pachaquis3s

2022-06-22

Let $C\ge 2$ and $L>0$ be fixed. Let $n\in \left\{1,2,\dots \right\}$. Does there exist a polynomial g of degree n such that

holds at least for n big enough?

rioolpijpgp

Step 1
For any fixed $L,C>0$ , pick n sufficient large so that $L<\frac{n}{C}$ . Now define the polynomial,
$p\left(x\right):=-{\left(-\frac{C}{2n}x+\frac{1}{2}\right)}^{2n}+1$
Note that p(x) satisfies the points $\left(-n/C,0\right)$ , $\left(0,1-\left(1/2{\right)}^{2n}\right)$ , $\left(n/C,1\right)$ , $\left(2n/C,1-\left(1/2{\right)}^{2n}\right)$ , which implies that p(x) has a vertex at $n/C$ (the absolute maximum of p), and that p(x) is increasing from $\left(-\mathrm{\infty },n/C\right]$ (moreover, p(x) has positive derivative and is in fact concave down on that same region); additionally we see that $0 for $x\in \left[0,L\right]$ . Thus $0<-\frac{C}{2n}x+\frac{1}{2}<1$1 for $x\in \left[0,L\right]$ . Additionally, we clearly see that $0 . So for any $x\in \left[0,L\right]$ we have
$\begin{array}{}{p}^{\prime }\left(x\right)& =& -2n{\left(-\frac{C}{2n}x+\frac{1}{2}\right)}^{2n-1}\cdot \frac{-C}{2n}\\ & =& C{\left(-\frac{C}{2n}x+\frac{1}{2}\right)}^{2n-1}\\ & >& C{\left(-\frac{C}{2n}x+\frac{1}{2}\right)}^{2n}\\ & >& C\left({\left(-\frac{C}{2n}x+\frac{1}{2}\right)}^{2n}-{\left(-\frac{C}{2n}L+\frac{1}{2}\right)}^{2n}\right)\\ & =& C\left(p\left(L\right)-p\left(x\right)\right)\end{array}$

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