I already showed that the function $\psi(\mu,\sigma)=\mathbb{E}U(X)$ is concave in $(\mu,\sigma)$, where $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$. $U$ is a nice concave increasing function.
Now I want to prove that $\psi$ increases with $\mu$ and decreases with $\sigma$ and if all returns (Consider $X$ as a contingent claim) are jointly Gaussian , an investor maximising the expected utility of his final wealth has to choose a mean-variance efficient portfolio.
According to the first point: This is not clear to me because when you draw a concave function and take the expectation (x-axis is variance, y-axis is mean) then $\psi$ grows as $\mu$ and $\sigma$ grows.
According to the second point: I do not know what I can do here.
Let $Y$ denote a standard normal random variable then $X$ is distributed like $\mu+\sigma Y$ hence $\psi(\mu,\sigma)=E[U(\mu+\sigma Y)]$. For every $y$, the function $\mu\mapsto U(\mu+\sigma y)$ is increasing because $U$ is increasing hence the function $\mu\mapsto\psi(\mu,\sigma)$ is increasing as a barycenter of increasing functions.
Likewise, $X$ is distributed like $\mu-\sigma Y$ hence $\psi(\mu,\sigma)=\frac12E[U(\mu+\sigma Y)+U(\mu-\sigma Y)]$. For every $y$ and $\mu$, the function $\sigma\mapsto U(\mu+\sigma y)+U(\mu-\sigma y)$ is nonincreasing on $\sigma\geqslant0$ because $U$ is concave hence the function $\sigma\mapsto\psi(\mu,\sigma)$ is nonincreasing on $\sigma\geqslant0$ as a barycenter of nonincreasing functions.
The key in the second paragraph is the fact that for every convex function $u$, the function $v:x\mapsto u(x)+u(-x)$ is nondecreasing on $x\geqslant0$. Can you show this?