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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.

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 $c$?

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.

Asaf Shachar
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1 Answers1

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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 $F$.
gerw
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  • Thank you so much! This is a great answer. Can you please elaborate on: (1) Why there exist $x,y$ and $\lambda \in [0,1]$ such that $c = \lambda , x + (1-\lambda), y$ and $\hat F(c) = \lambda , F(x) + (1-\lambda) , F(y)$? (2) Why $\inf F=\inf \hat F$? (This second fact is familiar to me in other more general Sobolev contexts, where the convex envelope is replaced by the quasiconvex envelope, but it doesn't quite fit the current setting. If there is an easy proof I would be happy to see it). Perhaps you have references for the relevant properties of convex envelopes? Thanks again. – Asaf Shachar Aug 28 '20 at 14:02
  • OK, I understand now why the infimums coincide (you may see my added argument in the question). So, the only remaining question is why there exist suitable $x,y,\lambda$... – Asaf Shachar Aug 29 '20 at 05:34
  • I asked about this separately here: https://math.stackexchange.com/questions/3808081/is-the-convex-envelope-always-equal-to-a-convex-combination-of-the-original-func. – Asaf Shachar Aug 30 '20 at 13:30