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. 

We intend to prove that the smallest value of the function $F$ defined by
\begin{equation*}
F({\alpha}) = \int_0^2f(x)f(x+{\alpha})\, dx\tag{1}
\end{equation*}
is $F(1)$.
At first we give $f$ the additional property to be differentiable with a continuous derivative. At the end we will fill that gap.
The function $F$ will be continuous and have the period $T = 2$. Thus there exists a smallest value of $F(\alpha).$
If $f$ is differentiable then $F$ will also be differentiable.
We intend to prove that 
\begin{equation*}
F'(\alpha) \le 0 \text{ for } 0 \le \alpha \le 1. \tag{2}
\end{equation*}
Since 
\begin{equation*}
f(x) = f(2-x) = f(-x)\tag{3}
\end{equation*}
then $F(\alpha) = F(2-\alpha)$. Then (2) implies that
\begin{equation*}
F'(\alpha) \ge 0 \text{ for } 1 \le \alpha \le 2. 
\end{equation*}
Then we know that $F(1)$ is the smallest value.
We start by studying (1). If we change $x$ to $x+1$ and then $x$ to $1-x$ we get
\begin{gather*}
\int_1^2f(x)f(x+\alpha)\, dx = \int_0^1f(x+1)f(x+1+\alpha)\, dx = [(3)]\\ = \int_0^1f(1-x)f(1-x -\alpha)\, dx = \int_0^1f(x)f(x-\alpha)\, dx.
\end{gather*}
Now we can rewrite $F(\alpha)$ as
\begin{equation*}
F(\alpha) = \int_0^1f(x)(f(x+\alpha)+f(x-\alpha))\, dx.
\end{equation*}
Via differentiation under the integral sign  (we now use that $f$ is differentiable) followed by integration by parts we get
\begin{gather*}
F'({\alpha}) = \int_0^1f(x)(f'(x+{\alpha})-f'(x-{\alpha}))\, dx = \left[f(x)(f(x+{\alpha})-f(x-{\alpha}))\right]_0^1 \\- \int_0^1f'(x)(f(x+{\alpha})-f(x-{\alpha}))\, dx = f(1)\underbrace{(f(1+{\alpha})-f(1-{\alpha}))}_{= 0}-f(0) \underbrace{(f({\alpha})-f(-{\alpha}))}_{= 0} \\- \int_0^1f'(x)(f(x+{\alpha})-f(x-{\alpha}))\, dx = - \int_0^1f'(x)(f(x+{\alpha})-f(x-{\alpha}))\, dx. 
\end{gather*}
Since $f$ is decreasing and differentiable on $[0,1]$ then $f'(x) \le 0$ in the integral above. It remains to prove that
\begin{equation*}
g(x,\alpha) = f(x+{\alpha})-f(x-{\alpha}) \le 0
\end{equation*}
on the square $0 \le x \le 1, \,  0 \le \alpha \le 1.$ But \begin{equation*}
g(\alpha,x) = f(\alpha +x) - f(\alpha-x) = [(3)] = f(x+{\alpha})-f(x-{\alpha}) = g(x,\alpha).
\end{equation*}
Consequently it is sufficient to examine $g$ on the triangle $0 \le \alpha \le x \le 1.$ To do that we study $g$ on that part of the straight line $\alpha = x-k$, that runs inside the triangle. Here $k$ is constant in $[0,1]$.
To be more precise look at
\begin{equation*}
g(x,x-k) = f(2x-k)-f(k), \quad 0 \le k \le 1, \: k \le x \le 1.
\end{equation*}
We split the $x$-interval into two. If $k \le x \le \dfrac{k+1}{2}$ then $k \le 2x - k \le 1$. Thus $2x-k \in[0,1]$ and $k\in[0,1]$. But on $[0,1]$ the function $f$ is decreasing. Since $2x-k \ge k$ then $f(2x-k) \le f(k)$ and $g(x,2x-k) \le 0.$
The second part of the $x$-interval is $\dfrac{k+1}{2} \le x \le 1$. Then $1 \le 2x-k \le 2-k \le 2$. Furthermore $f(k) = f(2-k)$ and $1 \le 2-k \le 2.$ But on the interval $[1,2]$ the function $f$ is increasing. Since $2x-k \le 2-k$ then $f(2x-k ) \le f(2-k)$ and $g(x,2x-k) \le 0.$
We have proved that $F(1)$ is the smallest value under the additional assumption that $f$ is differentiable with a continuous derivative.
