Proof of A Sum in Separated Variables of Semiconvex Functions is Semiconvex

Write $2=1+1$; then $0<2$ and $0<2^{-1}$ by claim 8 of Elementary Order Arithmetic in an Ordered Field, and $\tfrac{a}{2}$ abbreviates $a\cdot 2^{-1}$.

A nonnegativity remark. Let $t\in\mathbb{R}$ satisfy $0\le t$ and $t\le 1$, and let $\nu\in\mathbb{R}$ satisfy $0\le\nu$. Then $0\le 1-t$ by claim 3 of Elementary Arithmetic in an Ordered Field, and applying claim 5 of that lemma to $0\le 1-t$ with the nonnegative factor $t$, together with claim 1 of Zero Products and Elementary Identities in a Field, gives $0\le t(1-t)$. Applying claim 5 twice more, with the nonnegative factors $2^{-1}$ and $\nu$, gives

$$0\le\tfrac{1}{2}\,t(1-t)\,\nu .$$

Claim 1. Let $z,z'\in C$ and let $t\in\mathbb{R}$ satisfy $0\le t$ and $t\le1$. By the definition of $C$ there are $\xi,\xi'\in C_{1}$ and $\eta,\eta'\in C_{2}$ with $z=\iota(\xi,\eta)$ and $z'=\iota(\xi',\eta')$. By claim 2 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space,

$$t\,z=\iota(t\xi,t\eta),\qquad (1-t)\,z'=\iota\bigl((1-t)\xi',(1-t)\eta'\bigr),$$

and, adding these,

$$t\,z+(1-t)\,z'=\iota\bigl(t\xi+(1-t)\xi',\ t\eta+(1-t)\eta'\bigr).$$

Since $C_{1}$ and $C_{2}$ are convex, $t\xi+(1-t)\xi'\in C_{1}$ and $t\eta+(1-t)\eta'\in C_{2}$, so the point above lies in $C$. As $z,z'$ and $t$ were arbitrary, $C$ is convex by Convex Subset of $\mathbb{R}^n$.

Claim 2. From $0\le\mu_{1}$ and $\mu_{1}\le\mu$ we get $0\le\mu$ by transitivity of the order.

By Quadratic Increment Characterisation of Semiconvexity, applied to $w$ on the convex set $C$ with the constant $\mu$, it suffices to prove that

$$w\bigl(t\,z+(1-t)\,z'\bigr)\le t\,w(z)+(1-t)\,w(z')+\frac{\mu}{2}\,t(1-t)\,\lVert z-z'\rVert^{2}$$

for all $z,z'\in C$ and every $t\in\mathbb{R}$ with $0\le t$ and $t\le1$. Fix such $z,z',t$ and write $z=\iota(\xi,\eta)$, $z'=\iota(\xi',\eta')$ as in claim 1, and put $\zeta_{1}=t\xi+(1-t)\xi'\in C_{1}$ and $\zeta_{2}=t\eta+(1-t)\eta'\in C_{2}$. By claim 1, $t\,z+(1-t)\,z'=\iota(\zeta_{1},\zeta_{2})$, so by the defining property of $w$,

$$w\bigl(t\,z+(1-t)\,z'\bigr)=u_{1}(\zeta_{1})+u_{2}(\zeta_{2}).$$

Applying Quadratic Increment Characterisation of Semiconvexity to $u_{1}$ on $C_{1}$ and to $u_{2}$ on $C_{2}$,

$$u_{1}(\zeta_{1})\le t\,u_{1}(\xi)+(1-t)\,u_{1}(\xi')+\frac{\mu_{1}}{2}\,t(1-t)\,\lVert\xi-\xi'\rVert^{2},$$ $$u_{2}(\zeta_{2})\le t\,u_{2}(\eta)+(1-t)\,u_{2}(\eta')+\frac{\mu_{2}}{2}\,t(1-t)\,\lVert\eta-\eta'\rVert^{2}.$$

By claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square the numbers $\lVert\xi-\xi'\rVert$ and $\lVert\eta-\eta'\rVert$ are nonnegative, hence so are their squares by claim 5 of Elementary Arithmetic in an Ordered Field and claim 1 of Zero Products and Elementary Identities in a Field. So the nonnegativity remark applies with $\nu=\lVert\xi-\xi'\rVert^{2}$ and with $\nu=\lVert\eta-\eta'\rVert^{2}$, and claim 5 of Elementary Arithmetic in an Ordered Field, applied to $\mu_{1}\le\mu$ and to $\mu_{2}\le\mu$ with these nonnegative factors, gives

$$\frac{\mu_{1}}{2}\,t(1-t)\,\lVert\xi-\xi'\rVert^{2}\le\frac{\mu}{2}\,t(1-t)\,\lVert\xi-\xi'\rVert^{2},\qquad \frac{\mu_{2}}{2}\,t(1-t)\,\lVert\eta-\eta'\rVert^{2}\le\frac{\mu}{2}\,t(1-t)\,\lVert\eta-\eta'\rVert^{2}.$$

Adding the four inequalities just obtained — each is equivalent by claim 3 of Elementary Arithmetic in an Ordered Field to the nonnegativity of a difference, the sum of nonnegative differences is nonnegative by claim 2 there, and claim 3 converts back — and using $w(z)=u_{1}(\xi)+u_{2}(\eta)$ and $w(z')=u_{1}(\xi')+u_{2}(\eta')$ together with distributivity, we obtain

$$w\bigl(t\,z+(1-t)\,z'\bigr)\le t\,w(z)+(1-t)\,w(z')+\frac{\mu}{2}\,t(1-t)\,\Bigl(\lVert\xi-\xi'\rVert^{2}+\lVert\eta-\eta'\rVert^{2}\Bigr).$$

Finally, by claim 2 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space we have $z-z'=\iota(\xi-\xi',\eta-\eta')$, and then claim 3 of that lemma gives

$$\lVert z-z'\rVert^{2}=\lVert\xi-\xi'\rVert^{2}+\lVert\eta-\eta'\rVert^{2}.$$

Substituting this into the previous display yields exactly the required inequality, so $w$ is semiconvex on $C$ with constant $\mu$.