[proofplan] We prove both implications directly from the definitions. If $f$ is convex, then taking two points above the graph of $f$ and forming their convex combination preserves the required height inequality, so the convex combination remains in the epigraph. Conversely, if the epigraph is convex, then points just above $(x,f(x))$ and $(y,f(y))$ have convex combinations in the epigraph; sending the excess height to zero gives the convexity inequality for finite values, while cases involving $+\infty$ are immediate from the extended-real order. [/proofplan]

[step:Show that convexity of $f$ forces convexity of its epigraph]Assume that $f$ is convex. Let $(x,r),(y,s) \in \operatorname{epi} f$, where $x,y \in \mathbb{R}^n$ and $r,s \in \mathbb{R}$. Let $t \in [0,1]$. Since $(x,r),(y,s) \in \operatorname{epi} f$, we have $f(x) \le r$ and $f(y) \le s$. By convexity of $f$, \begin{align*} f(tx+(1-t)y) \le t f(x) + (1-t) f(y). \end{align*} Because $t \ge 0$ and $1-t \ge 0$, the inequalities $f(x) \le r$ and $f(y) \le s$ imply \begin{align*} t f(x) + (1-t) f(y) \le tr + (1-t)s. \end{align*} Therefore \begin{align*} f(tx+(1-t)y) \le tr + (1-t)s. \end{align*} Thus \begin{align*} t(x,r)+(1-t)(y,s) = (tx+(1-t)y, tr+(1-t)s) \in \operatorname{epi} f. \end{align*} Since the two epigraph points and the parameter $t \in [0,1]$ were arbitrary, $\operatorname{epi} f$ is convex.[/step]

[guided]Assume that $f$ is convex. To prove that $\operatorname{epi} f$ is convex, we must show that every convex combination of two points in $\operatorname{epi} f$ is again in $\operatorname{epi} f$. Let $(x,r),(y,s) \in \operatorname{epi} f$, with $x,y \in \mathbb{R}^n$ and $r,s \in \mathbb{R}$, and let $t \in [0,1]$. The condition $(x,r) \in \operatorname{epi} f$ means exactly that the height $r$ lies above the value $f(x)$: \begin{align*} f(x) \le r. \end{align*} Similarly, $(y,s) \in \operatorname{epi} f$ gives \begin{align*} f(y) \le s. \end{align*} Now use convexity of $f$ at the points $x$ and $y$ with weight $t$. This gives \begin{align*} f(tx+(1-t)y) \le t f(x) + (1-t) f(y). \end{align*} The right-hand side is bounded above by the corresponding convex combination of the heights $r$ and $s$, because $t$ and $1-t$ are nonnegative: \begin{align*} t f(x) + (1-t) f(y) \le tr + (1-t)s. \end{align*} Combining the two inequalities gives \begin{align*} f(tx+(1-t)y) \le tr + (1-t)s. \end{align*} This final inequality says precisely that the point \begin{align*} (tx+(1-t)y, tr+(1-t)s) \end{align*} belongs to $\operatorname{epi} f$. But this point is the convex combination \begin{align*} t(x,r)+(1-t)(y,s). \end{align*} Thus every convex combination of two points of $\operatorname{epi} f$ remains in $\operatorname{epi} f$, so $\operatorname{epi} f$ is convex.[/guided]

[step:Show that convexity of the epigraph forces convexity of $f$]Assume that $\operatorname{epi} f$ is convex. Let $x,y \in \mathbb{R}^n$ and let $t \in [0,1]$. We prove \begin{align*} f(tx+(1-t)y) \le t f(x) + (1-t) f(y). \end{align*} If $t=0$ or $t=1$, the inequality is an equality in the extended-real order. Hence assume $0<t<1$. If $f(x)=+\infty$ or $f(y)=+\infty$, then the right-hand side equals $+\infty$, so the desired inequality holds. It remains to consider the case $f(x),f(y) \in \mathbb{R}$. Let $\varepsilon>0$. Define the real heights \begin{align*} r_\varepsilon := f(x)+\varepsilon \end{align*} and \begin{align*} s_\varepsilon := f(y)+\varepsilon. \end{align*} Then $(x,r_\varepsilon),(y,s_\varepsilon) \in \operatorname{epi} f$. Since $\operatorname{epi} f$ is convex, \begin{align*} (tx+(1-t)y, tr_\varepsilon+(1-t)s_\varepsilon) \in \operatorname{epi} f. \end{align*} Therefore \begin{align*} f(tx+(1-t)y) \le tr_\varepsilon+(1-t)s_\varepsilon. \end{align*} Substituting the definitions of $r_\varepsilon$ and $s_\varepsilon$ gives \begin{align*} f(tx+(1-t)y) \le t f(x)+(1-t)f(y)+\varepsilon. \end{align*} Since this holds for every $\varepsilon>0$, the order completeness of $\mathbb{R}$ gives \begin{align*} f(tx+(1-t)y) \le t f(x)+(1-t)f(y). \end{align*} Thus $f$ is convex.[/step]

