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In the question being discussed here Prove that $|f(x)| \le \frac{3}{2}$ when $f(x)=ax^2+bx+c$ i get it how they used the triangle inequality and the bound here that is this : $$|f(x)|=|(\frac{f(1)}{2})(x^2+x)+(\frac{f(-1)}{2})(x^2-x)+(f(0))(1-x^2)|$$

$$\le |\frac{x^2+x}{2}|+|\frac{x^2-x}{2}|+|1-x^2|$$ $\leq5/4$ but i would like to know why this method too dont give the same maximum value as the above gives ? My method : $$|f(x)|$$ = |($$\frac{f(1)+f(-1)-2f(0)}{2} )x^2 + (\frac{f(1)-f(-1)}{2}) x + (f(0))| \leq \frac{1+1-2(-1)}{2} x^2 + \frac{1-(-1)}{2} |x| + 1$$ $$\leq 2x^2 + |x| + 1$$ whose maximum in that domain is even greater than $3/2$ , so whats wrong with this approach ? Triangle inequality like this we cannot apply ? Why so if thats the case?

Orion_Pax
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    There's nothing "wrong" here. Your final inequality is correct. It's just that what you've done is not enough to prove the desired inequality so this approach doesn't lead to a proof. – bjorn93 Apr 19 '22 at 03:03
  • @bjorn93 thanks , though why is it always better to do bounding like that , i mean how to see at glance that making changes changes a bit like what done in the original method leads to strong bound ? By experience? – Orion_Pax Apr 19 '22 at 07:16

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What you have done is correct. It's just that $|f(x)|\le 2x^2+|x|+1$ does not mean that $|f(x)|\le 3/2$ is ruled out. If you had proved, for instance $|f(x)|\le 2$ that too would be correct.

  • Thanks , so how do we think that making the progress like done in original method ? I mean i wouldnt have got any intuition behind making terms of uf(1) + vf(-1) + wf(0) , rather i would just apply triangle inequality at first sight in the quadratic in x^2 itself . – Orion_Pax Apr 19 '22 at 07:18
  • I guess the answer is that you learn how to think mathematically with time, although there may be exceptions. –  Apr 19 '22 at 12:30
  • Understood thanks – Orion_Pax Apr 19 '22 at 12:56
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You correctly identified the greatest possible value of $a,$ the greatest possible value of $b,$ and the greatest possible value of $c.$

What you have not accounted for is the fact that these greatest possible values cannot all occur simultaneously. The greatest $a$ requires $f(−1)=1$ but the greatest $b$ requires $f(−1) = -1.$ The greatest $a$ requires $f(0)=-1$ but the greatest $c$ requires $f(0)=1.$

These facts do not invalidate what you wrote; what you wrote is still true. But some ways of setting an upper bound are "tighter" and others are "looser". The answer to the question was a method with a tighter bound than yours.

David K
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  • How do we get to know which type of method would result in strong bounding rather than weak , is it just by experience? – Orion_Pax Apr 19 '22 at 07:14
  • Sir Is there a way to get a intuition behind converting it to uf(1)+vf(-1) +wf(0) rather than solving from x^2 ,x,1 itself ? I mean is it like a fact or something which will surely give a strong bound than the x^2... one always in other problems too? – Orion_Pax Apr 19 '22 at 07:20
  • I believe the other approach can be generalized to cases that have different intervals for the domain of the function and for the range of the three function values. I don't know how much farther the approach applies. At some point as you encounter new problems you just have to try different things, not knowing which will work. There are (much harder) math problems that nobody has solved yet, although many have tried. – David K Apr 19 '22 at 12:10
  • Understood thanks Sir – Orion_Pax Apr 19 '22 at 12:56