Let $\mathit{f}:\mathbb{R}\to\mathbb{R}$, $\mathit{f}(x)=ax^2+bx+c$ such that $|\mathit{f}(x)|\le1$ for any $x\in\mathbb{R}$ with $|x|\le1$. Find the maximum value of $|b|+|c|$. I presume that if $x=0$, then $|b|+|c|$ can't be determined, but I think I am missing an important point. Can anyone help me with a hint? It would be greatly appreciated.
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3$\begingroup$ There seems to be an implicit assumption in the question, namely that it makes sense at all, i.e. that we cannot have arbitrary large $b$ and $c$ and still expect $|f(x)|$ to stay below 1. So a good first step is to check if we are even willing to believe that example. What happens if we take $b$ and $c$ equal to a 1000 (everybody's definition of a 'large' number)? Do we indeed get that $f(x)$ gets too big? Why? That should get you started $\endgroup$– VincentCommented Mar 27, 2023 at 20:38
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3$\begingroup$ When it says "for any $x$" it doesn't mean you can pick one. You need your parabola to have all of its outputs for each $x \in [-1,1]$ to be at most $1.$ $\endgroup$– coffeemathCommented Mar 27, 2023 at 20:59
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2$\begingroup$ Note that the "standard problem" asks for $|a|+|b|+|c|$. Did you perhaps have a transcription error? $\endgroup$– Calvin LinCommented Mar 27, 2023 at 21:03
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2$\begingroup$ @user2661923 To me it means that one has to solve the problem for each real value of $a,$ and define some function $g(a)$ to denote the maximal value of $|b|+|c|$ satisfying $f(x) \le 1$ for $|x| \le 1.$ That done, one maximizes $g(a).$ $\endgroup$– coffeemathCommented Mar 27, 2023 at 21:06
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2$\begingroup$ @coffeemath Or you can just find the max of $|b| + |c|$ over all possible values of $a$ directly (Just pointing out that conditioning on $a$ isn't necessary, though it is a reasonable first step.) $\endgroup$– Calvin LinCommented Mar 27, 2023 at 21:16
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1 Answer
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Hints. If you're stuck, explain what you've tried.
Express $|b|, |c|$ in terms of some $ f(x_i)$.
EG $ |c| = |f(0) | \leq 1$, with maximum achieved when $f(0) = \pm 1$.
What's an expression for $ |b|$?
Hence, find the maximum for $|b|$. When does equality hold?
Hence, show that equality can hold throughout, while satisfying $|f(x)| \leq 1$ for $|x| \leq 1$.
Thus, $ \max \left( |b| + |c| \right) = \max |b| + \max |c|$.
Notes:
- As to why we choose those values of $x_i$, notice that we want to still allow for $|f(x)| \leq 1$ for $|x| \leq 1$.
- EG Requiring $f(x_1 ) = f(x_2) = f(x_3) = 1, f(x_4) = -1$ will not work since it's not a quadratic.
- EG Requiring $f(0.8) = 1, f(0) = -1, f(-0.5) = 1$ will not work since $f(1) > 1$.
- This suggests that we want to focus on (at least) $f(1)$ and $ f(-1)$, alongside $f(0)$.
- Yes, it's risky assuming that we can show $ \max \left( |b| + |c| \right) = \max |b| + \max |c|$, and in general I won't advice to do so.
- In this case, note that $ \max |b|, \max |c|$ basically boils down to specific constraints on particular values of $f(x_i)$, so we could see if equality can hold throughout (subject to the previous note).
- Alternatively, look at bounding $ |b+c| , |b-c|$ in terms of $f(x_i)$. This leads to an alternative solution.
$|b+c| = \frac{1}{2} | f(1) + 2f(0) - f(-1) | \leq 2$.
$|b-c| = \frac{1}{2} | f(1) - 2f(0) - f(-1) | \leq 2$.
$|b| + |c| \leq \max( |b+c| , |b-c|) \leq 2$.
- For the "standard problem" that I was referencing where we're maximizing $|a| + |b| + |c|$, we can extend this solution to show that $$ \max ( | a| + |b| + |c| ) = \max |a| + \max |b| + \max |c|. $$
$$|a| = \frac{1}{2} | f(1) + f(-1) - 2 f(0) | \leq 2.$$
Equality holds when $ f(1) = f(-1) = 1, f(0) = -1$ which occurs when $f(x) = 2x^2 - 1$ (and the negative of that).
We verify that this satisfies the original hypothesis.
answered Mar 27, 2023 at 21:05
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$\begingroup$ Are you saying that the problem is workable, as is, even though the value of the coefficient $~a~$ is unconstrained? $\endgroup$ Commented Mar 27, 2023 at 21:06
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$\begingroup$ @user2661923 Correct. I verified that this approach works. $\quad$ Note that even though $a$ appears unconstrained initially, we can actually do something similar to constrain $|a|$. $\endgroup$ Commented Mar 27, 2023 at 21:08
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$\begingroup$ @user2661923 FYI I included why $|a| \leq 2$. $\endgroup$ Commented Mar 27, 2023 at 21:38
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$\begingroup$ Thank you, this really helped me understand the problem as it is. $\endgroup$– David399Commented Mar 28, 2023 at 7:19