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$g\in C^2(\mathbb R^n)$ is called semi concave, if there exists $c>0$ such that for all $x,y\in\mathbb R^n$ the following holds: $$g(x+y) - 2g(x) + g(x-y) \leq c|y|^2$$

Now, in Evans "Partial Differential Equations", it is said that "it is easy to check that" $g$ is semi concave iff the mapping $x\mapsto \frac c 2 |x|^2 - g(x)$ is convex.

I do not find it that easy to check. I tried showing that $x\mapsto \frac C 2 |x|^2 -g(x)$ is convex by straigtforward calculation and estimation, but this did not get me very far.

Using $(a+b)^2 \leq 2a^2+2b^2$ I got this far:

\begin{align*} & \frac C 2 |\lambda x+(1-\lambda)y|^2 - g(\lambda x + (1-\lambda)y) \\ \leq & C\left(\lambda^2 |x|^2 + (1-\lambda)^2|y|^2\right) - g(\lambda x + (1-\lambda)y) \end{align*}

How can I do this?

dinosaur
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1 Answers1

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The starting point is the easily verified identity $$|x+y|^2 - 2|x|^2 + |x-y|^2 = 2|y|^2\tag1$$ Because of it, the inequality $g(x+y) - 2g(x) + g(x-y) \leq c|y|^2$ is equivalent to $$g(x+y)-\frac c2|x+y|^2 - 2\left(g(x)-\frac c2|x|^2\right) + g(x-y)-\frac c2|x-y|^2 \le 0\tag2$$ Write $h(x)= \frac c 2 |x|^2 - g(x)$ for brevity; then (2) becomes $$h(x+y)-2h(x)+h(x-y)\ge 0\tag3$$ Note that (3) is exactly the midpoint convexity of $h$. For continuous functions, it is equivalent to convexity: see Midpoint-Convex and Continuous Implies Convex. Or you can let $y\to 0$, relate (3) to second order derivative, and observe that the Hessian of $h$ is positive semidefinite.

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