The complete claim is,

Let $F$ be an ordered field and $\bar F=F\bigcup\{+\infty,-\infty\}$. Let $f:A \to \bar F$ be a convex map from affine space $(A,V)$ to affine space $(F,F)$, then $f(C)$ is a convex set if $C$ is convex.

Here $F$ can be treated as $\Bbb R$ except for that we do not use its topology and metric. $A$ is a point set, and $V$ is a vector space.

A convex function is defined using epigraph,

$f$ is a convex set if its epigraph $\text{epi} f=\{(x,y)\in A\times F:y\ge f(x)\}$ is a convex set.

I am not sure if this claim is right but it seems so. I find difficulty in proving it. The following is an attempt, but no contradiction is found.

If $f(C)$ is not convex, then there exists $y_1,y_2\in f(C)$ st. $y_1>y_2$ and $\theta y_1 + (1-\theta) y_2 \notin f(C)$ for some $\theta \in (0,1)$. Let $y_3=\theta y_1 + (1-\theta) y_2$, and suppose $f(x_1)=y_1, f(x_2) = y_2$ for some $x_1,x_2\in C$. Let $x_3 =\theta x_1 + (1-\theta) x_2)$, then $(x_1,y_1),(x_2,y_2)\in \text{epi} f \Rightarrow (\theta x_1 + (1-\theta) x_2, \theta y_1 + (1-\theta) y_2) = (x_3,y_3) \in \text{epi} f \Rightarrow y_3 > f(x_3)$

The last inequality $y_3>x_3$ does not contradict with Jensen's inequality.

I need help with this. If it is not comfortable to deal with affine space, we might just assume $f:\Bbb R^n \to \Bbb R$, and just try not to use topology information (such like continuity) and metric in the proof. I add "real analysis" as a tag of this question for this reason.

Again, note the claim might not be true. If it is not true, hope someone could come up with a counterexample.

Thank you!

  • 1
    $\begingroup$ The characteristic function $\chi_C$ of any convex set $C$ is convex and has image $\{0,\infty\}$, defined by $\chi_C(x) = 0$ for $x\in C$ and $=\infty$ otherwise. $\endgroup$
    – user251257
    Sep 1, 2016 at 22:51
  • $\begingroup$ @Tom in the question, $f$ is supposed to be scalar valued. $\endgroup$
    – user251257
    Sep 2, 2016 at 5:57
  • $\begingroup$ @Tom the epigraph is a convex set, not the graph. $\endgroup$ Sep 2, 2016 at 15:38
  • $\begingroup$ @MichaelGrant the question isn't about the epigraph, but the graph. $\endgroup$
    – user251257
    Sep 2, 2016 at 16:58
  • $\begingroup$ Ah, then really, I should be directing that comment to the OP :-) $\endgroup$ Sep 2, 2016 at 16:58

1 Answer 1


At least for $F=\mathbb R$ and $A=V=\mathbb R^n$ it is pretty simple.

If you allow $\infty$, then the statement is not true in general. The characteristic function $f=\chi_C$ of any convex set $C\subseteq A$, defined by $$f(x) = \begin{cases} 0, & x\in C, \\ \infty, & x\notin C, \end{cases}$$ is convex and has image $f(A)=\{0,\infty\}$. Obviously the image isn't convex.

If you only allow finite values, then it is true. On a finite dimensional space, a convex function $f$ is continuous on the interior of its proper domain. Since $\mathbb R^n$ is open, the image $f(C)$ must be connected for any connected $C\subseteq A$. Notice that convex sets are also connected. Now, the connected subsets of $\mathbb R$ are the intervals and thus are convex.


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