Prove that semiconvex function can be approximated by smooth functions with same semiconvexity constant

Question

On page 60 of the user's guide to viscosity solutions, proof of Lemma A.3, it is stated that given $\varphi$, a semiconvex function, we can find a smooth approximation $\varphi_\epsilon$ with the same semiconvexity constant as $\varphi$. Is there a reference to this fact? I tried proving it by taking $\varphi_\epsilon$ to be the convolution of $\varphi$ with an approximation to the identity, but it does not seem straightforward to prove that this satisfies the same semiconvexity constant as $\varphi$.

Answer 1

The definition of semi-convexity being used here is $\varphi$ is semi-convex with constant $\Lambda > 0$ if $\varphi = \psi + \eta$, where $\psi$ is a convex function and $\eta$ a smooth function with $D^{2} \eta \geq - \Lambda \text{Id}$ (pointwise). This is equivalent to the inequality $D^{2}\varphi \geq -\Lambda \text{Id}$ in the sense of distributions. (One direction comes from the fact that $D^{2} (\varphi + \frac{\Lambda}{2} \|\cdot\|^{2}) \geq 0$ in the sense of distributions, making $\varphi + \frac{\Lambda}{2} \|\cdot\|^{2}$ convex and $-\frac{\Lambda}{2} \|\cdot\|^{2}$ is certainly smooth with second derivative actually equal to $-\Lambda \text{Id}$. The other direction comes a direct computation and properties of convex functions.)

Either definition shows that if $\varphi$ is semi-convex, then the mollified family $\{\varphi^{\epsilon}\}_{\epsilon > 0}$ is also semi-convex. In either case, we get the same "semi-convexity constant" $\Lambda$.

(Throughout $\varphi^{\epsilon}$ is given by $\varphi^{\epsilon}(x) = \int_{\mathbb{R}^{d}} \varphi(y) \eta^{\epsilon}(x - y) \, dy$ and $\eta^{\epsilon} \in C^{\infty}_{c}(\mathbb{R}^{d})$ satisfies $\eta \geq 0$, $\eta(x) = \eta(-x)$, and $\int_{\mathbb{R}^{d}} \eta(x) \, dx = 1$. We set $\eta^{\epsilon}(\xi) = \epsilon^{-d} \eta(\epsilon^{-1} \xi)$. Most likely, I will not use the symmetry. I will assume, though, that $\eta(x) = 0$ if $\|x\| \geq 1$.)

Approach Using Definition 1: Here we have $\varphi = \psi + \eta$ so $\varphi^{\epsilon} = \psi^{\epsilon} + \eta^{\epsilon}$. Recall that $\psi$ is convex and $\eta$ is smooth. Hence $\eta^{\epsilon}$ is smooth and we see that all we have to prove is that $\psi$ convex implies $\psi^{\epsilon}$ convex. (Hence we only really needed to know that $\varphi$ convex implies $\varphi^{\epsilon}$ convex.)

In the next part of the answer, I will show mollification preserves convexity using the "distributional" approach. Hence in this part I'll prove it "by hand."

$\psi$ is convex so, for each $x, x' \in \mathbb{R}^{d}$ and $\lambda \in [0,1]$, we have \begin{align*} \psi^{\epsilon}((1 - \lambda)x + \lambda x') &= \int_{\mathbb{R}^{d}} \psi((1 - \lambda)x + \lambda x' - y) \eta^{\epsilon}(y) \, dy \\ &\leq \int_{\mathbb{R}^{d}} [(1 - \lambda) \psi(x) + \lambda \psi(x')] \eta^{\epsilon}(y) \, dy \\ &= (1 - \lambda) \psi^{\epsilon}(x) + \lambda \psi^{\epsilon}(x'). \end{align*} This proves $\psi^{\epsilon}$ is convex.

Similarly, it is not hard to show that $D^{2}\eta^{\epsilon} \geq -\Lambda \text{Id}$ holds.

Approach Using Definition 2: We want to show that if $f \in C^{\infty}_{c}(\mathbb{R}^{d})$ and $f \geq 0$, then \begin{equation*} \int_{\mathbb{R}^{d}} \varphi^{\epsilon}(x) D^{2}f(x) \, dx \geq -\Lambda \int_{\mathbb{R}^{d}} f(x) \, dx \cdot \text{Id}. \end{equation*}

Notice that if $f \in C^{\infty}_{c}(\mathbb{R}^{d})$ and $f \geq 0$, then \begin{align*} \int_{\mathbb{R}^{d}} \varphi^{\epsilon}(x) D^{2}f(x) \, dx &= \int_{\mathbb{R}^{d}} \int_{\mathbb{R}^{d}} \varphi(x - y) D^{2} f(x) \, dx \, \eta^{\epsilon}(y) \, dy \\ &= \int_{\mathbb{R}^{d}} \int_{\mathbb{R}^{d}} \varphi(\xi) D^{2} f(\xi + y) \, d\xi \, \eta^{\epsilon}(y) \, dy. \end{align*} Since $f(\cdot + y) \in C^{\infty}_{c}(\mathbb{R}^{d})$ for each $y \in \{\eta^{\epsilon} > 0\}$, we have \begin{equation*} \int_{\mathbb{R}^{d}} \varphi(\xi) D^{2} f(\xi + y) \, d\xi \eta^{\epsilon}(y) \geq - \Lambda \int_{\mathbb{R}^{d}} f(\xi + y) \, d \xi \, \eta^{\epsilon}(y) \cdot \text{Id}. \end{equation*} Thus, \begin{equation*} \int_{\mathbb{R}^{d}} \varphi^{\epsilon}(x) D^{2}f(x) \, dx \geq -\Lambda \int_{\mathbb{R}^{d}} \int_{\mathbb{R}^{d}} f(\xi + y) \, d \xi \, \eta^{\epsilon}(y) \, dy \cdot \text{Id} = -\Lambda \int_{\mathbb{R}^{d}} f(x) \, dx \cdot \text{Id}. \end{equation*}