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Let a function $f(x) = ax^2 + bx + c$, where $a, b, c \in R$, satisfy $-1 \leq f(x) \leq 1$ for all $x \in [0, 1]$ then which of the following conclusions can be made?

A) $|a| \leq 8$

B) $|b| \leq 8$

C) $|c| \leq 1$

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

I know that we can get $|c| \leq 1$ by substituting $x = 0$. Apart from this I don't have any idea about how to proceed.

Kremlin
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  • Tricky question! I think a profitable approach for A) is to consider on which interval of length $1$ a parabola attains the smallest range of values - and note that $a$ is invariant under horizontal and vertical translation. For B), you can consider $x = \tfrac 12$, and use your inequalities for $a$ and $c$ to reason about $|b|$. Then don't think too hard about D)! – Izaak van Dongen Apr 09 '22 at 13:05
  • So I tried the approach you suggested. I think the interval is $-b/2a \pm 1/2$.Surprisingly I am getting a stronger result for the values of a and b. I get $|a| \leq 4$ and $|b| \leq 6$. This makes me think that I may have done something wrong. – Kremlin Apr 10 '22 at 13:52
  • Good work! I agree with that interval - though I think you can have $|a| = 8$. For instance, take $8(x-\tfrac 12)^2 - 1$. Possibly you forgot that the range of values $f$ takes on $[0, 1]$ is assumed to have size at most $2$, not size at most $1$? – Izaak van Dongen Apr 10 '22 at 14:14
  • Yes that was my mistake. Taking the correct range I do get $|a| \leq 8$. I also get $|b| \leq 8$ now. So I am bit new to the site. What do I do now that the question has been solved. Is there any way to mark it as solved. – Kremlin Apr 10 '22 at 15:52
  • If you like, you can post an answer below and then accept it! It's fine to just leave it though. Maybe someone else would like to write an answer and then you can accept that. – Izaak van Dongen Apr 10 '22 at 16:04

2 Answers2

1

We can reduce the range of searching we will need to do by only considering $ \ a \ > \ 0 \ \ , \ $ since the case for $ \ a \ < \ 0 \ $ produces vertical reflections of those parabolas. As we are interested in the absolute-values of the coefficients, we can omit these redundant parabolas.

It will also be convenient to write the quadratic polynomial in "vertex form" as $ \ f(x) \ = \ a·(x - x_{v})^2 + y_{v} \ \ , \ $ where the coordinates of the vertex have the familiar expressions $ \ x_v \ = \ -\frac{b}{2a} \ $ and $ \ y_{v} \ = \ c - a· x_{v}^2 \ = \ c - \frac{b^2}{4a} \ \ . $ We will only need to examine $ \ 0 \ \le \ x_{v} \ \le \ \frac12 \ \ , \ $ since the parabolas for $ \ \frac12 \ \le \ x_{v} \ \le \ 1 \ \ $ are just reflections of the others about the line $ \ x \ = \ \frac12 \ \ . $

Using this form for the quadratic function, we can write function values in terms of the location of the parabola's vertex as $$ f(x_v) \ \ = \ \ y_{v} \ \ , \ \ f(0 ) \ \ = \ \ a· x_v^2 \ + \ y_{v} \ \ = \ \ f(x_v) \ + \ a· x_v^2 \ \ , $$ $$ f(1) \ \ = \ \ a· (x_v - 1)^2 \ + \ y_{v} \ \ = \ \ f(x_v) \ + \ a· (x_v - 1)^2 \ \ . $$

In this way, we can examine the values of $ \ f(x) \ $ at the endpoints of $ \ [0 \ , \ 1] \ $ simply as a function of the placement of the vertex. Because the parabolas "open upward", we get the greatest possible range for $ \ a \ $ by positioning the vertex "as low as permitted" and the function values at the interval endpoints "as high as permitted". Thus we have $$ \ f(x_v) \ \ = \ \ -1 \ \ \rightarrow \ \ f(0) \ \ = \ \ 1 \ \ = \ \ a· x_v^2 \ - \ 1 \ \ , \ \ f(1) \ \ = \ \ 1 \ \ = \ \ a· (x_v - 1)^2 \ - \ 1 \ \ . \ $$

Over $ \ 0 \ \le \ x_{v} \ \le \ \frac12 \ \ , \ $ both of these functions have their greatest value at $ \ x_v \ = \ \frac12 \ \ , \ $ that being $ \ \frac{a}{4} - 1 \ \ $ and that maximum is equal to $ \ 1 \ $ for $ \ \frac{a}{4} - 1 \ = \ 1 \ \Rightarrow \ a \ = \ 8 \ \ . \ $ (Using smaller values of $ \ f(0) \ $ or $ \ f(1) \ $ would give us smaller values for $ \ a \ \ . \ $ Should we decide to check the interval $ \ \frac12 \ \le \ x_{v} \ \le \ 1 \ \ , \ $ the function for $ \ f(0) \ $ would be the same and the function for $ \ f(1) \ $ becomes $ f(1) \ \ = \ \ a· (1 - x_v)^2 \ - \ 1 \ \ , \ $ which does not alter our conclusion.)

If we now include the neglected case of $ \ a \ < \ 0 \ \ , \ $ we have $ \ | \ a \ | \ \le \ 8 \ \ . $

As you noted, we must have for the $ \ y-$intercept $ \ -1 \ \le \ f(0) = c \ \le \ 1 \ \Rightarrow \ | \ c \ | \ \le \ 1 \ \ . \ $
EDIT (8/5) -- (Sorry, I wasn't happy with/convinced by the argument I had here earlier.) Since we set $ \ y_v \ = \ c - \frac{b^2}{4a} \ = \ -1 \ \ , \ $ then $ \ 4ac - b^2 \ = \ -4a \ \Rightarrow \ b^2 \ = \ 4a·(1 + c) \ \ $ for $ \ a \ > \ 0 \ \ . $ Then $$ -1 \ \le \ c \ \le \ 1 \ \ \Rightarrow \ \ 0 \ \le \ 1 + c \ \le \ 2 \ \ \Rightarrow \ \ 0 \ \le \ b^2 = 4a·(1 + c) \ \le \ 8a \ \ . $$ Hence $ \ b^2 \ \le \ 8·8 \ \Rightarrow \ | \ b \ | \ \le \ 8 \ \ . $

So we can immediately conclude that $ \ | \ a \ | + | \ b \ | + | \ c \ | \ \le \ 17 \ \ , \ $ as Izaak van Dongen remarks. All four statements are correct.

Note that these bounds for the coefficients cannot pertain all at once, so while we may say that $ \ | \ a + b + c \ | \ \le \ | \ a \ | + | \ b \ | + | \ c \ | \ \le \ 17 \ \ $ (by the triangle inequality), the bound on the absolute-value of the coefficient sum is substantially smaller (in fact, $ \ | \ f(1) \ = \ a + b + c \ | \ \le \ 1 \ ) \ . $

1

Assume first that $a\geq 0$ and apply , by continuty, the assumption to $x=\pm 1$. By continuity $a\pm b+c \leq 1$ so that $\vert a \vert +\vert b\vert +c \leq 1$ and $\vert a \vert +\vert b\vert - \vert c \vert \leq 1$ and the same is true if $a <0$ while changing $f$ to $-f$. As $\vert c\vert <1$ we get $\vert a \vert +\vert b\vert < 2$, so that $\vert a \vert <2$, $\vert b \vert <2$, and $\vert a \vert +\vert b\vert + \vert c \vert \leq 5$

Thomas
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