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Let $f(x)=ax^2+bx+c, a, b, c \in \mathbb R$. Suppose $|f(x)|\le1 \ \forall x\in [0,1]$ then prove that $|a|+|b|+|c|\le17$

$|f(0)|\le1\implies|c|\le1$

$|f(1)|\le1\implies|a+b+c|\le1$

Now, $|(a+b+c)-c|\le|a+b+c|+|c|\implies|a+b|\le2$

Also, $|f(\frac12)|\le1\implies|\frac{a}4+\frac{b}2+c|\le1\implies|a+2b+4c|\le4$

Now, $|(a+2b+4c)-(a+b+c)|\le|a+2b+4c|+|a+b+c|\implies|b+3c|\le5$

So, $|(b+3c)-3c|\le|b+3c|+|3c|\implies|b|\le8$

So, $|(a+b)-b|\le|a+b|+|b|\implies|a|\le10$

So, $|a|+|b+|c|\le19$

What's my mistake?

While typing this answer, I came across this thread. It has a nice answer but I still wonder where I went wrong (so that I avoid it in future).

Vishu
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aarbee
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    @Tavish, thanks for the edit. – aarbee May 18 '21 at 12:18
  • Why do you think there is a mistake? I don't see contradictions. – Арсений Кряжев May 18 '21 at 12:27
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    @AmanuelGetachew The result in the OP does not say what you claim... Saying that $|a|+|b|+|c| \leq 19$ does not exclude that $|a|+|b|+|c| \leq 17$. There is nothing wrong in the calculations, it is just that the inequalities (which are valid) are not strong enough to this purpose. – PierreCarre May 18 '21 at 12:40
  • @PierreCarre Maybe the linked answer can provide some hint. It says the bound is optimal when $a=8, b=-8. c=1$. Also, it asks to use inequalities by taking help of second differences. But I don't understand both the points. – aarbee May 18 '21 at 12:54

1 Answers1

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There is no mistake as yet.
Your current conclusion of $ | a | + |b| + |c| \leq 19 $ is a true statement.

However, all that you have shown is that 19 is an upper bound (and so you haven't completely answered the question as yet).

You have not shown that it is the least upper bound, which is the definition of the maximum.
One way of doing so is to show that equality can be acheived (which it cannot be, hence it isn't the maximum).

To fix this, in a similar manner to what you did for $b$, try to show that $ | a| \leq 8$, which strengthens the inequality that you found.

Then, show that equality holds under the conditions:

  1. $|f(1) | = 1$
  2. $| f ( 0.5) | = 1 $
  3. $| f(0) | = 1 $
  4. $ |a| = 8$
  5. $ |b| = 8 $
  6. $|c| = 1 $
    This can be achieved with $ f(x) = 8x^2 - 8x+1$ (via interpolation on $(0, 1), (0.5, -1), (1, 1) $ ), hence the maximum of $ |a| + |b| + |c|$ is 17.
Calvin Lin
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  • Hi. Instead of putting values, and then accepting or rejecting the maximum value one obtained, is there a way by which one can be sure of the maximum they obtained? Say, choosing those inequalities to solve that directly lead to the correct and final answer? I think the linked answer is trying to say something like this by emphasising on 'second differences'. But I don't understand that. – aarbee May 19 '21 at 03:55
  • @aarbee You check the conditions for the inequality to determine what is possible. EG you used $|f(0)| = 1 |f(1)| = 1 , | f(1/2) | = 1 $ to get the bound for $|b|, |c|$. This is satisfied (only) by $ f(x) = \pm(8x^2 - 8x + 1) $, which can be obtained via Lagrange interpolation. $\quad$ "second differences" refer to the Method of Differences, where $D_1 (n) = f(n+1) - f(n) $ and $D_2 (n) = D_1 (n+1) - D_1 (n) = f(n+2) - 2 f(n+1) + f(n)$ (which was what I used to show that $ |a| \leq 8$). – Calvin Lin May 19 '21 at 14:39
  • Calvin, which inequalities are you using to calculate $|b|$? Because different inequalities give different results. Without putting the values back in the given equation, is there a way we can determine which inequality to use so as to get optimal answer? e.g., I guess, to get $|a|$, we needed to use second difference. That would ensure optimal value of $|a|$. But how to get optimal value of $|b|$? Though I have obtained correct value of $|b|$ in my post above but that's just by accident. Because with different combinations, I am getting different results. – aarbee May 20 '21 at 18:34
  • @aarbee My $b$ comes from what you've written above. It involves $f|(0.5)|=1, |f(1)|=1$, along with $c$ (which involves $|f(0)| = 1$. $\quad$ "is there a way we can determine which inequality to use so as to get optimal answer" -> No, not often. It does take some experience and familiarity. Constantly check back against "can this be achieved". EG Even though one can show that $|b| \leq 9$, we only know if the max of $b$ is 9 if it can be achieved. – Calvin Lin May 20 '21 at 18:50