You can define the function $f(t)\cdot min_i|x_i-t|$ piecewise so that $f(t)\cdot min_i|x_i-t|=f(t)\cdot|x_i-t|$ when $t$ is on the interval $\left[\frac{x_{i-1}+x_i}{2},\frac{x_i+x_{i+1}}{2}\right]$. Here we have to define $x_0=-x_1$ (so $\frac{x_{0}+x_1}{2}=0$) and $x_{n+1}=2-x_n$ (so $\frac{x_n+x_{n+1}}{2}=1$). This makes the integral $$F(x_1,...,x_n)=\sum_{i=1}^n\int_\frac{x_{i-1}+x_i}{2}^\frac{x_i+x_{i+1}}{2}f(t)\cdot|x_i-t|dt$$ Now, $f$ isn't $0$ at the endpoints of these intervals and the intervals don't have length $1$, but they are concave and their average value is $\alpha/n$. You can consider f on each interval as the limit of a sequence of concave functions defined on that interval which are each $0$ at the endpoints, so you should be able to put a lower bound on each integral in the same way that you did it for $\int_0^1f(t)\cdot|x_1-t|dt$. I can't say what that lower bound might be, because I don't know how (or if) you used the length of the interval $[0,1]$ when you found the lower bound for $\int_0^1f(t)\cdot|x_1-t|dt$. If you didn't use the length of the interval at all, the integrals will have to be on average greater than or equal to $\alpha/6n$, implying that $F(x_1,...,x_n)$ is bounded below by $\alpha/6$, which would be kind of interesting. One more thing to note is that unlike your example when $n=1$, for the integrals in the sum above you have the additional condition that $f(x_i)$ is bounded away from $0$ by the value of $f$ at at least one of the endpoints of the intervals. Maybe this will make it possible to put stricter lower bounds on each of these integrals. I hope all that was clear (and correct).
You can define the function $f(t)\cdot min_i|x_i-t|$ piecewise so that $f(t)\cdot min_i|x_i-t|=f(t)\cdot|x_i-t|$ when $t$ is on the interval $\left[\frac{x_{i-1}+x_i}{2},\frac{x_i+x_{i+1}}{2}\right]$. Here we have to define $x_0=-x_1$ (so $\frac{x_{0}+x_1}{2}=0$) and $x_{n+1}=1+x_n$ x_{n+1}=2-x_n$(so$\frac{x_n+x_{n+1}}{2}=1$). This makes the integral $$F(x_1,...,x_n)=\sum_{i=1}^n\int_\frac{x_{i-1}+x_i}{2}^\frac{x_i+x_{i+1}}{2}f(t)\cdot|x_i-t|dt$$ Now,$f$isn't$0$at the endpoints of these intervals and the intervals don't have length$1$, but they are concave and their average value is$\alpha/n$. You can consider f on each interval as the limit of a sequence of concave functions defined on that interval which are each$0$at the endpoints, so you should be able to put a lower bound on each integral in the same way that you did it for$\int_0^1f(t)\cdot|x_1-t|dt$. I can't say what that lower bound might be, because I don't know how (or if) you used the length of the interval$[0,1]$when you found the lower bound for$\int_0^1f(t)\cdot|x_1-t|dt$. If you didn't use the length of the interval at all, the integrals will have to be on average greater than or equal to$\alpha/6n$, implying that$F(x_1,...,x_n)$is bounded below by$\alpha/6$, which would be kind of interesting. One more thing to note is that unlike your example when$n=1$, for the integrals in (1) the sum above you have the additional condition that$f(x_i)$is bounded away from$0$by the value of$f$at the endpoints of the intervals. Maybe this will make it possible to put stricter lower bounds on each of these integrals. I hope all that was clear (and correct). 1 You can define the function$f(t)\cdot min_i|x_i-t|$piecewise so that$f(t)\cdot min_i|x_i-t|=f(t)\cdot|x_i-t|$when$t$is on the interval$\left[\frac{x_{i-1}+x_i}{2},\frac{x_i+x_{i+1}}{2}\right]$. Here we have to define$x_0=-x_1$(so$\frac{x_{0}+x_1}{2}=0$) and$x_{n+1}=1+x_n$(so$\frac{x_n+x_{n+1}}{2}=1$). This makes the integral $$F(x_1,...,x_n)=\sum_{i=1}^n\int_\frac{x_{i-1}+x_i}{2}^\frac{x_i+x_{i+1}}{2}f(t)\cdot|x_i-t|dt$$ Now,$f$isn't$0$at the endpoints of these intervals and the intervals don't have length$1$, but they are concave and their average value is$\alpha/n$. You can consider f on each interval as the limit of a sequence of concave functions defined on that interval which are each$0$at the endpoints, so you should be able to put a lower bound on each integral in the same way that you did it for$\int_0^1f(t)\cdot|x_1-t|dt$. I can't say what that lower bound might be, because I don't know how (or if) you used the length of the interval$[0,1]$when you found the lower bound for$\int_0^1f(t)\cdot|x_1-t|dt$. If you didn't use the length of the interval at all, the integrals will have to be on average greater than or equal to$\alpha/6n$, implying that$F(x_1,...,x_n)$is bounded below by$\alpha/6$, which would be kind of interesting. One more thing to note is that unlike your example when$n=1$, for the integrals in (1) you have the additional condition that$f(x_i)$is bounded away from$0$by the value of$f\$ at the endpoints of the intervals. Maybe this will make it possible to put stricter lower bounds on each of these integrals. I hope all that was clear (and correct).