# Minimum value of M such that the cubic modulus value is always less than M for x in between -1 and 1 both included

Minimum value of $$M$$ such that $$\exists a, b, c \in \mathbb{R}$$ and $$\left|4 x^{3}+a x^{2}+b x+c\right| \leq M ,\quad \forall|x| \leq 1$$

What i considered was that putting x= 0, 1 and -1 we get c should be between M and -M and similarily $$|\pm4 + a \pm b +c| \leq M$$ , from this i was not able to conclude anything about M , Or is it possible to solve it using the derivative to be 0 at minima ?

Update : i made a graph so that to try doing from that . I can make a observation that minimum will be when the M ,-M line is touching the both maxima , minima occuring there . (Not sure though if its correct ) does it help ?

Let $$P[x]$$ be the set of all cubic polynomials with leading coefficient $$4$$, and let $$P_M[x]$$ be the subset of $$P[x]$$ consisting of polynomials $$p(x)$$ satisfying the required property that $$-M\leq p(x) \leq +M,$$ for all $$x\in [-1,+1]$$.

Clearly, if $$p(x) \in P_M[x]$$, then also $$q(x)={p(x)-p(-x)\over 2} \in P_M[x]$$. This implies that if $$P_M[x]$$ is nonempty, then so is $$Q_M[x] \subset P_M[x]$$ consisting only of odd functions. Hence we can restrict our attention to polynomials of the form $$f_b(x)=4x^3+bx$$ over the range $$x\in[0,1]$$.

In other words, the problem to solve becomes to compute

$$M^* = \min_b \max_{x\in[0,1]}\left|f_b(x)\right|$$

where $$f_b(x)=4x^3+bx$$.

Note $$f'_b(x) = 12x^2+b$$. If $$b\geq0$$, then $$f_b(x)$$ is nondecreasing over the range $$[0,1]$$. While if $$b<0$$, then $$f_b(x)$$ has a single local extremum in the range $$(0, \infty)$$ at $$x=\sqrt{{-b \over 12}}$$. This local extremum is a minimum that falls in the range $$[0,1]$$ iff $$-12\leq b\leq 0$$.

For $$b\not \in (-12,0)$$ and $$x\in[0,1]$$, then, $$\left|f_b(x)\right|$$ is maximized at one of the endpoints of the interval $$[0,1]$$, and so we have:

$$\begin{eqnarray} \min_{b\not\in(-12,0)} \max_{x\in[0,1]}\left|f_b(x)\right| &=& \min_{b\not\in(-12,0)} \left|f_b(1)\right| \\ &=& \min_{b\not\in(-12,0)} \left|4+b\right| \\ &=& 4 \end{eqnarray}$$

For $$b\in[-12,0]$$, on the other hand, $$\left|f_b(x)\right|$$ is maximized at either $$x=\sqrt{{-b \over 12}}$$ or $$x=1$$, so

$$\begin{eqnarray} \min_{b\in[-12,0]} \max_{x\in[0,1]}\left|f_b(x)\right| &=& \min_{b\in[-12,0]} \max_{x\in\bigl\{\sqrt{{-b \over 12}}, 1\bigr\}}\left|f_b(x)\right| \\ &=& \min_{b\in[-12,0]} \max\left(\left|\sqrt{{-b^3 \over 27}}\right|, \left|4+b\right|\right) \end{eqnarray}$$

The $$\max$$ is minimized at $$b=-3$$; this can be seen by solving the equation

$$\left|\sqrt{{-b^3 \over 27}}\right| = \left|4+b\right|,$$

which yields a cubic equation when squaring both sides, and by factoring that cubic.

The final answer then is $$M^*=1$$, attained by the cubic polynomial $$4x^3-3x$$.

• Sir i understood most part of it , few queries : 1. we restricted the domain to [0,1] because odd functions are symmetric about origin , thats why when we take modulus it will not effect anyhow if we consider just the x lying in [0,1]? 2. We can say for any M we are sure to be having odd functions in the subset , so thats why we restricted right ? 3. why cannot we consider even functions too that will also be surely there we can just restrict ourselves too ax^2 +b too Sir right ? And find min a,b of max |fa,b (x)| right?We just ignored it because its two variable minimization,compared to one? Apr 19 at 15:52
• Odd functions are symmetric about the origin, so the max of their absolute value over $[-a, +a]$ is equal to the max of their absolute value over $[0, +a]$. Apr 19 at 15:59
• In the beginning of my solution, I proved that if $P_M[x]$ is nonempty, then $P_M[x]$ contains at least one odd function. So the minimum $M$ we seek can be attained by some odd function. This is why the restriction to odd functions is justified. Apr 19 at 16:04
• Thanks a lot understood Sir , finally can this method of odd function/even function restriction can be done in case of this too : math.stackexchange.com/q/288152/922054 , as such |ax^2 + bx + c| <= 3/2 means this set contains both odd and even functions so we can focus either on odd or even whichever better for convenience to solve Sir ? Or it will not work Apr 19 at 16:15
• @Orion_Pax I constructed a function $q(x)$ that is an element of $P_M[x]$. Take a look at that function. Is it clear to you that if you changed a minus sign to a plus sign, or if you changed the $2$ to a different value, that the resultant $q(x)$ does not work? Apr 19 at 16:22

Remarks: It is related to the Chebyshev polynomials of the first kind. I used it before e.g. in Minimizing the maximum of $|x^2 - x - k|$

The condition is written as $$|x^3 + (a/4)x^2 + (b/4)x + c/4| \le M/4, \quad \forall -1 \le x \le 1.$$

Recall the well-known result:

For any given $$n \ge 1$$, among the polynomials of degree $$n$$ with leading coefficient $$1$$, $$f(x) = \frac{1}{2^{n - 1}}T_n(x)$$ is the one of which the maximal absolute value on the interval $$[-1, 1]$$ is minimal, where $$T_n(x)$$ is the Chebyshev polynomials of the first kind. This maximal absolute is $$\frac{1}{2^{n - 1}}$$, and $$|f(x)|$$ reaches this maximum exactly $$n + 1$$ times at $$x = \cos \frac{k\pi}{n}, \quad 0 \le k \le n.$$ See: see "Minimal $$\infty$$-norm" https://en.wikipedia.org/wiki/Chebyshev_polynomials, or https://handwiki.org/wiki/Chebyshev_polynomials

Here, $$n = 3$$, $$T_3(x) = 4x^3 - 3x$$, and $$\frac{M^\ast}{4} = \frac{1}{2^2}$$ i.e. $$M^\ast = 1$$.

• Sir that well known result high school level proof is there anywhere? Apr 20 at 4:00
• @Orion_Pax Perhaps Chebyshev polynomials are unkind to high school students. A proof is given in math.stackexchange.com/questions/259428/… Apr 20 at 6:11
• Understood sir thanks Apr 20 at 8:27