[guided]Assume that $\operatorname{epi} f$ is convex. We want to recover the convexity inequality for $f$, so fix arbitrary points $x,y \in \mathbb{R}^n$ and a weight $t \in [0,1]$. The desired inequality is \begin{align*} f(tx+(1-t)y) \le t f(x) + (1-t) f(y). \end{align*} When $t=0$, this reads $f(y) \le f(y)$, and when $t=1$, it reads $f(x) \le f(x)$. Hence the endpoint cases hold. We now assume $0<t<1$. If $f(x)=+\infty$ or $f(y)=+\infty$, then because $t>0$ and $1-t>0$, the extended-real expression $t f(x)+(1-t)f(y)$ is $+\infty$. The inequality then holds automatically in $(-\infty,+\infty]$. Thus the only substantive case is \begin{align*} f(x), f(y) \in \mathbb{R}. \end{align*} Why do we use heights slightly above $f(x)$ and $f(y)$ rather than the heights $f(x)$ and $f(y)$ themselves? The epigraph is a subset of $\mathbb{R}^n \times \mathbb{R}$, so its second coordinate must be real. In the present finite-valued case the exact heights are real, but using a positive excess $\varepsilon$ also makes the limiting argument explicit and matches the extended-real formulation. Let $\varepsilon>0$. Define \begin{align*} r_\varepsilon := f(x)+\varepsilon \end{align*} and \begin{align*} s_\varepsilon := f(y)+\varepsilon. \end{align*} These are [real numbers](/page/Real%20Numbers), and they lie above the corresponding function values. Hence \begin{align*} (x,r_\varepsilon),(y,s_\varepsilon) \in \operatorname{epi} f. \end{align*} Since $\operatorname{epi} f$ is convex, the convex combination of these two epigraph points also belongs to $\operatorname{epi} f$: \begin{align*} t(x,r_\varepsilon)+(1-t)(y,s_\varepsilon) \in \operatorname{epi} f. \end{align*} Computing this convex combination in $\mathbb{R}^n \times \mathbb{R}$ gives \begin{align*} t(x,r_\varepsilon)+(1-t)(y,s_\varepsilon) = (tx+(1-t)y, tr_\varepsilon+(1-t)s_\varepsilon). \end{align*} Membership in the epigraph now translates back into the height inequality \begin{align*} f(tx+(1-t)y) \le tr_\varepsilon+(1-t)s_\varepsilon. \end{align*} Substitute the definitions of $r_\varepsilon$ and $s_\varepsilon$: \begin{align*} f(tx+(1-t)y) \le t(f(x)+\varepsilon)+(1-t)(f(y)+\varepsilon). \end{align*} Since $t+(1-t)=1$, this becomes \begin{align*} f(tx+(1-t)y) \le t f(x)+(1-t)f(y)+\varepsilon. \end{align*} This inequality holds for every $\varepsilon>0$. Therefore the left-hand side is bounded above by every number strictly larger than $t f(x)+(1-t)f(y)$. By the order completeness of $\mathbb{R}$, \begin{align*} f(tx+(1-t)y) \le t f(x)+(1-t)f(y). \end{align*} Since $x$, $y$, and $t$ were arbitrary, $f$ is convex.[/guided]

[step:Conclude the equivalence] The first step proves that convexity of $f$ implies convexity of $\operatorname{epi} f$. The second step proves that convexity of $\operatorname{epi} f$ implies convexity of $f$. Therefore $f$ is convex if and only if $\operatorname{epi} f$ is a convex subset of $\mathbb{R}^{n+1}$. [/step]