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.
To make reflection symmetry more obvious let us define a shifted function $g: \mathbb{R} \to \mathbb{R}$ as $$g(x)~:=~f(x\!+\!1)+k,\tag{1}$$ where $k\in\mathbb{R}$ is a constant, introduced for later convenience. Hence we have an even continuous & periodic function $g$ with period 2, $$g(-x)~=~g(x)~=~g(x+2),\tag{2}$$ which is monotonically weakly increasing in the interval $[0,1]$.
Define a function $G: \mathbb{R} \to \mathbb{R}$ as $$ G(x)~:=~\int_{-1}^1 \!dy ~g(y)g(y\!+\!x)~=~\int_{-1}^1\!dy ~g(y\!-\!x)g(y)~=~G(-x).\tag{3} $$ Clearly the function $G$ is also an even continuous & periodic function with period 2.
Moreover the function $G$ has an additional reflection symmetry around the $x=1$ axis: $$ G(2\!-\!x)~=~\int_{-1}^1 \!dy ~g(y)g(y\!+\!2\!-\!x)~=~\int_{-1}^1 \!dy ~g(y)g(y\!-\!x)~=~G(x).\tag{4} $$ Hence it is enough to study the function $G$ in the interval $[0,1]$.
Note that the constant $k\in\mathbb{R}$ only shifts the function $G$ up and down in a trivial manner $$ G(x) ~=~F(x) + 2k^2 + 2 k\int_0^2 \!dy ~f(y) . \tag{5} $$
It is easy to see that $x=0$ is a global maximum point: $$G(0)~=~\frac{1}{2}\int_{-1}^1 \!dy \left[g(y)^2+g(y\!+\!x)^2\right]~\geq~\int_{-1}^1 \!dy ~g(y)g(y\!+\!x)~=~G(x).\tag{6} $$
We next prove that the function $G$ is weakly decreasing in the interval $\left[0,\frac{1}{2}\right]$. Choose a point $x_0\in \left[0,\frac{1}{2}\right]$. Choose the constant $k$ such that $g(x_0)=0$. Then for $h\in[0,1]$, we calculate $$ G(x_0\!+\!h)-G(x_0\!-\!h) ~=~ \int_{-1}^1 \!dy \left[g(y\!-\!h)-g(y\!+\!h)\right]g(y\!+\!x_0)$$ $$~=~ \int_{-1}^{-2x_0} \!dy \underbrace{\left[g(y\!-\!h)-g(y\!+\!h)\right]}_{\geq 0}~\underbrace{g(y\!+\!x_0)}_{\geq 0} +\int_{-2x_0}^0 \!dy \underbrace{\left[g(y\!-\!h)-g(y\!+\!h)\right]}_{\geq 0}~\underbrace{g(y\!+\!x_0)}_{\leq 0}$$ $$+\int_0^{2x_0} \!dy \underbrace{\left[g(y\!-\!h)-g(y\!+\!h)\right]}_{\leq 0}~\underbrace{g(y\!+\!x_0)}_{\geq 0} +\int_{2x_0}^1 \!dy \underbrace{\left[g(y\!-\!h)-g(y\!+\!h)\right]}_{\leq 0}~\underbrace{g(y\!+\!x_0)}_{\geq g(y\!-\!x_0)\geq 0}$$ $$~\leq~\int_{-1}^{-2x_0} \!dy \underbrace{\left[g(y\!-\!h)-g(y\!+\!h)\right]}_{\geq 0}~\underbrace{g(y\!+\!x_0)}_{\geq 0} +\int_{2x_0}^1 \!dy \underbrace{\left[g(y\!-\!h)-g(y\!+\!h)\right]}_{\leq 0}~\underbrace{g(y\!-\!x_0)}_{\geq 0}$$ $$~=~0. \tag{7} $$ The last equality follows via changing the sign of the integration variable $y$ in one of the two integrals. Hence the function $G$ is weakly decreasing in the interval $\left[0,\frac{1}{2}\right]$.
We finally prove that the function $G$ is weakly decreasing in the interval $\left[\frac{1}{2},1\right]$. Choose a point $x_0\in \left[\frac{1}{2},1\right]$. Choose the constant $k$ such that $g(x_0)=0$. Then for $h\in[0,1]$, we calculate $$ G(x_0\!+\!h)-G(x_0\!-\!h) ~=~ \int_{-1}^1 \!dy \left[g(y\!-\!h)-g(y\!+\!h)\right]g(y\!+\!x_0)$$ $$~=~ \int_{-1}^{2x_0-2} \!dy \underbrace{\left[g(y\!-\!h)-g(y\!+\!h)\right]}_{\geq 0}~\underbrace{g(y\!+\!x_0)}_{\leq g(y\!-\!x_0)\leq 0} +\int_{2x_0-2}^0 \!dy \underbrace{\left[g(y\!-\!h)-g(y\!+\!h)\right]}_{\geq 0}~\underbrace{g(y\!+\!x_0)}_{\leq 0}$$ $$+\int_0^{2-2x_0} \!dy \underbrace{\left[g(y\!-\!h)-g(y\!+\!h)\right]}_{\leq 0}~\underbrace{g(y\!+\!x_0)}_{\geq 0} +\int_{2-2x_0}^1 \!dy \underbrace{\left[g(y\!-\!h)-g(y\!+\!h)\right]}_{\leq 0}~\underbrace{g(y\!+\!x_0)}_{\leq 0}$$ $$~\leq~\int_{-1}^{2x_0-2} \!dy \underbrace{\left[g(y\!-\!h)-g(y\!+\!h)\right]}_{\geq 0}~\underbrace{g(y\!-\!x_0)}_{\leq 0} +\int_{2-2x_0}^1 \!dy \underbrace{\left[g(y\!-\!h)-g(y\!+\!h)\right]}_{\leq 0}~\underbrace{g(y\!+\!x_0)}_{\leq 0}$$ $$~=~0. \tag{8} $$ The last equality follows via changing the sign of the integration variable $y$ in one of the two integrals. Hence the function $G$ is also weakly decreasing in the interval $\left[\frac{1}{2},1\right]$.
Altogether, it follows from the above that $x=1$ is a global minimum point, which in turn answers OP's question.
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 $\mathbb{R}$.
