Proof of, and requirements for, the reverse of Jensen's Inequality for concave functions

Question

As I understand it, Jensen's Inequality states

$$\int_{U}f_{V}\left(h(u)g(u)\right)du\geq f_{V}\left(\int_{U}h(u)g(u)du\right)$$

For a convex function $f_{V}$, a probability distribution $g(u)$ on the space $U$, and a linear combination of some $u$-dependent probabilities $h(u)$. Which would give a lower bound as a function of some total probabilities where $u$ is unknown. (I realize that perhaps this is not the most general form but it is the one that is relevant to me.)

My question then is what if the function $f$ is not convex but concave - $f_{C}$. My understanding now is that I use the above equation with the inequality reversed, that is

$$\int_{U}f_{C}(h(u)g(u))du\leq f_{C}\left(\int_{U}h(u)g(u)du\right)$$

But this is based on nothing more than an intuition and single vague sentence on Wikipedia. Can anyone refer me to, or provide, a proof? Also, would there be any further requirements on the probability space $U$, measure $g$, or linear combination of inequalities $h$?

Any help is greatly appreciated.

Answer 1

Hint: $f(x)$ is convex if and only if $-f(x)$ is concave.

If $$ f\left(\int_\Omega g(x)\,\mathrm{d}\mu(x)\right)\le\int_\Omega f(g(x))\,\mathrm{d}\mu(x) $$ then negating both sides, we get $$ -f\left(\int_\Omega g(x)\,\mathrm{d}\mu(x)\right)\ge\int_\Omega-f(g(x))\,\mathrm{d}\mu(x) $$

Answer 2

The proof of Jensen's Inequality in both cases is very simple. Let $f_C$ be a concave function. Consider a number (a point) $x_0 = \int_U {h(u)g(u)du}$ and a tangent to $f_C$ at $x_0$. Let its equation be $y(x)=a+bx$. Then it is obvious that $f_C(x) \le y(x)$ for all $x$ and $$\int_U f_C(h(u))g(u) du \le \int_U y(h(u))g(u) du = a+b\int_U h(u)g(u) du = y(x_0) = f_C(x_0),$$ so we get $$\int_U f_C(h(u))g(u) du \le f_C\left(\int_U {h(u)g(u)}du\right).$$