Proof of Mollification Preserves Semiconvexity

Write $g=f*\rho$, a real-valued function on $\Omega^{\delta}$. By Quadratic Increment Characterisation of Semiconvexity it suffices to prove that for all $x_{1},x_{2}\in\Omega^{\delta}$ and every $t\in\mathbb{R}$ with $0\le t$ and $t\le 1$,

where $\frac{s}{2}$ has the meaning fixed in Quadratic Increment Characterisation of Semiconvexity.

Fix such $x_{1}$, $x_{2}$ and $t$, put $z=t\,x_{1}+(1-t)\,x_{2}$, which lies in $\Omega^{\delta}$ because that set is convex by The $\delta$-Interior of a Convex Subset of $\mathbb{R}^n$ is Convex, and put $c=\frac{\mu}{2}\,t(1-t)\,\lVert x_{1}-x_{2}\rVert^{2}$.

For $x\in\Omega^{\delta}$ let $h_{x}:\mathbb{R}^{n}\to\mathbb{R}$ be the integrand of Convolution of a Continuous Function with a Compactly Supported Continuous Kernel, that is, $h_{x}(w)=f(x-w)\rho(w)$ for those $w\in\mathbb{R}^{n}$ with $x-w\in\Omega$ and $h_{x}(w)=0$ for all other $w$; thus $g(x)=\int_{\mathbb{R}^{n}}h_{x}\,d\lambda_{n}$, and $h_{x}$ is integrable with respect to $\lambda_{n}$ by claim 1 of The Convolution Integrand is Continuous, Compactly Supported and Integrable. Here $\mathbb{R}^{n}$ together with its Borel $\sigma$-algebra and $\lambda_{n}$ is a measure space, by Lebesgue Measure on $\mathbb{R}^n$.

Step 1 (a pointwise inequality). We claim that

Suppose first that $\delta<\lVert w\rVert$. Then $\rho(w)=0$ by clause 3 of Mollifier Kernel of Radius $\delta$ on $\mathbb{R}^n$, and consequently $h_{x}(w)=0$ for every $x\in\Omega^{\delta}$: either $x-w\notin\Omega$, in which case $h_{x}(w)=0$ by definition, or $x-w\in\Omega$, in which case $h_{x}(w)=f(x-w)\cdot 0=0$ by claim 1 of Zero Products and Elementary Identities in a Field. By the same annihilation claim both sides of the asserted inequality equal $0$, so it holds.

Suppose now that $\lVert w\rVert\le\delta$. For any $x\in\Omega^{\delta}$ we have $x-w=x+(-w)$ by claim 3 of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space and $-w=(-1)\,w$ by claim 2 of the same result, so claims 2 and 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ give

whence $x-w\in\bar B(x,\delta)\subseteq\Omega$. In particular $x_{1}-w$, $x_{2}-w$ and $z-w$ all lie in $\Omega$, so

In the real vector space $\mathbb{R}^{n}$ of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, distributivity of scalar multiplication over vector addition and over addition of scalars, together with associativity and commutativity of addition, give

and likewise $(x_{1}-w)-(x_{2}-w)=x_{1}-x_{2}$.

Since $\Omega$ is convex and $f$ is semiconvex on $\Omega$ with constant $\mu$, Quadratic Increment Characterisation of Semiconvexity applied to the points $x_{1}-w$ and $x_{2}-w$ of $\Omega$ and to the scalar $t$ yields

By clause 2 of Mollifier Kernel of Radius $\delta$ on $\mathbb{R}^n$ we have $0\le\rho(w)$, so multiplying this non-strict inequality by $\rho(w)$, which is permitted by claim 5 of Elementary Arithmetic in an Ordered Field, and expanding by distributivity in the field of real numbers gives precisely the asserted inequality at $w$. This proves the claim.

Step 2 (integration). By clause 4 of Mollifier Kernel of Radius $\delta$ on $\mathbb{R}^n$ the kernel $\rho$ is integrable with respect to $\lambda_{n}$ and $\int_{\mathbb{R}^{n}}\rho\,d\lambda_{n}=1$. Let $u:\mathbb{R}^{n}\to\mathbb{R}$ be given by $u(w)=t\,h_{x_{1}}(w)+(1-t)\,h_{x_{2}}(w)+c\,\rho(w)$. Applying claim 2 of Linearity and Monotonicity of the Lebesgue Integral first to the integrable functions $h_{x_{1}}$ and $h_{x_{2}}$ with the scalars $t$ and $1-t$, and then to the resulting integrable function and $\rho$ with the scalars $1$ and $c$, shows that $u$ is integrable with

By Step 1, $h_{z}(w)\le u(w)$ for every $w\in\mathbb{R}^{n}$, so the monotonicity assertion of claim 2 of Linearity and Monotonicity of the Lebesgue Integral, applied to the integrable functions $h_{z}$ and $u$, gives

This is the required inequality. Since $x_{1},x_{2}\in\Omega^{\delta}$ and $t$ with $0\le t\le 1$ were arbitrary, Quadratic Increment Characterisation of Semiconvexity shows that $f*\rho$ is semiconvex on $\Omega^{\delta}$ with constant $\mu$.