$\int_{-1}^{1}|f(t)|dt \geq C\left(\int_{0}^{2}|f(t)|^2\right)^{1/2}$ for polynomials 
Prove that there exists constant $C>0$ that for all $f \in P_n$ we have:
  $$\int_{-1}^{1}|f(t)|dt \geq C\left(\int_{0}^{2}|f(t)|^2\right)^{1/2}$$
  Where $P_n$ is space of polynomials with degree less than or equal to $n$.

There is one solution using functional analysis: first we note that functions $\|\|_1$ and $\|\|_2$ are norms on $P_n$:
$$\|f\|_1=\int_{-1}^{1}|f(t)|dt$$
$$\|f\|_2=\left(\int_{0}^{2}|f(t)|^2\right)^{1/2}$$
Now $P_n$ is finite vector space and all norms on finite vector spaces are equivalent, so the inequality holds.
In my opinions it's a suprising result, so I'm looking for elementary proof of this inequality.
 A: Assume the contrary, that for every $k$ there exists a non zero polynomial $p_k$ of degree $n$ such that 
$$ \int_{-1}^1 |p_k(t)|dt \leq \frac{1}{k} \left( \int_0^2 |p_k(t)|^2dt \right)^{1/2}$$
We can assume that $\int_0^2|p_k|^2=1$, because the inequality is scale invariant. This means that $\int_{-1}^1 |p_k(t)|dt \leq \frac{1}{k}$ so $p_k \to 0$ in $L^1([-1,1])$. We can extract a subsequence such that $p_k \to 0$ almost everywhere. We note by $N$ the negligible set where the convergence does not take place.
Choose $n+1$ distinct points $x_1,...,x_{n+1}$ in $[0,1]\setminus N$. Then $p_k(x_i)$ is convergent to $0$ for $i=1..n+1$. If we do a Lagrange interpolation procedure for each $p_k$ with the values at the points $x_1,...,x_{n+1}$, we see that all coefficients in the Lagrange basis go to $0$. 
Thus $p_k$ converges uniformly to $p=0$ on compact intervals. But this contradicts $\int_0^2|p_k|^2=1$

This is not really elementary. It would be if you could get rid of the part that the convergence in $L^1$ implies convergence almost everywhere of a subsequence.
A: I can show the claim when the interval is the same for each integral, let's say it is $[-1,1]$, and if there is a restriction on the roots of $f(t)$. I thought about this for a while and figured I might as well post it. I'm slightly hopeful with some more thinking a full solution will come.
First, suppose $f(t)=\prod (t-\alpha_i)$ with $|\alpha_i| > 3$. We will prove a helpful inequality. For $|x|<1$, $|t|<1$, $|\alpha|>3$, we have
$$ \frac{|t-\alpha|}{|x-\alpha|} \geq \frac{||t|-|\alpha||}{|x|+|\alpha|} \geq \frac{|\alpha|-1}{|\alpha|+1} = 1 - \frac{2}{|\alpha|+1} \geq \frac{1}{2} $$
This means $\sup_{|x|<1} |x-\alpha_i| \leq 2 |t-\alpha_i|$ for all $|t|<1$ and hence
$$\prod_{i=1}^n \sup_{|x|<1} |x-\alpha_i| \leq 2^n |f(t)| \quad \mbox{ or } \quad \prod_{i=1}^n \sup_{|x|<1} |x-\alpha_i| \leq 2^{n+1} \int_{-1}^1 |f(t)| dt.$$
Second, we write 
$$\int_{-1}^1 f(t)^2 dt = \int_{-1}^1 \prod (t-\alpha_i)^2 dt \leq \left( \prod_{i=1}^n \sup_{x \in [-1,1]} ||x-\alpha_i|| \right) \int_{-1}^1 \prod |t-\alpha_i|dt$$
Because $2xy \leq x^2 +y^2$ we can write
$$2 \int_{-1}^1 f(t)^2 dt \leq \left( \prod_{i=1}^n \sup_{x \in [-1,1]} ||x-\alpha_i|| \right)^2 + \left( \int_{-1}^1 \prod |t-\alpha_i|dt \right)^2.$$
Finally, we use one of the inequalities we proved earlier to write
$$\frac{2}{1+2^{n+1}}\int_{-1}^1 f(t)^2 dt  \leq  \left( \int_{-1}^1 |f(t)|dt \right)^2.$$
