# Is $f(C)$ a convex set if $f$ and $C$ are convex?

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!

• 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. Sep 1, 2016 at 22:51
• @Tom in the question, $f$ is supposed to be scalar valued. Sep 2, 2016 at 5:57
• @Tom the epigraph is a convex set, not the graph. Sep 2, 2016 at 15:38
• @MichaelGrant the question isn't about the epigraph, but the graph. Sep 2, 2016 at 16:58
• Ah, then really, I should be directing that comment to the OP :-) Sep 2, 2016 at 16:58

## 1 Answer

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.