Find the smallest value of the function $F:\alpha\in\mathbb R\rightarrow \int_0^2 f(x)f(a+x)dx$ 
Let $f -$ fixed continuous on the whole real axis function which is periodic with period $T = 2$, and it is known that the function $f$ decreases monotonically on the segment $[0, 1]$ increases monotonically on the segment $[1, 2]$ and for all $ x \in \mathbb R $ satisfies the equality $f(x)=f(2-x)$. Find the smallest value of the function
  $$F:\alpha\in\mathbb R\rightarrow \int_0^2 f(x)f(\alpha+x)dx$$

I have no clue how to start. Any kind of help will be appreciated. 

Let $F(\alpha)=\int_0^2{f(x)f(\alpha+x)dx}$, changing in variable by letting $\alpha+x=u$, we get $F(\alpha)=\int_{\alpha}^{\alpha+2}{f(u-\alpha)f(u)du}$. Since $f$ is continuous then $F$ is differentiable and 
\begin{align*}
F'(\alpha)=f(2)f(\alpha+2)-f(0)f(\alpha)
\end{align*}
but since $f(0)=f(2)$ and $f(x+2)=f(x)$, for all $x\in \mathbb{R}$. Then
\begin{align*}
F'(\alpha)=0,\,\,\,\, \forall \alpha
\end{align*}
which means that $F(\alpha)$ is constant over all $\mathbb{R}$. 
