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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

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

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  • $\begingroup$ 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! $\endgroup$ – John Polcari Jul 7 '18 at 4:20
  • $\begingroup$ @Angle It is not. Do you happen to know the source of the problem? $\endgroup$ – Carl Schildkraut Jul 12 '18 at 22:07

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