Proof of A Function on a Convex Subset of $\mathbb{R}^n$ Is Convex If and Only If Its Epigraph Is Convex

Throughout, convexity of a subset of $\mathbb{R}^{n+1}$ is understood as in Segments and Convex Subsets of $\mathbb{R}^n$ (i) and Segments and Convex Subsets of $\mathbb{R}^n$ (ii), with $n+1$ in place of $n$, membership in $\operatorname{epi}f$ is read off from Epigraph of a Real-Valued Function on a Subset of $\mathbb{R}^n$ (i), and pairs are combined and compared as in Convex Functions on $\mathbb{R}^n$: Standing Notation (ii): a point of $\mathbb{R}^{n+1}$ determines its pair uniquely, and for $x,y\in\mathbb{R}^{n}$ and $\mu,\nu,t\in\mathbb{R}$,

$$(1-t)(x,\mu)+t(y,\nu)=\bigl((1-t)x+ty,\ (1-t)\mu+t\nu\bigr).$$

Suppose first that $f$ is convex. Let $(x,\mu)$ and $(y,\nu)$ be points of $\operatorname{epi}f$ and let $t\in[0,1]$. Then $x,y\in C$, $\mu\ge f(x)$ and $\nu\ge f(y)$. The point $(1-t)x+ty$ lies in $C$ because $C$ is convex, and the convexity inequality of Convex Function on a Convex Subset of $\mathbb{R}^n$ (i), together with $1-t\ge0$ and $t\ge0$, gives

$$f\bigl((1-t)x+ty\bigr)\le(1-t)f(x)+tf(y)\le(1-t)\mu+t\nu .$$

By the displayed formula for combinations of pairs, this says that $(1-t)(x,\mu)+t(y,\nu)$ lies in $\operatorname{epi}f$. Hence $\operatorname{epi}f$ is convex.

Conversely, suppose that $\operatorname{epi}f$ is convex, and let $x,y\in C$ and $t\in[0,1]$. The points $(x,f(x))$ and $(y,f(y))$ lie in $\operatorname{epi}f$, and therefore so does their combination

$$(1-t)\bigl(x,f(x)\bigr)+t\bigl(y,f(y)\bigr)=\bigl((1-t)x+ty,\ (1-t)f(x)+tf(y)\bigr).$$

Since a point of $\mathbb{R}^{n+1}$ determines its pair uniquely, membership in the epigraph says that the second entry is at least the value of $f$ at the first entry:

$$f\bigl((1-t)x+ty\bigr)\le(1-t)f(x)+tf(y).$$

This is the inequality of Convex Function on a Convex Subset of $\mathbb{R}^n$ (i), so $f$ is convex.