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$|ax^2+bx+c|\le1$ for $0\le x\le1$ for real $a,b,c$. Prove the strict inequality $|a|+|b|+|c|\le17$. The best i could do was $|a|+|b|+|c|\le11+7\sqrt2$.

Random guy
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1 Answers1

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Plugging in $x=0$ one gets $|c| \leq 1$.

To extract $a$, we look at second differences and observe that $$|f(0) - 2 f(1/2) + f(1)| \leq |f(0)| + 2|f(1/2)| + |f(1)|$$ $$\leq 4$$ But $$f(0) - 2 f(1/2) + f(1) = c - 2(a/4) - 2(b/2) - 2c + a + b + c$$ $$={a \over 2}$$ So we have $|a| \leq 8$.

To find bounds for $b$, plugging in $x = 1/2$ suffices. We have $$|a/4 + b/2 + c| \leq 1$$ As a result using the triangle inequality we have $$|b/2| \leq |a/4 + b/2 + c| + (1/4)|a| + |c|$$ $$\leq 1 + 2 + 1 = 4$$ Hence $|b| \leq 8$ as well. Putting these all together, we see that $|a| + |b| + |c| \leq 8 + 8 + 1 \leq 17$ as needed. This bound is optimal, as can be seen from the function $8(x - 1/2)^2 - 1 = 8x^2 - 8x + 1$.

Zarrax
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  • Understood . Thanks zarrax. Could you explain just the term 'second differences' as i have never heard of it before. – Random guy Mar 20 '16 at 01:35
  • If the second derivative of a function is constant on some interval, you can bound the maximum of this function's absolute value from below in terms of this constant. Rather than prove this theorem, I used the difference of differences (the "second difference") to get at this second derivative, and then show that $|a/2| \leq 4$. – Zarrax Mar 20 '16 at 05:08
  • I can safely assume $|f(\frac 14)| \leq 0$ which implies $|a + 4b + 16c| \leq 16$. Now, $3|a| \leq 4|a+b+c| + |a+4b+16| + 12|c| = 4 + 16 + 12 = 32$. Or $|a| \leq \frac{32}{3}$. Similarly I can find a different bound if I use $x=\frac 18$ as my analysis point. However, for $x=\frac 13$ it does give $|a| \leq 8$. What am I missing? – Sandeep Deb May 08 '18 at 18:56
  • Similarly, if we take the double difference for $|f(\frac 12)|, |f(\frac 14)|, |f(0)|$, we end up with $|a| \leq 32$ – Sandeep Deb May 08 '18 at 19:29
  • Hi Zarrax, when you say 'If the second derivative of a function is constant on some interval, you can bound the maximum of this function's absolute value from below in terms of this constant.', what does that mean? Can you elaborate? – aarbee May 18 '21 at 18:10
  • Hi Zarrax, can I request you again to answer my question? Thanks. – aarbee Jun 20 '21 at 11:21
  • @aarbee Sorry about the delay, I don't come here much anymore. So there's something called the Van der Corput lemma for sublevel sets (https://en.m.wikipedia.org/wiki/Van_der_Corput_lemma_(harmonic_analysis)) that gives such statements. Look at the last line of that wiki page. The apply it for $\phi^{(k) }(x) > 1$ but you can scale it to hold for $\phi^{(k)}(x) > c$ for any constant $c$. So you can use it for $k = 2$ here. – Zarrax Jul 11 '21 at 15:50
  • @Zarrax In our question, $f''(x)=c\implies |f(x)|>c?$ Is this what you are saying? – aarbee Jul 11 '21 at 15:56
  • @aarbee It will imply that the maximum of $|f(x)|$ on the interval is at least some constant $d > 0$, where $d$ will depend on both $c$ and the length of the interval (assuming $c \neq 0$ of course). – Zarrax Jul 11 '21 at 16:48
  • Thankyou for your comments. But I am sorry I am not understanding much. My main doubt is why you took second differences to calculate $a$? And why you calculated $b$ the way you did? By plugging any other value, we could have obtained different bounds. How do we ensure that we get optimal bounds? My questions are same as that of @SandeepDeb above. – aarbee Jul 11 '21 at 17:23
  • Also, I had asked the same question here. If you want to reply there, that would be great. Once again thankyou for your time and engagement. – aarbee Jul 11 '21 at 17:24
  • The idea is that if $f(x)$ is a polynomial $ax^2 +bx + c$, then $f(x + 1/2) - f(x)$ is a linear polynomial $ax + ...$. Then doing the same operation again on $f(x + 1/2) - f(x)$, one gets $f(x + 1) - 2 f(x + 1/2) - f(x) = a$. When $x = 0$ this is a symmetric way of getting an expression for $a$, which one might hope leads to good bounds on $a$. And in this case it did. – Zarrax Jul 11 '21 at 17:31