TITLE: Show that there is a root in [0,1]
QUESTION [0 upvotes]: suppose that there are $P_0,P_1,P_2$ where $P_i$ is a polynomial of degree $i$ with the following properties :
$1) P_0(x)=1$
$2) n \neq m ; \int_0^1 P_n(x)P_m(x)dx=\frac{1}{3}$
$3) \int_0^1 P_n(x)P_n(x)dx \geq 0$
show that there is a $c \in [0,1] $ that $P_1(c)=0$ .

$\int_0^1 P_1(x)P_2(x)dx =\frac{1}{3}$ and $\int_0^1 P_0(x)P_1(x)dx =\frac{1}{3}$ so :
$$ \int_0^1 P_1(x)(1- P_2(x)) dx =0 $$
therefore there is a $c \in [0,1]$ that $P_1(c)=0$ or $P_2(c)=1$ .
but I can not prove that $P_2(c) \neq 1$.
For sure, maybe I am entirely on the wrong path!

REPLY [1 votes]: I was trying to prove the statement, but on the way I found a counterexample. You might check the last paragraph directly, where I mention a counterexample. Although, the procedure to get that counterexample is given in detail here. Also, are you sure condition 3) is given? because condition 3) is useless since it is tautologically true. 
You are not using one important fact from the statement, $P_i$ is a polynomial of degree $i$. So, you could write $P_1=ax+b$, $P_2=mx^2+nx+p$. We have that $c=-b/a$ is the root of $P_1(x)$. We need to show that $-b/a\in[0,1]$.
You have that $$\frac{1}{3}=\int_0^1P_0(x)P_1(x)dx=\int_0^1 ax+b dx=\frac{a}{2}+b.$$ So, $$b=\frac{1}{3}-\frac{a}{2}.$$ Hence, $c=-b/a=\frac{a/2-1/3}{a}=\frac{1}{2}-\frac{1}{3a}.$ 
We want to show that 
$0\leq \frac{1}{2}-\frac{1}{3a}\leq 1$, i.e. $-\frac{1}{2}\leq  \frac{1}{3a}\leq \frac{1}{2}$. i.e. $|\frac{1}{3a}|\leq \frac{1}{2}$. 
Or equivalently, what we want to show is that $|a|\geq \frac{2}{3}$.
Now, the other two pieces of information that we have is that 
\begin{align}\label{con}
\frac{1}{3}=\int_0^1P_0(x)P_2(x)dx=\int_0^1 mx^2+nx+p dx=\frac{m}{3}+\frac{n}{2}+p.
\end{align}
and
\begin{align*}\frac{1}{3}&=\int_0^1P_1(x)P_2(x)dx=\int_0^1 (ax+b)P_2(x)dx\\
&=a\int_0^1 xP_2(x)dx+b\int_0^1 P_2(x)dx\\
&=a\int_0^1 xP_2(x) dx+\frac{b}{3}\\
&=a\int_0^1 xP_2(x) dx+\frac{1}{9}-\frac{a}{6}
\end{align*}
Arranging terms, we have that 
$$a\left(\int_0^1 xP_2(x) dx-\frac{1}{6}\right)=\frac{2}{9}$$
So, in order to prove $|a|\geq \frac{2}{3}$, we need to show that $\left|\int_0^1 xP_2(x) dx-\frac{1}{6}\right|\leq \frac{1}{3}$, or equivalently, we need to show that 
$-\frac{1}{6}\leq\int_0^1 xP_2(x) dx\leq \frac{1}{2}$
We have that $\int_0^1 xP_2(x) dx=\int_0^1 mx^3+nx^2+px dx=\frac{m}{4}+\frac{n}{3}+\frac{p}{2}$.
So, the problem is reduced in showing $\frac{-1}{6}\leq \frac{m}{4}+\frac{n}{3}+\frac{p}{2}\leq \frac{1}{2}$ from the condition $\frac{m}{3}+\frac{n}{2}+p=\frac{1}{3}$. 
This is clearly not true: A counter example would be $m=-5$, $n=0$, $p=2$. For this particular case, we have $a=-\frac{8}{15}$, $b=\frac{3}{5}$.
So, the counterexample I am finding is that  $P_1(x)=-\frac{8}{15}x+\frac{3}{5}$ and $P_2(x)=-5x^2+2$ satisfy all your conditions, although the root of $P_1$ which is $c=\frac{9}{8}$ is not in $[0,1]$.
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