a-b+c\le 1$$ Multiplying $$-1\le a+b+c\le 1$$ by $$(-1)$$ we get

$$-1\le -a-b-c\le 1$$ adding this to $$-1\le a+b+c\le 1$$ we get

$$-2\le -2b\le 2$$ so $$|b|\le 1$$ The last $$|a|$$ is for you! Second solution: Plugging $$x=0,x=-1,x=1$$ we get $$|c|\le \epsilon,|a+b+c|\le \epsilon,|a-b+c|\le \epsilon$$ we get $$-2\epsilon \le -2c\le 2\epsilon$$ $$-\epsilon \le a+b+c\le \epsilon$$ $$-\epsilon\le a-b+c\le \epsilon$$ doing the same like above and using that $$|a|+|b|=|a-b|$$ we get $$|a|+|b|+|c|\le 4\epsilon$$.

• @Sonnhard How did you use $-1\leq a-b+c\leq1$ and $|b|\leq1$ in your solution? Also, what is your answer? Also, if you'll see some down-voting so it's not mine. Mar 3, 2019 at 12:09
• I computed $$f(-1)=a-b+c$$ so $$-1\le a-b+c\le 1$$ Mar 3, 2019 at 12:22
• How did you use this in your solution? Mar 3, 2019 at 12:22
• multiplying by $-1$ we get $$1\ge -a+b-c\geq -1$$ and then we add the in equality $$-1\le a+b+c\le 1$$ and we get $$-1\le b\le 1$$ Mar 3, 2019 at 12:24
• How did you use it in the solution of the starting problem? Mar 3, 2019 at 12:25

Let $$f(x)=ax^2+bx+c$$.

Thus, $$f(1)=a+b+c$$, $$f\left(-1\right)=a-b+c$$ and $$f(0)=c,$$ which gives $$a=\frac{1}{2}f(1)+\frac{1}{2}f(-1)-f(0),$$ $$b=\frac{1}{2}f(1)-\frac{1}{2}f(-1).$$ We can assume that $$a\geq0$$ and since we can always change $$x$$ at $$-x$$, we can assume also that $$b\geq0.$$

Now, if $$c\geq0$$ we obtain: $$|a|+|b|+|c|=a+b+c=f(1)\leq1.$$

Let $$c<0$$.

Thus, $$|a|+|b|+|c|=a+b-c=f(1)-2f(0)\leq3.$$ The equality occurs for the Chebyshov's polynomial $$f(x)=2x^2-1,$$

which says that $$3$$ is a maximal value.