pachaquis3s

2022-06-22

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

$0<g(0)<g(L),\phantom{\rule{2em}{0ex}}{g}^{\prime}(x)>(g(L)-g(x))C,\text{}\text{}\text{for all}x\in [0,L]$

holds at least for n big enough?

$0<g(0)<g(L),\phantom{\rule{2em}{0ex}}{g}^{\prime}(x)>(g(L)-g(x))C,\text{}\text{}\text{for all}x\in [0,L]$

holds at least for n big enough?

rioolpijpgp

Beginner2022-06-23Added 19 answers

Step 1

For any fixed $L,C>0$ , pick n sufficient large so that $L<\frac{n}{C}$ . Now define the polynomial,

$p(x):=-{(-\frac{C}{2n}x+\frac{1}{2})}^{2n}+1$

Note that p(x) satisfies the points $(-n/C,0)$ , $(0,1-(1/2{)}^{2n})$ , $(n/C,1)$ , $(2n/C,1-(1/2{)}^{2n})$ , which implies that p(x) has a vertex at $n/C$ (the absolute maximum of p), and that p(x) is increasing from $(-\mathrm{\infty},n/C]$ (moreover, p(x) has positive derivative and is in fact concave down on that same region); additionally we see that $0<p(x)<1$ for $x\in [0,L]$ . Thus $0<-\frac{C}{2n}x+\frac{1}{2}<1$1 for $x\in [0,L]$ . Additionally, we clearly see that $0<p(0)<p(L)<1$ . So for any $x\in [0,L]$ we have

$\begin{array}{}{p}^{\prime}(x)& =& -2n{(-\frac{C}{2n}x+\frac{1}{2})}^{2n-1}\cdot \frac{-C}{2n}\\ & =& C{(-\frac{C}{2n}x+\frac{1}{2})}^{2n-1}\\ & >& C{(-\frac{C}{2n}x+\frac{1}{2})}^{2n}\\ & >& C({(-\frac{C}{2n}x+\frac{1}{2})}^{2n}-{(-\frac{C}{2n}L+\frac{1}{2})}^{2n})\\ & =& C(p(L)-p(x))\end{array}$

For any fixed $L,C>0$ , pick n sufficient large so that $L<\frac{n}{C}$ . Now define the polynomial,

$p(x):=-{(-\frac{C}{2n}x+\frac{1}{2})}^{2n}+1$

Note that p(x) satisfies the points $(-n/C,0)$ , $(0,1-(1/2{)}^{2n})$ , $(n/C,1)$ , $(2n/C,1-(1/2{)}^{2n})$ , which implies that p(x) has a vertex at $n/C$ (the absolute maximum of p), and that p(x) is increasing from $(-\mathrm{\infty},n/C]$ (moreover, p(x) has positive derivative and is in fact concave down on that same region); additionally we see that $0<p(x)<1$ for $x\in [0,L]$ . Thus $0<-\frac{C}{2n}x+\frac{1}{2}<1$1 for $x\in [0,L]$ . Additionally, we clearly see that $0<p(0)<p(L)<1$ . So for any $x\in [0,L]$ we have

$\begin{array}{}{p}^{\prime}(x)& =& -2n{(-\frac{C}{2n}x+\frac{1}{2})}^{2n-1}\cdot \frac{-C}{2n}\\ & =& C{(-\frac{C}{2n}x+\frac{1}{2})}^{2n-1}\\ & >& C{(-\frac{C}{2n}x+\frac{1}{2})}^{2n}\\ & >& C({(-\frac{C}{2n}x+\frac{1}{2})}^{2n}-{(-\frac{C}{2n}L+\frac{1}{2})}^{2n})\\ & =& C(p(L)-p(x))\end{array}$