Is the infimum $\int_{X} F(g)$ over all $g$ such that $ \int_X g=c $ obtained

Question

Let $F:[0,\infty) \to [0,\infty)$ be a $C^1$ function satisfying $F(1)=0$, which is strictly increasing on $[1,\infty)$, and strictly decreasing on $[0,1]$. Suppose also that $F|_{(1-\epsilon,1+\epsilon)}$ is convex for some $\epsilon>0$.

Let $c\in (0,1)$. Let $X$ be a probability space and let $g:X \to [0,\infty)$ be measurable.

Question:
Is
$
\,E=\inf \{ \int_{X} F(g)\, | \, \int_X g=c \} \,$ always a minimum? i.e. does there exist a $g$ such that $\int_X g=c$ and $\int_{X} F(g)=E$?

It is proved here that $E>0$.

Here is a necessary condition for $g$ to be a minimizer:

Let $h:X \to [0,\infty)$ be measurable with $\int_X h=0$. Set $f(t)=\int_{X} F(g+th)$; $f$ has a minimum at $t=0$. Differentiating, we get $\int_{X} F'(g) \cdot h=0$. Since this holds for any $h$ with $\int_X h=0$, $F'(g)$ must be constant, or more explicitly the function $x \to F'(g(x))$ is a constant function on $X$.

Actually, there might be a subtle issue of integrability here- it is not clear that $f(t)<\infty$, but I think that it's OK to ignore it.

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If $F$ is convex at $c$, i.e. for any $x_1,x_2>0, \alpha \in [0,1]$ satisfying $\alpha x_1 + (1- \alpha)x_2 =c$, we have
$$
F(c)=F\left(\alpha x_1 + (1- \alpha)x_2 \right) \leq \alpha F(x_1) + (1-\alpha)F(x_2), \tag{1}
$$
then Jensen inequality implies that $\int_{X} F(g) \ge F(\int_{X} g)=F(c)$, so $E=F(c)$ is realized by the constant function $g=c$.

What happens when $E$ is not convex at $c$?

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Edit:

Let $\hat F$ denote the (lower) convex envelope of $F$, i.e.
$$
\hat F(x) = \sup \{ h(x) \mid \text{$h$ is convex on $[0, \infty)$}, h \le F \} \, .
$$
$\hat F$ is a non-negative convex function.

We have
$$
\int_{X} F(g) \ge \int_{X} \hat F(g) \ge \hat F(\int_{X} g)=\hat F(c).
$$

So, if $X$ is non-atomic, it suffices to find $x,y$ and $\lambda \in [0,1]$ such that $c = \lambda \, x + (1-\lambda)\, y$ and $\hat F(c) = \lambda \, F(x) + (1-\lambda) \, F(y)$.

If we have these, then we can choose $g$ which takes the value $x$ with probability $\lambda$ and $y$ with probability $1-\lambda$. Then, $g$ minimizes $F$:

$\int_{X} g=c$, and $\int_{X} F(g)=\lambda \, F(x) + (1-\lambda) \, F(y)=\hat F(c) \le E$ by the previous argument.

Answer 1

To long for a comment. This answers the case of a non-atomic measure space.

Let $\hat F$ be the convex envelope of $F$. Then, one can check that:

• the infimum over $\hat F$ equals the infimum over $F$.
• the infimum over $\hat F$ is attained at $c$
• there are $x,y$ and $\lambda \in [0,1]$ such that $c = \lambda \, x + (1-\lambda)\, y$ and $\hat F(c) = \lambda \, F(x) + (1-\lambda) \, F(y)$
• then, you can take a function $g$ which takes the value $x$ with probability $\lambda$ and $y$ otherwise. Then, $g$ minimizes $F$.