I came across this problem as part of a recreational mathematics challenge on university:

Suppose $a, b, c$ are real numbers where for all $ -1 \le x \le 1 $ we have $|ax^2 + bx + c| \le 1$.

Prove that for all $-1 \le x \le 1 ,$

$$ |cx^2 + bx + a| \le 2$$

Was interested to know how you can approach this as I have not so far been successful, especially frustrating as it does not appear to be that hard of a problem?

The inequality to prove is presented accurately by professors and I am sure it holds.

edit: I have tried with a graph, as I only have to worry about $x=1, -1$ and at the maxima/minima of where the quadratic will be at most 2 but I don't know if a graph can constitute a 'prove' and in any case it hasn't worked.

I have tried a lot of algebraic manipulation like completing the square but again, nothing.

I have noticed the symettrey in the inequalities, specifically, if we add them then we get

$$|(a+c)x^2 + 2bx + (a+c)| = |dx^2 + 2b + d|$$

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    $\begingroup$ The title and body don't state the same inequalities at all. $\endgroup$ – Arnaud Mortier Aug 25 '18 at 22:00
  • $\begingroup$ Apologies @Arnaud let me add the modulus to the equations in the title. Thanks. $\endgroup$ – Skunk Works Aug 25 '18 at 22:00
  • $\begingroup$ It's not only that : $\leq $ versus $<$, $1$ versus $2$ $\endgroup$ – Arnaud Mortier Aug 25 '18 at 22:01
  • $\begingroup$ What does it mean by "for all $|x|\leq 1$, though? Does it mean "for all real numbers $x$ with $|x|\leq 1$? Or does it mean "for all complex numbers $x$ with $|x|\leq 1$? And yes, there are many discrepancies in the question as Arnaud Mortier said. $\endgroup$ – Batominovski Aug 25 '18 at 22:06
  • $\begingroup$ I didn't have the symbol for ≤, thanks @Arnaud, will update. So I've tried algebraic manipulation but no success. I have tried with a graph, as I only have to wory about x=1, -1 and at the maxima/minima of where the quadratic will be at most 2 but I don't know if a graph can constitute a 'prove' and in any case it hasn't worked. $\endgroup$ – Skunk Works Aug 25 '18 at 22:07

Let $f(x):=ax^2+bx+c$ and $g(x):=cx^2+bx+a$ for all $x\in\mathbb{R}$. Note that $$g(x)=\frac{1+x}{2}\,f(+1)-\left(1-x^2\right)\,f(0)+\frac{1-x}{2}\,f(-1)\text{ for all }x\in\mathbb{R}\,.$$ By the Triangle Inequality, we see that $$\big|g(x)\big|\leq \frac{1+x}{2}\,\big|f(+1)\big|+\left(1-x^2\right)\,\big|f(0)\big|+\frac{1-x}{2}\,\big|f(-1)\big|$$ for all $x\in[-1,+1]$. Since $\big|f(t)\big|\leq 1$ for all $t\in[-1,+1]$, we obtain $$\big|g(x)\big|\leq \frac{1+x}{2}+\left(1-x^2\right)+\frac{1-x}{2}=2-x^2\leq 2\text{ for each }x\in[-1,+1]\,.$$ The inequality becomes an equality iff $(a,b,c)=\pm(2,0,-1)$ and $x=0$.

Let $\mathbb{D}$ be the open unit disc $\big\{z\in\mathbb{C}\,\big|\,|z|< 1\big\}$ in the complex plane, and write $\mathbb{U}$ for the boundary $\partial\mathbb{D}=\big\{z\in\mathbb{C}\,\big|\,|z|=1\big\}$ of $\mathbb{D}$ (that is, $\mathbb{U}$ is the unit circle centered at $0$). Then, $\bar{\mathbb{D}}$ denote the topological closure $\mathbb{D}\cup\partial\mathbb{D}=\big\{z\in\mathbb{C}\,\big|\,|z|\leq 1\big\}$ of $\mathbb{D}$.

Now, suppose that $a_0$, $a_1$, $a_2$, $\ldots$, $a_n$ are complex numbers such that the entire function $f$ defined by $$f(z):=a_0+a_1z+a_2z^2+\ldots+a_{n-1}z^{n-1}+a_nz^n\text{ for all }z\in\mathbb{C}$$ satisfies the inequality $\big|f(z)\big|\leq M$ for all $z\in\bar{\mathbb{D}}$, where $M$ is a fixed positive real number. I shall prove that, if the entire function $g$ is given by $$g(z):=a_n+a_{n-1}z+a_{n-2}z^2+\ldots+a_1z^{n-1}+a_0z^n\text{ for every }z\in\mathbb{C}\,,$$ then $\big|g(z)\big|\leq M$ for all $z\in\bar{\mathbb{D}}$ as well.

To prove this, we note by the Maximum Modulus Principle that the maximum of $\big|g(z)\big|$ for $z\in\bar{\mathbb{D}}$ is attained on $\mathbb{U}$. Since $g(z)=z^n\,f(\bar{z})$ for every $z\in\mathbb{U}$, we conclude that $$\big|g(z)\big|=\big|f(\bar{z})\big|\leq M\text{ for every }z\in\mathbb{U}\,,$$ and the claim follows immediately.

