# Showing that constant term of the polynomial satisfy the below inequality

If a polnomial $$R(x)$$ of ninth degree satisfies $$|a_9x^9 + a_8x^8+..a_0|$$ $$\leq$$ $$1$$ $$\forall$$ x $$\in$$ $$[-1,1]$$ - {$${0}$$} , then show that $$a_0$$ satisfies $$|a_0| \leq 1$$. And does equality every achieved ?
This was originally from a problem: Amongst all polynomials $$p(x)=c_{0}+c_{1} x+\cdots+c_{10} x^{10}$$ with real, coefficients satisfying $$|p(x)| \leq|x|$$ for all $$x \in[-1,1]$$. What is the maximum possible value of $$\left(2 c_{0}+c_{1}\right)^{10 ?}$$. where its easy to see $$c_0$$ is $$0$$ when we put $$x= 0$$ , so we can write $$p(x)$$ = $$x(R(x))$$ and then $$|p(x)| = |xR(x)|$$ $$\leq |x|$$ , cancelling mod x for x $$\neq$$ 0 we get the new question which i posted above.

My method was :
We can let $$x$$ to approach zero and put it in the expression we get for very very close values to $$0$$ that $$a_0 <1$$ , it seems like this is sufficient to conclude the result but not sure if this is correct way to judge it or not. ( That is $$|R(0^+)|$$ = $$|a_9(0^+)^9 .... +a_0 |$$ , all the high order terms leaving the $$a_0$$ can be assumed to approach zero , hence $$|a_0| < 1$$)

• By my method if we plug x= h in the expression itself , where h is very very close to 0 we get that @user2661923 May 12 at 17:48
• I am not disagreeing with your approach. I am asking you to explicitly show your work. If this was a test problem, what work would you have to show in order to have your teacher mark your work as correct? For more information concerning the protocol for posting questions at MathSE, see this article. May 12 at 17:54
• I see i will make it more clear then May 12 at 17:55

Assume that $$a_0 > 1.$$ The demonstration, if $$a_0 < -1$$ will be similar.

Let $$f(x) = a_9x^9 + \cdots + a_1x^1.$$

Let $$M = \max(|a_9|, |a_8|, \cdots, |a_1|).$$

Construct $$g(x) = -(1) \times M \times (x^9 + x^8 + \cdots + x^1).$$

Since you are permitted to choose $$x > 0$$ as small as you want, you can certainly find $$x_0 > 0$$ such that $$g(x_0) > 1 - a_0.$$ By the construction of the function $$g$$, this implies that $$f(x_0) > 1 - a_0$$.

This implies that $$f(x_0) + a_0 > 1,$$ which violates the constraint that $$|f(x) + a_0| \leq 1,$$ for all $$x$$ in the $$[-1,1] - \{0\}$$ interval.

Now, consider what happens if $$a_0 = 1$$. Since it is assumed that $$f(x)$$ is a 9th degree polynomial, you must have that $$a_9 \neq 0.$$ Reading the coefficients $$a_9, a_8, \cdots, a_1$$, from left to right, let $$a_k$$ denote the rightmost non-zero coefficient. For example, if $$k = 5$$, this implies that $$a_4, a_3, a_2, a_1$$ are all $$0$$.

First, I will present a hand-waving argument, that you can not have $$a_0 = 1.$$ Then, I will justify it. Assume that $$a_k > 0.$$ As $$x$$ approaches $$0$$ from above (i.e. from the positive side), $$a_kx^k$$ will dwarf each of $$a_9x^9, a_8x^8, \cdots, a_{k+1}x^{k+1}$$. This is because, you have total freedom to choose $$x > 0$$ as small as you want, and you have that $$x^k$$ is $$\frac{1}{x}$$ times greater than $$x^{k+1}.$$

This guarantees, that you will be able to find an $$x_0 > 0$$ such that $$f(x_0) > 0.$$ This will yield a contradiction, because, since $$a_0 = 1,$$ the result will be that $$|f(x_0) + a_0| > 1.$$

Formal argument:

With $$a_0 = 1$$ (rather than $$-1$$), assume that $$a_k > 0.$$ Here, the corresponding argument if $$a_k < 0$$ will be parallel, since you will be allowed to choose $$x_0 < 0$$ as large as you want.

Without loss of generality, $$k < 9.$$ That is, if $$a_9 > 0$$ is the rightmost non-zero coefficient, then you can obviously find $$x_0$$ such that $$f(x_0) > 0.$$

Let $$M$$ denote $$~\displaystyle \max\left(\frac{|a_9|}{a_k}, \frac{|a_8|}{a_k}, \cdots, \frac{|a_{k+1}|}{a_k}\right).$$

Choose $$x_0 > 0$$ small enough so that

$$\displaystyle (9-k) \times \frac{1}{x_0} > M.$$

This will guarantee that

$$\displaystyle |a_r(x_0)^r| < \frac{a_k(x_0)^k}{9-k} ~: r \in \{9,8,\cdots,k+1\}.$$

This implies that $$f(x_0)$$ is strictly greater than $$0$$.

Then, since $$a_0 = 1,$$ you will have the contradiction that $$|f(x_0) + 1| > 1.$$

• Excellent answer to the problem Sir , in original question actually there were four options given for max (2c_0 +c_1)^10 value to be A) 1 , B)2^10 , C) 3^10 , D ) 4^10 , will it be right to choose option A as answer as such it will never be achieved isnt ? So i think all options are techincally wrong May 13 at 4:13
• @ProblemDestroyer I suggest that you make the question that you are asking in your comment into a separate posting, and provide a link it to this posting, so reviewers can read both postings separately. I also suggest that you add the link to the new posting into this posting. Off the top of my head, the question seems complicated. May 13 at 4:24
• Sir you mean maximum value of c_1 ^2 is a complicated question ? May 14 at 4:48
• @ProblemDestroyer No, your comment discussed the problem of computing the maximum value of $(2c_0 + c_1)^{10}.$ This is the problem that I am referring to, as the one that you should present in a separate posting. May 14 at 4:51
• Hmm i see , by the way Sir i think i got how the maximum can be achievable , if a_9 = 0 then we can have that situation possible that a_1 = 1 isnt ? May 15 at 0:17