Proof of Quadratic Increment Characterisation of Semiconvexity

Let $g:C\to\mathbb{R}$ be the function $g(x)=f(x)+\frac{\mu}{2}\,\lVert x\rVert^{2}$. By Semiconvex Function on a Convex Subset of $\mathbb{R}^n$, $f$ is semiconvex on $C$ with constant $\mu$ if and only if $g$ is convex on $C$, that is, if and only if

for all $x,y\in C$ and every $t\in\mathbb{R}$ with $0\le t$ and $t\le 1$. It therefore suffices to show, for each such $x$, $y$ and $t$, that this inequality is equivalent to the asserted quadratic inequality; the stated equivalence then follows by taking the conjunction over all admissible $x$, $y$ and $t$.

So fix $x,y\in C$ and $t\in\mathbb{R}$ with $0\le t$ and $t\le 1$, and put $z=t\,x+(1-t)\,y$, a point of $C$ because $C$ is convex. Put

both real numbers.

Step 1 (rewriting $g$ at $z$). By The Squared Norm of a Convex Combination of Two Points,

Multiplying by $\frac{\mu}{2}$ and using distributivity in the field of real numbers gives $\frac{\mu}{2}\lVert z\rVert^{2}=A-B$, hence

Step 2 (rewriting the right-hand side). Again by distributivity,

Step 3 (cancelling and rearranging). By claim 3 of Elementary Arithmetic in an Ordered Field, for real numbers $p$ and $q$ one has $p\le q$ if and only if $0\le q-p$. Apply this twice, first to

and then to

In the field of real numbers both differences $q-p$ and $q'-p'$ are equal to $t\,f(x)+(1-t)\,f(y)+B-f(z)$, so $p\le q$ holds if and only if $p'\le q'$, that is,

By Steps 1 and 2 the first of these inequalities is exactly $g(z)\le t\,g(x)+(1-t)\,g(y)$, and the last is exactly the asserted quadratic inequality for $x$, $y$ and $t$. This is the required equivalence.