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Given a function $f(\boldsymbol{x})$ defined on the hypercube $\boldsymbol{x} \in [0,1]^n$. Suppose $f(\boldsymbol{x})$ can be expressed as $f(\boldsymbol{x})=c(\boldsymbol{x})+g(\boldsymbol{x})$, where $c(\boldsymbol{x})$ is a convex function and $g(\boldsymbol{x})$ is arbitrary. Suppose $c$ and $g$ are both infinitely differentiable at the interior of the hypercube.

Let the operator $E$ take a function and return its convex envelope. Then does the following identity hold?

$E[f(\boldsymbol{x})]=c(\boldsymbol{x})+E[g(\boldsymbol{x})]$

Thank you for your help!

Golabi
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1 Answers1

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In general no. Let's look at $n = 1$, and the function $f(x) = (x-\frac 12)^2$. Then $f$ is convex, hence $E(f) = f$. On the other hand, let $c = 2f$ and $g = -f$. Then $c + g = f$ and $c$ in convex. But $E(g) = 0$, and $$ c + E(g) = 2f + 0 = 2f \ne f = E(f). $$

martini
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