Picking up from anton's answer, notice that by Cauchy-Schwartz:
$\int_0^1 u \cdot v \cdot dx \leq \sqrt{ \int_0^1 u^2 \cdot dx \cdot \int_0^1 v^2 \cdot dx}$
So a bound in the case $u=v$ is sufficient. We wish to show:
$\int_0^1 u^2 dx \leq C \left( \int_0^1 u dx\right)^2$
Fix a triangle with the same average value of $u$, and that attains its maximum over the same $x$-coordinate. If $u$ is equal to that triangle, a trivial calculation gives a value of $C$ of $4/3$. We want to show that it not being a triangle makes $\int u^2$ no bigger. In fact, letting $t$ be the triangle, we have $\int u^2 \leq \int ut\leq \int t^2$.
This is because $(t-u)$ is positive closer to the maximum value, and negative further, so $u$ and $t$ are both higher where $(t-u)$ is positive than where it is negative, so $\int u(t-u)\geq 0$, $\int t(t-u)\geq 0$.
Therefore $C=4/3$ is a bound, obtained by setting $u$ and $v$ to triangles. One can check that this does indeed come from a polytope in $\mathbb R^3$ - an octahedron, for instance.
Edit: This proof isn't quite complete, because the area on one side of the triangle is not necessarily the area on one side of the function. We can remedy this by choosing an unbalanced "triangle" with a discontinuity where the maximum should be, which still satisfies $C=4/3$.
