Let $g:\mathbb R_+\to\mathbb R$ be a measurable function (which could be supposed to be bounded and Lipschitz if required). Let $\mathcal P$ be the collection of probability measures $\mu$ on $\mathbb R_+$ with finite first moment, i.e.

$$\int_{\mathbb R_+}|x|d\mu(x)~~<~~+\infty.$$

For any fixed $\epsilon\in [0,1)$, denote

$$\mathcal P_y^{\epsilon}~~:=~~\left\{\mu\in\mathcal P:~ \big(1-\epsilon\big)y~\le~\int_{\mathbb R_+}xd\mu(x)~\le~\big(1+\epsilon\big)y\right\} \mbox{ for all } y\in\mathbb R_+.$$

Define further $G^{\epsilon}:\mathbb R_+\to\mathbb R$ by

$$G^{\epsilon}(y)~~:=~~\sup_{\mu\in\mathcal P_y^{\epsilon}}~ \int_{\mathbb R_+}g(x)d\mu(x).$$

It is well known that, if $\epsilon=0$, then $G^{0}$ turns to be the concave envolope of $g$, and thus the left and right derivatives of $G^{0}$ exist. Consiser the following dual problem defined by

$$D^{\epsilon}(y)~~:=~~\inf_{(z,h)\in \mathcal D^{\epsilon}(y)}~ z,$$

where

$$\mathcal D^{\epsilon}(y)~~:=~~\left\{(z,h)\in\mathbb R^2:~ z~+~h(x-y)~-~\epsilon|h|~\ge~g(x) \mbox{ for all } x\in\mathbb R_+\right\}.$$

The aim is to show the duality $G^{\epsilon}(y)=D^{\epsilon}(y)$. A straightforward computation yields $G^{\epsilon}(y)\le D^{\epsilon}(y)$. My question is how to show $G^{\epsilon}(y)\ge D^{\epsilon}(y)$. Indeed, if $\epsilon=0$, it is very easy: Since $G^{0}$ is concave, then

$$g(x)~~\le~~G^{0}(x)~~\le~~G^{0}(y)~+~(x-y)(G^{0})'(y).$$

Thus one has $\big(G^{0}(y),(G^{0})'(y)\big)\in \mathcal D^{\epsilon}(y)$ and thus $G^{\epsilon}(y)\ge D^{\epsilon}(y)$. I wonder how to generalize to $\epsilon\in [0,1)$.

Let $g:\mathbb R_+\to\mathbb R$ be a measurable function (which could be supposed to be bounded and Lipschitz if required). Let $\mathcal P$ be the collection of probability measures $\mu$ on $\mathbb R_+$ with finite first moment, i.e.

$$\int_{\mathbb R_+}|x|d\mu(x)~~<~~+\infty.$$

For any fixed $\epsilon\in [0,1)$, denote

$$\mathcal P_y^{\epsilon}~~:=~~\left\{\mu\in\mathcal P:~ \big(1-\epsilon\big)y~\le~\int_{\mathbb R_+}xd\mu(x)~\le~\big(1+\epsilon\big)y\right\} \mbox{ for all } y\in\mathbb R_+.$$

Define further $G^{\epsilon}:\mathbb R_+\to\mathbb R$ by

$$G^{\epsilon}(y)~~:=~~\sup_{\mu\in\mathcal P_y^{\epsilon}}~ \int_{\mathbb R_+}g(x)d\mu(x).$$

It is well known that, if $\epsilon=0$, then $G^{0}$ turns to be the concave envelope of $g$, and thus the left and right derivatives of $G^{0}$ exist. Consiser the following dual problem defined by

$$D^{\epsilon}(y)~~:=~~\inf_{(z,h)\in \mathcal D^{\epsilon}(y)}~ z,$$

where

$$\mathcal D^{\epsilon}(y)~~:=~~\left\{(z,h)\in\mathbb R^2:~ z~+~h(x-y)~-~\epsilon|h|~\ge~g(x) \mbox{ for all } x\in\mathbb R_+\right\}.$$

The aim is to show the duality $G^{\epsilon}(y)=D^{\epsilon}(y)$. A straightforward computation yields $G^{\epsilon}(y)\le D^{\epsilon}(y)$. My question is how to show $G^{\epsilon}(y)\ge D^{\epsilon}(y)$. Indeed, if $\epsilon=0$, it is very easy: Since $G^{0}$ is concave, then

$$g(x)~~\le~~G^{0}(x)~~\le~~G^{0}(y)~+~(x-y)(G^{0})'(y).$$

Thus one has $\big(G^{0}(y),(G^{0})'(y)\big)\in \mathcal D^{\epsilon}(y)$ and thus $G^{\epsilon}(y)\ge D^{\epsilon}(y)$. I wonder how to generalize to $\epsilon\in [0,1)$.

Let $g:\mathbb R_+\to\mathbb R$ be a measurable function (which could be supposed to be bounded and Lipschitz if required). Let $\mathcal P$ be the collection of probability measures $\mu$ on $\mathbb R_+$ with finite first moment, i.e.

$$\int_{\mathbb R_+}|x|d\mu(x)~~<~~+\infty.$$

For any fixed $\epsilon\in [0,1)$, denote

$$\mathcal P_y^{\epsilon}~~:=~~\left\{\mu\in\mathcal P:~ \big(1-\epsilon\big)y~\le~\int_{\mathbb R_+}xd\mu(x)~\le~\big(1+\epsilon\big)y\right\} \mbox{ for all } y\in\mathbb R_+.$$

Define further $G^{\epsilon}:\mathbb R_+\to\mathbb R$ by

$$G^{\epsilon}(y)~~:=~~\sup_{\mu\in\mathcal P_y^{\epsilon}}~ \int_{\mathbb R_+}g(x)d\mu(x).$$

I'm interested in the characterization of the function $G^{\epsilon}$. It is well known that, if $\epsilon=0$, then $G^{0}$ turns to be the concave envolope of $g$, and thus the left and right derivatives of $G^{0}$ exist. My question is that could we have a more detailed analysis on $G^{\epsilon}$? Many thanks!

