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Let $ax^2+bx+c$ be a quadratic polynomial with real coefficients such that $$|ax^2+bx+c| \leq 1,$$ for $ 0\leq x\leq 1$. Prove that $$|a|+|b|+|c|\leq 17$$

How to proceed in this particular question. Sorry I can't show any work because I really not getting how to initiate.

StubbornAtom
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H.P. Das
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3 Answers3

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Let $f(x) = ax^2 + bx + c$ We know that $$ \left|\frac{a}2\right| = |f(0) + f(1) - 2f(0.5)| = |[f(0) - f(0.5)] - [f(0.5) - f(1)]|\\ \leq |f(0) - f(0.5)| + |f(0.5) - f(1)| \leq 2+2 = 4 $$ so $|a| \leq 8$. Clearly, $|c| = |f(0)| \leq 1$. That leaves $b$. We get $$ |b| = |4f(0.5) - f(1) - 3f(0)| \leq 3|f(0.5)-f(0)| + |f(0.5) - f(1)| \leq 3\cdot 2 + 2 = 8 $$ which is what we need.

It's also worth noting that $$ f(x) = 8x^2 - 8x + 1 $$ demonstrates that $17$ is a strict bound, so we cannot do any better.

Arthur
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  • Furthermore, $f(x) = 8x^2 - 8x + 1$ satisfies the conditions, so 17 is the best possible bound. – Michael Lugo Jun 17 '16 at 18:26
  • @MichaelLugo Yes, that is what I meant when I said "(...) demonstrates that $17$ is a strict bound, so we cannot do any better." – Arthur Jun 17 '16 at 19:45
  • Why can't we do $|\frac a8| = |f(0) + f(\frac 12) - 2f(\frac 14)|$. This would give us $|a| \leq 32$. What is special about 0, 1/2 and 1 as reference points for doing double difference? – Sandeep Deb May 08 '18 at 19:34
  • @SandeepDeb But $|a|\leq32$ clearly isn't good enough to get $17$ in the end. The special thing about $0,\frac12,1$ is that it lets us take full advantage of $|f(x)|\leq 1$. If you use $1,\frac14,\frac12$ instead, then you're not using the fact that $|f(x)|\leq1$ for $x>\frac12$. Consequently, the range of possible $a$ you get will allow for, say, $f(1)=7$, if $f(0)=f(1/2)=1$ and $f(1/4)=-1$. This is not restrictive enough. – Arthur May 08 '18 at 20:45
  • @Arthur - thank you for the response. I am still not able to wrap myself around the justification of taking 1/2. Double difference should work on any three points. I can take two (or multiple) sets of points to cover the $x \leq 1$ domain. I have presented a different approach to justifying the importance of $x=\frac 12$ in the answer below. Can you please share your thoughts? – Sandeep Deb May 10 '18 at 12:45
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Hint: Let $P$ your polynomial. Put $P(0)=u$, $P(1/2)=v$ and $P(1)=w$. Then $|u|, |v|, |w|$ are $\leq 1$. Find $a,b,c$ in function of $u,v,w$, and bound them.

Kelenner
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There are a few additional posts like this quora post link, which try to solve this problem via double difference, clever manipulation of triangle inequality and translating the inequalities at $x=0,\frac 12, 1$ to arrive at $a \leq 8$.

Unfortunately I have been finding it difficult to defend the approach against the following arguments:

  • Why can't we use double difference on sample points like $x=0, 1/4, 1/2$ as a viable approach
  • Why does the method of triangle inequality give us different results when we use points like $x=\frac 14$

Following is an alternate approach to solve the problem. Would appreciate if this can be verified/challenged.

The problem can be translated to finding a bounding condition for the set of quadratics which can pass through the bounding box defined by $y=\pm 1, x=0,1$, without intersecting top and bottom edges ($y=\pm 1$). This takes care of the $|\ |$ condition too.

Image 1 - Possible sets

To find the bounding condition, we can simplify the ask by translating the bounding box to $y=0,-2; x=0,1$. We can see that quadratics whose difference in roots is less than 1 and $V_y \lt -2$ won't qualify.

Hence, we can conclude that all quadratics which have distance between roots > 1 and $V_y \leq 2$ can be translated to fit the bounding box.

Image 2 - Bounding limits

Therefore, the limiting quadratic is $8x^2 - 8x$, which when translated to the original bounding box $(y=\pm 1; x=0,1)$ becomes $8x^2 - 8x + 1$ and $-8x^2 + 8x -1$.

Thus $|a| \leq 8$, $|b| \leq 8$ and $|c| \leq 1$. Hence, proving the ask:

$|a| + |b| + |c| \leq 17$

Sandeep Deb
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