  • $\begingroup$ @Batominoskski, thank you for your answer and for displaying your work, you actually proved it! I covered the rest of the answer after reading your 'Note' and managed to similary get the answer. I was keen that I not completley fail in solving the problem! I just have to ask, how did you manage to observe that 'Note'? I can see it as a transformation to the original inequality but it is such an elegant solution, I admit I would not have thought to transform it like that esp. with naming the equations f & g and dividing along the points 1, 0 and -1. Any tips on how to think like this? $\endgroup$ – Skunk Works Aug 26 '18 at 0:06
  • $\begingroup$ @Batominoskski, thank you for the second solution! love that we can include complex numbers for another type of proof. Just going through your answer and trying to use it as hints as I cover my screen with my hand so I can try and solve with second way you have provided lol. Give me some time, will get back to you. looks very interesting! $\endgroup$ – Skunk Works Aug 26 '18 at 0:13
  • $\begingroup$ @SkunkWorks I decided to add this. When I saw your problem, at first, I thought you wanted $x$ to be complex, as the upper bound $1$ follows directly from the Maximum Modulus Principle. That is why we were very confused when you gave two upper bounds $1$ and $2$. But anyhow, it is also interesting to include this solution. $\endgroup$ – Batominovski Aug 26 '18 at 0:17
  • $\begingroup$ @Batominoskski, I see, that's why you wanted to know. It's so cool the Maxium Mod Principle can be used if the numbers are complex. Mulling over your answer and keen to see how the prinicple can be applied in this case without fully reading your answer just yet! Again, many thanks for taking the time to share your expertise on this problem of mine. If you have any insights on how I should have realised to transform g(x) like that, feel free to let me know! I get quite frustrated at how math proofs can be so unintuitive at times. $\endgroup$ – Skunk Works Aug 26 '18 at 0:26
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    $\begingroup$ @dxiv OK. Fixed. Thanks for the suggestion. $\endgroup$ – Batominovski Aug 26 '18 at 0:52

Denote $$f(x)=ax^2+bx+c.$$ Hence $$f(0)=c,~~~f(1)=a+b+c,~~~f(-1)=a-b+c.$$ Further $$a=\frac{f(1)+f(-1)}{2}-f(0),~~~b=\frac{f(1)-f(-1)}{2},~~~c=f(0).$$ Thus \begin{align*}|cx^2+bx+a|&=\big|f(0)\cdot x^2+\frac{f(1)-f(-1)}{2}\cdot x+\frac{f(1)+f(-1)}{2}-f(0)\big|\\&=\big|f(0)\cdot (x^2-1)+f(1)\cdot\frac{x+1}{2}+f(-1)\cdot\frac{1-x}{2}\big|\\ &\leq(1-x^2)\cdot|f(0)|+\frac{x+1}{2}\cdot|f(1)|+\frac{1-x}{2}\cdot|f(-1)|\\ &\leq(1-x^2)+\frac{x+1}{2}+\frac{1-x}{2}\\ &=2-x^2\\ &\leq 2. \end{align*}

  • $\begingroup$ @mengie1982, thank you for your answer and for showing your working! you proved it! a very elegant solution and I admit, I would not have considered to partition and split the function as such along 1, -1 and 0 but after reading that hint, i managed to also complete the proof like you showed, that was a brillant and very helpful hint. How would you say people like me can better gain such realisations which seem so important to more formal mathematics and proofs? How did you come to realize to split $cx^2 + bx + a$ as such? Any tips on getting better at this? Thanks a million! $\endgroup$ – Skunk Works Aug 26 '18 at 0:18

Based on the Extreme value theorem, the given conditions: $|ax^2 + bx + c| \le 1, -1\le x\le 1,a,b,c\in \mathbb R$ imply (max/min at borders or critical points): $$\begin{cases} -1\le c\le 1\\ -1\le a-b+c \le 1\\ -1\le a+b+c\le 1\\ -1\le c-\frac{b^2}{4a}\le 1\end{cases}.$$ Based on the Extreme value theorem, the conditions to be proved: $|cx^2 + bx + a| \le 2, -1\le x\le 1,a,b,c\in \mathbb R$ imply (max/min at borders or critical pointd): $$\begin{cases} -2\le a\le 2\\ -2\le c-b+a \le 2\\ -2\le c+b+a\le 2\\ -2\le a-\frac{b^2}{4c}\le 2\end{cases}.$$ Note: If $c=0$, then the 4th condition disappears as it is a straight line (meaning no turning point).

The first three conditions are consistent. Indeed, the 2nd and 3rd are straightforward. For the 1st, note that: $$\begin{cases}-1\le c\le 1 \\ -1\le a-b+c\le 1\\ -1\le a+b+c\le 1\end{cases} \Rightarrow \begin{cases}-1\le c\le 1 \\ -1\le a+c\le 1\end{cases} \Rightarrow -2\le a\le 2.$$

We show the 4th condition. First note that: $$-1\le c-\frac{b^2}{4a}\le 1 \Rightarrow -1\le \frac{4ac-b^2}{4a}\le 1 \Rightarrow |4ac-b^2|\le |4a| \ \ \ (1)$$ Hence: $$-2\le a-\frac{b^2}{4c}\le 2 \iff -2\le \frac{4ac-b^2}{4c}\le 2 \\ \iff |4ac-b^2|\le |8c| \stackrel{(1)}{\iff} |4ac-b^2|\le |4a|\le |8c| \iff |a|\le 2|c|,$$ which is true.


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