# Let $P(x)\in \mathbb{R}[x]$ be of degree $n$ and for any $x \in \left(0,1\right]$, we have $x\cdot P^2(x) \le 1$. Calculate $\max P(0)$.

Let $$P(x)$$ be a polynomial with real coefficient and with degree $$n$$ such that for any $$x \in \left(0,1\right]$$, we have $$x\cdot P^2(x) \le 1$$ Find the maximum of $$P(0)$$.

Note: $$P^2(x) = (P(x))^2$$. Any idea how to start?

I suppose we can write $$P(x) = xQ(x)+c$$ and plug that into inequality and try to say something about $$c$$. But what then?

Source: KoMaL A.654

• As a start I'd let $P = a$ and may be $P = ax+b$ to get a feel – rsadhvika Jul 6 '18 at 19:36
• Yes, i did that with $p(x) =1-x$. But what then? – Aqua Jul 6 '18 at 19:37
• Why not trying to write inequations obtained when $x=\dfrac{k}{n}$ with $k \in \{1,...,n\}$? Maybe this gives enough constraints on $P$... – paf Jul 6 '18 at 19:50
• It might be easier to handle the inequality rewritten as $|P(x)| \leq 1/\sqrt x.$ – md2perpe Jul 6 '18 at 20:30

Let $x=\sin^2(\theta)$. Then, as

$$x=\frac{1-\cos(2\theta)}{2},$$

we can write $x^k$ in terms of $\cos(2\theta),\cos(4\theta),\cdots,\cos(2k\theta)$, so we may write

$$P(x)=\sum_{k=0}^n a_k\cos(2k\theta).$$

We need

$$|P(x)\sin(\theta)|\leq 1$$

for all $\theta$, and we see that $P(0)=\sum_{k=0}^n a_k$. We claim the maximum value of this quantity is reached with $a_k=2$ everywhere except $a_0=1$. First, we see that this indeed works:

\begin{align}P(x)\sin(\theta) &= \sum_{k=-n}^n \cos(2k\theta)\sin(\theta)\\ &=\frac{1}{2}\sum_{k=-n}^n \big(\sin((2k+1)\theta)-\sin((2k-1)\theta)\big)\\ &=\frac{1}{2}\big[\sin((2n+1)\theta)-\sin((-2n-1)\theta)\big]\\ &=\sin((2n+1)\theta),\end{align}

which obviously has magnitude $\leq 1$. On the other hand, consider the polynomial

$$Q(x)=(x+1)P\left(\frac{x+1}{2}\right)^2-1$$

of degree $2n+1$. For $-1\leq x\leq 1$, we have

$$0\leq \frac{Q(x)+1}{2}\leq 1 \implies |Q(x)|\leq 1,$$

so by the Markov brothers' inequality we have

$$Q'(-1)\leq \max_{-1\leq x\leq 1}|Q'(x)|\leq (2n+1)^2\max_{-1\leq x\leq 1}|Q(x)|\leq (2n+1)^2.$$

Finally, we see

$$Q'(x)=(x+1)\frac{1}{2}\left[2P\left(\frac{x+1}{2}\right)P'\left(\frac{x+1}{2}\right)\right]+P\left(\frac{x+1}{2}\right)^2.$$

$$Q'(-1)=P(0)^2,$$

so

$$P(0)^2\leq (2n+1)^2 \implies P(0)\leq 2n+1,$$

finishing the proof.

• Numerical evaluation at n=1 for P(0)=3 suggests P=-4x+3, which achieves xP^2=1 at two points, x=0.25 and x=1. Very nice! – John Polcari Jul 7 '18 at 4:20
• @Angle It is not. Do you happen to know the source of the problem? – Carl Schildkraut Jul 12 '18 at 22:07