# help complete this attempt (real analysis)

Given 2 quadratic functions $f$ and $g$, such that $f(x)=ax²+bx+c$ and $g(x)=cx²+bx+a$ and the absolute value of $f(x)$ is less than $1$, for each $x\in[-1,1]$, prove that the absolute value of $g(x)$ is less than $2$, for each $x\in[-1,1]$.

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And where is the attempt? – t.b. Jun 28 '11 at 14:04
@Theo: $\varnothing$ :-P – Asaf Karagila Jun 28 '11 at 14:29
Are you sure the question states that the absolute value of $f(x)$ is less than 1 or less than or equal to 1? – user38268 Jun 28 '11 at 14:58

Here is a complete solution; but note that a lot of thinking went into reducing the number of cases to consider.

Multiplying through by $-1$, if necessary, we may assume $a\geq c$, and replacing $x$ by $-x$, if necessary, we may assume $b\geq0$.

We note first that $c=f(0)\geq -1$ and $a+b+c=f(1)\leq 1$, therefore $$a\leq 1 -c-b\leq 2-b\ .\qquad(1)$$

Now $g(x)-f(x)=(a-c)(1-x^2)\geq0$; whence we already now that $g(x)\geq f(x)\geq-1$. If $\arg\max_x g(x)\in\{\pm 1\}$ then for all $x$ we have $g(x)\leq\max g(\pm1)=\max f(\pm1)\leq 1$.

So it remains the case that $\arg\max_x g(x)=:\tau \in\ ]{-1},1[\$. In this case we have a local maximum at $\tau$, whence necessarily $g'(\tau)=0$ and $c<0$. It follows that $${b\over 2|c|}=\tau<1\ .\qquad (2)$$ Using (1) and (2) we now obtain $$\max_x\ g(x)=g(\tau)= a-{b^2\over 4c}=a+{b^2\over 4|c|} \leq 2 -b +{b\over2}\leq 2\ ,$$ as stated.

The pair $f(x):=2x^2-1$, $\ g(x):=2-x^2$ shows that the proven inequality is sharp.

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Really I did not understand why c is ncessarily negative in the second case, can you explain more – Mopzer Moreena Jun 29 '11 at 15:16
@Mopzer: $g$ has a local maximum, so its leading coefficient $c$ is negative. – TonyK Jun 29 '11 at 17:01
thanks Tony, yeah it does by using the theorem of parabola extremum – Mopzer Moreena Jun 30 '11 at 10:54

Your two functions are parabolas. The only way the extreme values are inside of $[-1,1]$ is if the vertex $x_0$ of the parabola satisfies $|x_0|<1$ . Then the maximum is reached at the vertex of the parabola if $a<0$ —parabola opens downwards —and the minimum is reached inside when $a>0$. Otherwise, if the vertex of the parabola is not in $[-1,1]$, i.e., if the x-coordinate satisfies $|x|>1$ ,the maximum values will happen at the boundaries, i.e., at either $x=1$ or at $x=-1$. The case of the parabola opening downwards (when $a<0$) comes down to finding the "standard form" $y-y_0=c(x- x_0)^2$, and determining the values of $x_0$ and $y_0$. Check what happens to $f'(x)$ left- and right- of the vertex of the parabola.

Otherwise, the extreme values happen at the endpoints $\{-1,1\}$ , so that the extreme values will be $a+b+c$ (at $x=1$), and/or $a-b+c$ (at $x=-1$) unless $a<0$ and, when you express the parabola in standard form, $|x_0|<1$.

So if $a<0$ and $|x_0|<1$ , then $y_0$ is the maximum value; otherwise, the max will be reached at either of the endpoints, and will be either $a+b+c$ , or $a-b+c$

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Hey, you're making progress :) Only one fix this time! To typeset $\{-1,1\}$ use $\{-1,1\}$ – t.b. Jun 28 '11 at 16:03
Thanks, Theo, even I did not want to read my old messages :). Just hope the content is also good. – gary Jun 28 '11 at 16:07
No complaints contentwise. – t.b. Jun 28 '11 at 16:09
Thanks again for your help and patience, Theo. – gary Jun 28 '11 at 16:11
@: D Lim; you are right, less or equal to 1 for f(x), and less or equal to 2 for g(x) – Mopzer Moreena Jun 28 '11 at 19:51