Given real polynomial $P\left ( x \right )$ so that $P\left ( 0 \right )= P\left ( 1 \right )= 0, \int\limits_{0}^{1}\left | {P}'\left ( x \right ) \right |{\rm d}x= 1.$ For $x\in\left ( 0, 1 \right )$ prove or disprove that $\left | P\left ( x \right ) \right |< \frac{1}{2}.$
I read on Twitter a fleet that proves the following inequality with strictly decreasing $f\left ( x \right )$ for all $x\in \left [ 0, 1 \right ]$ by Cauchy-Schwarz $$\int_{0}^{1}f\left ( x \right ){\rm d}x+ \left ( \int_{0}^{1}xf\left ( x \right ) \right )^{2}+ 1> 0$$ I'm trying to find a strictly decreasing function from the Original Problem.. of course there exists no such $P\left ( x \right ), {P}'\left ( x \right ).$ Then replace it for $f\left ( x \right ).$
As $P(0) = 0$, we get $P(x) = \int_0^x P'(t) \, \mathrm{d}t$. Moreover, $0 = P(1) = P(x) + \int_x^1 P'(t) \, \mathrm{d}t$. Taking absolute values and adding these two equations up, we get $$2|P(x)| \leq \int_0^x |P'(t)| \, \mathrm{d}t + \int_x^1 |P'(t)| \, \mathrm{d}t = 1$$ and therefore $|P(x)| \leq \frac{1}{2}$.