Extreme points of a subset of density functions Let us consider the set of real density functions with fixed mean and variance:
$$S(\mu, \sigma^2)=\{f: \Bbb R\to \Bbb R: f\geq 0, f \textrm{ continuous, }\\ \int_{\Bbb R}f(x)=1, \int_{\Bbb R}xf(x)=\mu,\; \int_{\Bbb R}(x-\mu)^2f(x)=\sigma^2\}.$$ 
I am curious about the following questions:
What is the set of extreme points of S? Is the normal distribution with corresponding mean and variance an extreme point?
Any pointer to the literature for either this or similar problems is also appreciated. 
 A: $S(\mu,\sigma^2)$ has no extreme points.  We can always write $f$ as an average of two variants, which we get by making adjustments at a few points.
To see this, find three adjacent triangles below the graph of $f$, e.g.:

More formally, by the continuity of $f$, choose $b,u,v$ such that $f(x)>v>0$ for $x$ in $[b-3u,b+3u]$. Let
\begin{align}
f_1 &= v \max(0, 1-|x-b+2u|/u)\\
f_2 &= v \max(0, 1-|x-b+0u|/u)\\
f_3 &= v \max(0, 1-|x-b-2u|/u)\\
f_4 &= f-f_1-f_2-f_3\\
\end{align}
The picture above shows $f_1,f_2,f_3$ under a standard normal $f$, with $b=0,$ $u=1/2,$ $v=1/8.$
Now we seek functions in $S(\mu,\sigma^2)$ of the form $\sum w_i f_i$, so that the $w_i$ satisfy
$$\sum w_i\! \int\! f_i = 1,\ \ \sum w_i\! \int\! f_i x = \mu,\ \ \sum w_i\! \int\! f_i x^2 = \mu^2+\sigma^2.$$
When we integrate and put the $w_4$'s on the right hand side, this gives:
$$
\begin{pmatrix}
uv &uv(b-2u) &uv(b^2 - 4bu + \frac{25}{6} u^2)\\
uv &uv(b+0u) &uv(b^2 + 0bu + \frac{1}{6} u^2)        \\
uv &uv(b+2u) &uv(b^2 + 4bu + \frac{25}{6} u^2)\\
\end{pmatrix}^T
\begin{pmatrix}
w_1 \\
w_2 \\
w_3 \\
\end{pmatrix}
=
\begin{pmatrix}
1 - (1-3uv)w_4 \\
\mu - (\mu-3buv)w_4\\
\mu^2 + \sigma^2 - kw_4  \\
\end{pmatrix}
$$
where $k=\mu^2+\sigma^2-uv(3b^2+\frac{51}{6}u^2).$
The matrix has determinant $16u^6v^3$, which is non-zero, so any value of $w_4$ yields a unique solution for $w_1,w_2,w_3$.  When $w_4=1$, the solution is $w_1=w_2=w_3=1.$
Now choose $\epsilon$ such that $w_4=1\pm\epsilon$ both yield positive values of $w_1, w_2, w_3$. Let $g$ be the value of $\sum w_i f_i$ corresponding to $w_4 = 1+\epsilon$, and let $h$ be the value corresponding to $w_4 = 1-\epsilon$.  Then $f$ is an even mixture of $g$ and $h$.
In the example above, choosing $\epsilon=1/4$ gives $f,g,h$ as in the blue, green and orange of the graph below, and all are in $S(0,1)$.