Let $g:\mathbb R_+\to\mathbb R$ be a measurable function (which could be supposed to be bounded and Lipschitz if required). Let $\mathcal P$ be the collection of probability measures $\mu$ on $\mathbb R_+$ with finite first moment, i.e.

$$\int_{\mathbb R_+}|x|d\mu(x)~~<~~+\infty.$$

For any fixed $\epsilon\in [0,1)$, denote

$$\mathcal P_y^{\epsilon}~~:=~~\left\{\mu\in\mathcal P:~ \big(1-\epsilon\big)y~\le~\int_{\mathbb R_+}xd\mu(x)~\le~\big(1+\epsilon\big)y\right\} \mbox{ for all } y\in\mathbb R_+.$$

Define further $G^{\epsilon}:\mathbb R_+\to\mathbb R$ by

$$G^{\epsilon}(y)~~:=~~\sup_{\mu\in\mathcal P_y^{\epsilon}}~ \int_{\mathbb R_+}g(x)d\mu(x).$$

It is well known that, if $\epsilon=0$, then $G^{0}$ turns to be the concave envolope of $g$, and thus the left and right derivatives of $G^{0}$ exist. Consiser the following dual problem defined by

$$D^{\epsilon}(y)~~:=~~\inf_{(z,h)\in \mathcal D^{\epsilon}(y)}~ z,$$

where

$$\mathcal D^{\epsilon}(y)~~:=~~\left\{(z,h)\in\mathbb R^2:~ z~+~h(x-y)~-~\epsilon|h|~\ge~g(x) \mbox{ for all } x\in\mathbb R_+\right\}.$$

The aim is to show the duality $G^{\epsilon}(y)=D^{\epsilon}(y)$. A straightforward computation yields $G^{\epsilon}(y)\le D^{\epsilon}(y)$. My question is how to show $G^{\epsilon}(y)\ge D^{\epsilon}(y)$. Indeed, if $\epsilon=0$, it is very easy: Since $G^{0}$ is concave, then

$$g(x)~~\le~~G^{0}(x)~~\le~~G^{0}(y)~+~(x-y)(G^{0})'(y).$$

Thus one has $\big(G^{0}(y),(G^{0})'(y)\big)\in \mathcal D^{\epsilon}(y)$ and thus $G^{\epsilon}(y)\ge D^{\epsilon}(y)$. I wonder how to generalize to $\epsilon\in [0,1)$.

Let $g:\mathbb R_+\to\mathbb R$ be a measurable function (which could be supposed to be bounded and Lipschitz if required). Let $\mathcal P$ be the collection of probability measures $\mu$ on $\mathbb R_+$ with finite first moment, i.e.

$$\int_{\mathbb R_+}|x|d\mu(x)~~<~~+\infty.$$

For any fixed $\epsilon\in [0,1)$, denote

$$\mathcal P_y^{\epsilon}~~:=~~\left\{\mu\in\mathcal P:~ \big(1-\epsilon\big)y~\le~\int_{\mathbb R_+}xd\mu(x)~\le~\big(1+\epsilon\big)y\right\} \mbox{ for all } y\in\mathbb R_+.$$

Define further $G^{\epsilon}:\mathbb R_+\to\mathbb R$ by

$$G^{\epsilon}(y)~~:=~~\sup_{\mu\in\mathcal P_y^{\epsilon}}~ \int_{\mathbb R_+}g(x)d\mu(x).$$

I'm interested in the regularity of the function $G^{\epsilon}$. It is well known that, if $\epsilon=0$, then $G^{0}$ turns to be the concave envolope of $g$, and thus the left and right derivatives of $G^{0}$ exist. My question is that could we show that the left and right derivatives of $G^{\epsilon}$ exist as well? Or maybe a more detailed analysis on $G^{\epsilon}$ could be provided? Many thanks!

Let $g:\mathbb R_+\to\mathbb R$ be a measurable function (which could be supposed to be bounded and Lipschitz if required). Let $\mathcal P$ be the collection of probability measures $\mu$ on $\mathbb R_+$ with finite first moment, i.e.

$$\int_{\mathbb R_+}|x|d\mu(x)~~<~~+\infty.$$

For any fixed $\epsilon\in [0,1)$, denote

$$\mathcal P_y^{\epsilon}~~:=~~\left\{\mu\in\mathcal P:~ \big(1-\epsilon\big)y~\le~\int_{\mathbb R_+}xd\mu(x)~\le~\big(1+\epsilon\big)y\right\} \mbox{ for all } y\in\mathbb R_+.$$

Define further $G^{\epsilon}:\mathbb R_+\to\mathbb R$ by

$$G^{\epsilon}(y)~~:=~~\sup_{\mu\in\mathcal P_y^{\epsilon}}~ \int_{\mathbb R_+}g(x)d\mu(x).$$

I'm interested in the characterization of the function $G^{\epsilon}$. It is well known that, if $\epsilon=0$, then $G^{0}$ turns to be the concave envolope of $g$, and thus the left and right derivatives of $G^{0}$ exist. My question is that could we have a more detailed analysis on $G^{\epsilon}$? Many thanks!