1
$\begingroup$

Given quadratic polynomial $f(x)$ satisfies $\lvert ax^2 + bx + c \rvert \leq \lvert x \rvert$ for all $x \in [-1,1]$. Show that $\lvert a \rvert + \lvert b \rvert \leq 1$.

My approach was graphical , when we make a quadratic and try satisfying the given inequality, we get few points to notice from the graph itself. Suppose $a \geq 0$ then we will have a upward opening parabola. For it to satisfy the inequality that is graph of |quadratic| to lie below the $\lvert x \rvert$ graph on $[-1,1]$.

(1) It must have equal roots in $[-1,1]$ (that is, it should be touching the $x$-axis only), consider it intersects at two points on $x$-axis then the graph of |quadratic| will have that negative curve to be reflected above, making it always intersect the $|x|$ twice on $[-1,1]$. But that we don't want to.

(2) We need the equal roots to occur at $x=0$. If not then because of symmetric nature about the minima point of the parabola, we would be having in one part that is either $[0, 1]$ or $[-1, 0]$ to be satisfied by the $\lvert x \rvert$ inequality, but in other half its not or partially satisfied in both regions.

From this we can conclude $f(x) = +x^2$ as the possible solution to it. Same we can argue for $a \leq 0$ we would get $f(x) = -x^2$ as possible solution.

Another case would be of a linear polynomial that is degree one, then in that case by similar analogy we can say only $f(x) = \pm x$ satisfy it. From these all we can observe $\lvert a \rvert + \lvert b \rvert = 1$ is being satisfied so and less than case would be the $f(x) = cx$ or $cx^2$ where $-1 < c < 1$? Hence proved? If its all correct please confirm and any another method after checking whether this all is correct can be also appreciated.

Curious thought: can this graphical approach be applicable for cubic, fourth degree ones? If there were some past research being done in this method, can it be shown here?

$\endgroup$
2
  • 2
    $\begingroup$ Very interesting if somewhat bizarre posting. The problem is directly conquered by analysis. That is, after concluding that $c = 0$, you know that for any two real numbers $r$ and $s$, if they are both positive or both negative, then $|r + s| = |r| + |s|$. So, examining $x = 1, x = -1$ conquers the problem. Therefore, the underlying question is whether you can construct a similar problem, involving a higher degree polynomial, that is more readily conquered by graphical examination rather than analysis. $\endgroup$ May 12 at 19:09
  • $\begingroup$ Understood Sir thanks $\endgroup$ May 13 at 4:05

2 Answers 2

2
$\begingroup$

You can establish this inequality, by only evaluating the polynomial at $x \in \{-1, 0, 1\}$: \begin{align} x &= -1 &&\Longrightarrow &\lvert a - b \rvert &\leq 1 \tag{$-1$}\\ x &= 0 &&\Longrightarrow &\lvert c \rvert &\leq 0 \tag{$0$}\\ x &= 1 &&\Longrightarrow &\lvert a + b \rvert &\leq 1 \tag{$1$} \end{align}

Immediately from Equation $(0)$, we see that $c=0$.

Since the polynomial is quadratic, $a \neq 0$. If $b=0$, then $\lvert b \rvert = 0$, and by Equation $(1)$, $$ \lvert a \rvert + \lvert b \rvert = \lvert a \rvert = \lvert a + b \rvert \leq 1, $$ as desired.

Now, assume that $b \neq 0$. There are two cases: $ab > 0$ (same signs) or $ab < 0$ (opposite signs).

If the coefficients have same signs, then $\lvert a \rvert + \lvert b \rvert = \lvert a + b \rvert$, so by Equation $(1)$,
$$ \lvert a \rvert + \lvert b \rvert = \lvert a + b \rvert \leq 1. $$ On the other hand, if the coefficients have opposite signs, then $\lvert a \rvert + \lvert b \rvert = \lvert a - b \rvert$, so by Equation $(-1)$,
$$ \lvert a \rvert + \lvert b \rvert = \lvert a - b \rvert \leq 1. $$


Can you prove the equations used in the final two paragraphs?

Lemma. For any two nonzero real numbers $a$ and $b$:

  • If $a$ and $b$ have same signs, then $\lvert a \rvert + \lvert b \rvert = \lvert a + b \rvert$.
  • If $a$ and $b$ have opposite signs, then $\lvert a \rvert + \lvert b \rvert = \lvert a - b \rvert$.

Hint: there are $4$ cases to consider.

$\endgroup$
1
  • $\begingroup$ Nice approach thanks Sammy Black $\endgroup$ May 13 at 4:07
1
$\begingroup$

I'll describe a very different approach.

Firstly, $x=0 \implies |c| \leq 0 \implies c=0$.

Now, $x=1 \implies |a+b| \leq 1$ and $x=-1 \implies |-a+b| = |a-b| \leq 1$.

Now, $|a|=\pm a, |b|=\pm b$. So, it's easy to see (just check $4$ possible cases) that $|a|+|b|$ is either $|a+b|$ whenever $ab \geq 0$ and $|a-b|$ whenever $ab \leq 0$.

So, we're done!

Just noticed that @user2661923 has made a very similar comment. If you want to post it as an answer, I'm happy to delete mine.

$\endgroup$
1
  • $\begingroup$ No its fine as both are in all solutions to the problem , i do not mind if both have the same ideas :) thanks for this $\endgroup$ May 13 at 4:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.