# Prove $|P(0)|\leq 2n+1$

Let $P(x)$ be a polynomial with degree $\leq n$ and $|P(x)|\leq\frac{1}{\sqrt{x}}$ for $x\in(0,1]$. Prove that $|P(0)|\leq 2n+1$. The idea should be that if $|P(0)|$ is too large, then the polynomial cannot change values fast enough to avoid intersecting the curve $\pm 1/\sqrt{x}$, but I don't see how to formalize this.

• Perhaps you could say that because $P(x)$ is continuous? – Chad Shin Feb 1 '16 at 7:04
• @ChadShin Continuity is not enough. One can easily draw a continuous function $f$ on $[0,1]$ so that $|f(x)|\le \frac{1}{\sqrt x}$ and $f(0)$ is arbitrarily large. Indeed, one can also choose a polynomial to do that, but the degree might be very large. – user99914 Feb 1 '16 at 7:26
• What's the source of this problem, please? – Gerry Myerson Feb 4 '16 at 10:30
• It would be great if someone can translate the solution in the above link and post as an answer below. – user99914 Feb 6 '16 at 3:09

Since $-\frac{1}{\sqrt{x}}\le P(x)\le\frac{1}{\sqrt{x}}$ for $0<x\le 1$, $\;\;\;-1\le \sqrt{x}P(x)\le1\;$ for $0\le x\le1$.

If we let $t=\sqrt{x}$ and $h(t)=tP(t^2)$, then $h$ is a polynomial of degree $2n+1$ with $\big|h(t)\big|\le1$ for $|t|\le1$.

By Bernstein's inequality for polynomials, $\;\;\displaystyle\big|h^{\prime}(t)\big|\le\frac{2n+1}{\sqrt{1-t^2}}\;$ for $|t|<1$; so $\;\big|P(0)\big|=\big|h^{\prime}(0)\big|\le 2n+1$.

What follows is more or less the translation of the solution given here.

Part A. First, we show that a polynomial $F(x)$ exists for which the conditions are satisfied and equality in the estimate holds, i.e. $F(0)=2n+1$. Let us denote by $U_{2n}$ the $2n$-th Chebyshev polynomial of the second kind: $$U_{2n}(\cos t)=\frac{\sin(2n + 1)}{\sin t}.$$ This polynomial has degree $2n$. It is known that $U_{2n}(x)$ is even: the reason is that it contains only terms having even degree. This fact can be obtained from the recursive formula for the Chebyshev polynomial of the second kind $U_n(\cos t)=\frac{\sin(n+1)t}{\sin t}$, which is $U_{n+1}(x) = 2x U_n(x)-U_{n-1}(x)$, starting from $U_0(x)=1$, $U_1(x)=2x$.

So we can write $U_{2n}(x)=u(x^2)$ for some polynomial $u$. Now let $$F(x) = u(1-x) = U_{2n}(\sqrt{1-x})$$ and write any $x\in (0,1]$ as $x=\sin^2 t$ (with $t\in(0,\frac{\pi}{2}]$). We get $$F(x) = u(1-\sin^2 t) = u(\cos^2 t) = U_{2n}(\cos t) = \frac{\sin (2n + 1)t}{\sin t} = \frac{\sin (2n + 1)t}{\sqrt{x}}.$$ So for $x\in(0,1]$ we have $|F(x)|\le\frac{1}{\sqrt{x}}$, with equality occurring when $\sin (2n + 1)t = \pm 1$, i.e. at $x_k:=\sin^2\frac{(k + \frac{1}{2})\pi}{2n+1}$ (for $k = 0,1,\dots,n$).
Moreover, $F(0)$ can be found by taking the limit as $t\to 0$: $$F(0) = U_{2n}(1) = \lim_{t\to 0}U_{2n}(\cos t) = \lim_{t\to 0}\frac{\sin (2n + 1)t}{\sin t} = 2n+1.$$

Part B. Now we show the inequality $|P(0)|\le 2n+1$ for a generic polynomial $P$ satisfying the hypotheses. Suppose by contradiction that $|P(0)|>2n+1$; wlog we can assume $P(0)>2n+1$. Choose $c>1$ such that $P(0)>c(2n+1)$ and consider the polynomial $$Q(x) := P(x)-cF(x).$$ The polynomial $Q$ takes alternatively positive and negative values at the points $x_{-1}:=0<x_0<x_1<\dots<x_n$: indeed $Q(x_{-1})=Q(0)>0$ and, if $k$ is even, $Q(x_k)\le\frac{1}{\sqrt{x}}-\frac{c}{\sqrt{x}}<0$; similarly, if $k\ge 1$ is odd, $Q(x_k)>0$. So it possesses at least $n+1$ roots, precisely at least one in each of the intervals $(x_{-1}, x_0), (x_0, x_1), \dots, (x_{n-1}, x_n)$. But this is impossible as $Q$ is not identically zero and has degree $\le n$, and so it cannot have more than $n$ roots.

• did you find the "optimal" polynomial of degree $\le n$ maximizing $P(0)$ and being bounded by $1/x$ on $]0;1]$ ? isn't it of the form $1 + C (1-x)^n$ ? – reuns Feb 7 '16 at 2:06
• Yes, but the condition is $|P(x)|\le\frac{1}{\sqrt{x}}$. I think $F(x)$ does not have an explicit formula as the one you wrote. – Mizar Feb 7 '16 at 10:59