Optimization of a integral function I have a function $h(y,x_1,x_2,\ldots,x_n)$. It is known that the minimum value of $h$ for any $y$ is attained when $x_1 = x_n$ and $x_2 = x_3 = \cdots = x_{n-1}$. Now consider the following function
\begin{equation}
  g(x_1,\ldots,x_n) = \int_{y\in\Theta}h(y,x_1,x_2,\ldots,x_n)f(y)dy
\end{equation}
where $f$ is some probability density function and $\Theta$ is appropriate space for $y$.
Numerically, I am getting that $g$ is also minimised when $x_1 = x_n$ and $x_2 = x_3 = \cdots = x_{n-1}$. However, analytically it is difficult to prove. Is there any result which ensures the optimal symmetry of solution even after taking the integration?
Edit 1: It is know that $h$ is a concave function of $(x_1,\ldots,x_n)$ and the vector $(x_1,\ldots,x_n)$ belongs to a convex set. Moreover, the density function is continuous (not discrete).
Edit 2: It is given that $(x_1,\ldots,x_n)\in\{(x_1,\ldots,x_n):x_i\geq 0,\sum_{i=1}^{n}x_i = 1\}$.
 A: With no additional structure, no.
Let $y$ take the value $0$ and $1$ with probability $1/2$ each. That has no density, but densities sufficiently close will do too.
Let $h(y,x_1,x_2)=\sqrt{|y-x_1|}+(y-x_2)^2$. Clearly, for each $y$ it is optimal to have $x_1=x_2=y$. It is straightforward to calculate that the minimal $x_2$ for the integral
$$1/2 \Big(\sqrt{|x_1|}+x_2^2\Big)+1/2\Big(\sqrt{|1-x_1|}+(1-x_2)^2\Big)$$
is $x_2=1/2$, but $x_1=1$ gives a smaller value than $x_1=1/2$.
If $h$ is convex, this problem should not occur.
A: EDIT: The answer is based on the stronger assumption that there is a value $h_{\min}$ such that for all $y\in \Theta$ and $a,b$ it holds
$$
\min_{x_1,\dots,x_n} h(y,x_1,\dots,x_n) = h(y,a,b,\dots,b)=: h_{\min}.
$$
In the OPs setting the minimizes $x_1,\dots,x_n$ might be $y$ dependent.
For $y$-independent minimizer: On the one hand
$$\begin{align*}
  \min_{x_1,\dots,x_n} g(x_1,\dots,x_n) &= \min_{x_1,\dots,x_n} \int_\Theta h(y,x_1,\dots,x_n) f(y) \, dy \\
&\geq \int_{\Theta} \min_{x_1,\dots,x_n} h(y,x_1,\dots,x_n) f(y) \, dy \\
&= \int_{\Theta} h_{\min} f(y) \, dy = h_{\min},
\end{align*}
$$
since $f$ is a probability density on $\Theta$.
On the other hand for the choice $x_1=x_n=a$ and $x_2=x_3=\dots=x_{n-1}=b$ it also holds
$$
  g(a,b,\dots,b,a) = \int_\Theta h(y,a,b,\dots ,b,a) f(y) \,dy =  \int_{\Theta} h_{\min} f(y) \, dy = h_{\min},
$$
by assumption.
