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 $|a|+|b|+|c|\le17$

Question

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).

Answer 1

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.