Now we will prove that the same is true if $f$ only is continuous. Define a smooth substitute $\tilde{f}$ for $f$ via
\begin{equation*}
\tilde{f}(x) = \dfrac{1}{2h}\int_{x-h}^{x+h}f(t)\, dt = \dfrac{1}{2h}\int_{-h}^{h}f(x+t)\, dt.\tag{4}
\end{equation*}
Then $\tilde{f}$ will be differentiable with the continuous derivative $\dfrac{f(x+h)-f(x-h)}{2h}$. Furthermore it inherits a lot of properties from  $f$. It is obvious that $\tilde{f}$ will be periodic with $T=2$. It will also fulfil the condition $\tilde{f}(x) = \tilde{f}(2-x) = \tilde{f}(-x).$ It will also be decreasing on $[0,1]$. To realize that we need a little argument. Assume that $0 < x_1 < x_2 <1$. If $h$ in (4) is small enough then $|t|$ will be so small that
\begin{equation*}
0 <x_1+t<x_2+t <1.
\end{equation*}
From (4) we then get that $\tilde{f}(x_1) \ge \tilde{f}(x_2)$. We have proved that $\tilde{f}$ is decreasing on the open interval $0 < x < 1$. But since $\tilde{f}$ is differentiable it will also be decreasing on the closed interval $0 \le x \le 1$ (use the intermediate value theorem).
Analogously we prove that $\tilde{f}$ is increasing on $1\le x\le 2.$
Since $f$ is continuous and since we work on a compact interval $f$ will be uniformly continuous. Thus to a given $\varepsilon $ there exists a $\delta $ such that
\begin{equation*}
 |f(x_1)-f(x_2)| < \varepsilon \text{ if } |x_1-x_2|< \delta .
 \end{equation*}
 With these $\varepsilon$ and $\delta$ and with $0<h<\delta$ we get
 \begin{equation*}
 |f(x)-\tilde{f}(x)| \le \dfrac{1}{2h}\int_{-h}^{h}|f(x)-f(x+t)|\, dt \le  \dfrac{1}{2h}\int_{-h}^{h}\varepsilon\, dt = \varepsilon .\tag{5}
 \end{equation*}
We are now prepared to study the function
\begin{equation*}
\tilde{F}(\alpha) = \int_0^2\tilde{f}(x)\tilde{f}(x+\alpha)\, dx.
\end{equation*}
According to what we have done above we know that
\begin{equation*}
\tilde{F}(\alpha) \ge \tilde{F}(1)
\end{equation*}
and that $\tilde{F}(1)$ is the smallest value.
From (5) we get
\begin{gather*}
|F(\alpha) - \tilde{F}(\alpha)| \le \int_0^2|f(x)f(x+\alpha)-\tilde{f}(x)\tilde{f}(x+\alpha)|\, dx \\= \int_0^1|(f(x)-\tilde{f}(x)) f(x+{\alpha})+\tilde{f}(x)(f(x+{\alpha})-\tilde{f}(x+{\alpha}))|\, dx \\
\le \int_0^2(\epsilon C + C\varepsilon)\, dx = 2C\varepsilon
\end{gather*}
where  $\displaystyle C = \max_{0 \le x\le 2}|f(x)|$.
Consequently
\begin{equation*}
F({\alpha}) = F({\alpha}) -\tilde{F}({\alpha}) +\tilde{F}({\alpha}) \ge -2C\varepsilon +\tilde{F}(1).
\end{equation*}
But $\varepsilon$ can be arbitrarily small.
 In the limit we have that
 \begin{equation*}
 F({\alpha})\ge F(1)
 \end{equation*}
 i.e. $F(1)$ is the smallest value of $F(\alpha).$
I welcome shorter solutions.
