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Let $f(x)\in \mathbb{R}[x]$ be a quadratic polynomial with $|f(-1)|\leq 1, |f(0)| \leq 1, |f(1)|\leq 1.$ Prove that for any $|x|\leq 1$, $|f(x)|\leq 5/4$.

I'm not sure how to solve this problem, and below is my attempt. One example quadratic is obviously $f(x)=x^2$. Write $f(x) = ax^2 + bx+c.$ Note that we may assume WLOG that $a>0$ because for $a<0,$ we can replace $f(x)$ with $-f(x)$ and obtain the same conclusion ($|f(x)|=|-f(x)|$). Then $|a-b+c|, |c|, |a+b+c|\leq 1.$ Suppose $|x|\leq 1.$ We need to show that $|ax^2 + bx+c|\leq 5/4.$ Note that since $f$ is convex, on any interval $[s,t]$, the maximum value of $f(x)$ is attained at either $a$ or $b$. Since f is continuous, the minimum of $f(x)$ is attained in $[s,t]$ for any $s<t\in \mathbb{R}.$ We also know that $|f(x)|$ attains its maximum either when f attains its minimum or maximum. Indeed suppose $f(x)$ is neither a min nor a max. WLOG, suppose $f(x)$ is negative. Then $f(x)$ can be decreased, which increases $|f(x)|$. Perhaps Lagrange interpolation might be useful?

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    $\begingroup$ You probably meant "attained at either $s$ or $t$". $\endgroup$
    – jjagmath
    Oct 23, 2023 at 2:33
  • $\begingroup$ Note that we only need to look at the graph of $f$ in the square $[-1, 1]\times [-1, 1]$, which is symmetric w.r.t both $x$ and $y$ axis. Hence WLOG, we can only consider the case where the leading polynomial coefficient is positive and the axis of $f(x)$ is between $0$ and $1$. If we narrow our sight to this problem, the worst case scenario(i.e when $f(x)$ has the smallest minimum) will be when $f(-1)=1$, $f(0)=f(1)=-1$. I believe tackling this case will give some hint/insight. $\endgroup$
    – Joshua Woo
    Oct 23, 2023 at 3:49
  • $\begingroup$ The max/min will be at f(-b/2a) = c - b^2/4a . I would guess something with the triangle inequality $\endgroup$
    – qwr
    Oct 23, 2023 at 4:45
  • $\begingroup$ In my opinion: you have $|a+c-b| \le1$ and $|a+c+b| \le 1$, this yields a maximum value for $|b|$. In top of that, you know that $|c| \le 1$. You might combine this for finding a maximum value for $|a|$. $\endgroup$
    – Dominique
    Oct 23, 2023 at 7:13

2 Answers 2

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We can express the coefficients of $f(x)$ in terms of $f(-1), f(0), f(1)$: $$f(x) = \left(\frac12 f(1) + \frac12 f(-1) - f(0)\right) x^2 + \left(\frac12f(1) - \frac12f(-1)\right) x + f(0).$$ (The proof is easy.)

Thus, we have, for all $x\in [-1, 1]$, \begin{align*} |f(x)| &= \left|(1 - x^2)f(0) + \frac{x^2 + x}{2}f(1) + \frac{x^2 - x}{2}f(-1)\right|\\[6pt] &\le |1 - x^2|\cdot |f(0)| + \left|\frac{x^2 + x}{2}\right|\cdot |f(1)| + \left|\frac{x^2 - x}{2}\right|\cdot |f(-1)|\\[6pt] &\le |1 - x^2| + \left|\frac{x^2 + x}{2}\right| + \left|\frac{x^2 - x}{2}\right|\\[6pt] &= (1 - x^2) + \frac{|x|(1 + x)}{2} + \frac{|x|(1 - x)}{2}\\[6pt] &= -x^2 + |x| + 1\\ &= - (|x| - 1/2)^2 + 5/4\\ &\le 5/4. \end{align*}

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By the triangle inequality $$|ax^2+bx+c|=\left|\left(\frac{1}{2}f(1)+\frac{1}{2}f(-1)-f(0)\right)x^2+\frac{1}{2}\left(f(1)-f(-1)\right)x+f(0)\right|\leq$$ $$\leq\frac{1}{2}|f(1)|\cdot|x^2+x|+\frac{1}{2}|f(-1)|\cdot|x^2-x|+|f(0)|\cdot|x^2-1|\leq$$ $$\leq\frac{1}{2}|x^2+x|+\frac{1}{2}|x^2-x|+|x^2-1|\leq\frac{1}{2}\cdot\frac{1}{4}+\frac{1}{2}\cdot\frac{1}{4}+1=\frac{5}{4}.$$

The last inequality is true by the following reasoning.

Let $g(x)=\frac{1}{2}|x^2+x|+\frac{1}{2}|x^2-x|+|x^2-1|.$

Thus, $g$ is an even function and we can assume $0\leq x\leq1.$

In this case by AM-GM we obtain: $$g(x)=\frac{1}{2}(x^2+x)+\frac{1}{2}(x-x^2)+1-x^2=1+x-x^2=$$ $$=\frac{5}{4}-\frac{1}{4}-x^2+x\leq\frac{5}{4}-2\sqrt{\frac{1}{4}\cdot x^2}+x=\frac{5}{4}.$$ The equality occurs for $x=\frac{1}{2},$ which gave a legitimation to write $$f(x)\leq \frac{1}{2}\cdot\frac{1}{4}+\frac{1}{2}\cdot\frac{1}{4}+1=\frac{5}{4}.$$

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    $\begingroup$ @MichaelRozenberg It is nice. (+1) In my opinion, $\frac{1}{2}|x^2+x|+\frac{1}{2}|x^2-x|+|x^2-1|\leq\frac{1}{2}\cdot\frac{1}{4}+\frac{1}{2}\cdot\frac{1}{4}+1$ is easy to misunderstand. I think you just $\frac{1}{2}|x^2+x|+\frac{1}{2}|x^2-x|+|x^2-1|\leq \frac54$ then your proof of the last inequality. $\endgroup$
    – River Li
    Oct 23, 2023 at 8:48
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    $\begingroup$ @RiverLi, Michael: The question is a duplicate. It took me less than a minute to find an identical question with Approach0 $\endgroup$
    – Martin R
    Oct 23, 2023 at 17:57
  • $\begingroup$ @MartinR Yes, it is. $\endgroup$
    – River Li
    Oct 23, 2023 at 22:54

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