# Extension of convex function to boundary

Let $C\subseteq\mathbb{R}^n$ be an (open) convex set and suppose $f:C\rightarrow \mathbb{R}$ is a convex function. We can extend $f$ to the boundary as follows: For $x\in\partial C$, define $$\tag{1} f(x)=\left\{\begin{array}{ll} \lim_{x'\rightarrow x} f(x') & \text{if the limit exists}\\ +\infty & \text{otherwise.}\end{array}\right.$$ So we have an extend real-valued function $f:\overline{C}\rightarrow \mathbb{R}\cup\{+\infty\}$. This extension is not necessarily continuous even if $f$ is bounded, e.g. see here.

However, my question is the following: Is this extension necessarily convex?

The function in (1) is defined by extending the function continuously if the limit exists. If the extended function were not convex, then there would exist points $x,y\in\partial C$ on the boundary and $t\in(0,1)$ such that $$f(tx+(1-t)y)>tf(x)+(1-t)f(y)$$ This would necessarily mean that $f(x),f(y)<\infty$ but $f(tx+(1-t)y)=\infty$, but I don't think this can ever happen.

I believe that this extended function must indeed be convex, but I cannot prove it.

(1) Assume that $C$ is not open, but convex in $\mathbb{R}^2$.

Define $$C :=C_1\cup C_2,\ C_1:=\{ (x,y)| y>0 \} ,\ C_2:= \{ (x,0) |x<0 \}$$

Define $f|C_1=0$ and $f|C_2=1$.

Clearly $f((x,0))=0$ for all $x>0$ so that $f|\{ y=0\}$ is not convex.

(2) Assume that $C$ is open

Partial result : The case where $C$ is one-dimension is trivial. Assume that $C$ is two-dimensional.

To show that extension of $f$ is convex, the following example is considered. So it will be proved that the case arising in OP can not happen.

Assume that $C=\{ (x,y)| y>0\}$ in $\mathbb{R}^2$, $$(x_{n},\frac{1}{n})\rightarrow (0,0),\ a_n:=(x_{2n},\frac{1}{2n} ),\ \lim_n\ f(a_n)=a,$$ $$b_n:= (x_{2n+1},\frac{1}{2n+1}),\ \lim_n\ f (b_n)=b <a$$ and $f(\pm 1,0)$ is finite.

Step 1 : WLOG $$a_k\in {\rm conv}\ \{ (-1,\epsilon_1),\ (-1,\epsilon_2),\ b_l,\ b_m\}$$ for some $k,\ l,\ m$ where $\epsilon_i>0$ are small.

Consider $p\in [(-1,\epsilon_1)b_m]$ s.t. $a_k\in [pb_l]$

Here indexing is not important. And $[xy]$ is a line segment between $x$ and $y$. Consider four points whose convex hull is 2-dimensional 4-gon and contains $a_k$. Consider a ray starting point at $b_l$ passing through $a_k$. Hence the ray meets a side of the 4-gon. So $[(-1,\epsilon_1) b_m]$ is denoted as the side and $p$ is the intersection point between ray and the side.

Step 2 : Let $$s=d(a_k,b_l),\ t=d(a_k,p),\ u=d((-1,\epsilon_1),p),\ v= d(p, b_m)$$

So $$f(p) \geq \frac{(t+s)f(a_k)-tf(b_l) }{s}$$ and $$\frac{vf(-1,\epsilon_1)+u f(b_m) }{u+v} \geq f(p)$$

That is $$f(-1,\epsilon_1) \geq \frac{1}{v} \{ (u+v)\frac{t}{s} [f(a_k)-f(b_l)] + v f(a_k) +u [f(a_k)-f(b_m) ] \}$$

If $k,\ l,\ m$ is large, it can be assumed that $s,\ d(b_l,b_m)<\delta$. If $t=Cs$, then $$\bigg(\frac{1}{v} (u+v)\frac{t}{s} +\frac{u}{v} \bigg)[f(a_k)-f(b_l)] \approx \frac{u+v}{C\delta+2\delta} C (a-b) +\frac{u}{C\delta+2\delta} (a-b)$$

So $f(-1,0)=\infty$ and it is a contradiction.

[Add] I just suggests the some case However it rule out other cases

If $f(0,0)=\infty$, there are sequences :

1) $a_n\rightarrow (0,0)$ s.t. $f(a_n)\rightarrow \infty$

2) $a_n,\ b_n\rightarrow (0,0)$ and $f(a_n),\ f(b_n) \rightarrow a,\ b,\ a>b$

3) $a_n \rightarrow (0,0)$ and $f(a_n)\rightarrow -\infty$

In the first case, if $a_n=(x_n,y_n)$, then $\max\ \{ f(-1,y_n), f(1,y_n) \} =\infty$

In the second case, if $b_n=(x_n,y_n)$, then we can assume that $$y_n>y_{n+1}$$

Then it goes down strictly. And define $$R_n^\pm :={\rm conv}\ \{ (x_n,y_n), ( x_{n+1},y_{n+1} ), (\pm 1,y_n),(\pm 1,y_{n+1}) \}$$

Each $R_{n}^\pm$ is a closed 2-dimensional 4-gon. In further $$\bigcup_n\ R_n^+ \cup \bigcup_n\ R_n^-$$ contains ${\rm conv}\ \{ (-1,y_1),(1,y_1),(-1,\epsilon ),(1,\epsilon )\}$ for any $\epsilon >0$

By reindexing, we can assume that $a_n$ is in $R_n^-$. If $a_n$ is in the interior of $R_n^-$, then the above argument holds

If $a_n\in [b_n (-1,y_n)]$, then note that $$\frac{ f(-1,y_n) |a_n-b_n| + f(b_n) |(-1,y_n)-a_n| }{|-1-x_n|} \geq f(a_n)$$

So if $\lim_n\ f(-1,y_n) =C$ is finite, then $C\cdot \lim_n\ |a_n-b_n| + b \geq a$. Hence we have a contradiction.

If $a_n \in [b_nb_{n+1}]$, then note that $\max\ \{ f(b_n),f(b_{n+1}) \} <\frac{a+b}{2}< f(a_n)$ for sufficiently large $n$. It contradicts to convexity of $f$.

In third case, assume that $a_n=(x_n,y_n),\ f(a_n) <-n$ Then $f(-1,y_n)\rightarrow a$ Then $$\frac{ f(-1- \epsilon,y_n)\cdot |-1-x_n| + f(a_n)\cdot \epsilon }{|-1-x_n|+\epsilon } \geq f(-1,y_n)$$

If $\epsilon = \frac{-1}{f(a_n)}$, then we have $a-1\geq a$ by limiting. So it is a contradiction.

• I don't get your first example. It looks like you are trying to show what can go wrong if $C$ is not open. But, convex functions must be defined over convex sets, and your set $C$ is not convex. – Michael Apr 4 '17 at 22:32
• @ Michael The $C$ in the first example is indeed convex. It is the open upper half plane with the positive $x$ axis added in. – luftbahnfahrer Apr 5 '17 at 1:48
• @luftbahnfahrer : I add more explanation – HK Lee Apr 5 '17 at 1:56
• I'm trying to understand your proof, and I think it boils down to this: We can suppose that $C$ is the open upper-half plane. Suppose the continuous limits exist at $(\pm1,0)$, but not at $(0,0)$. Then find two different sequences $\{a_n\}$ and $\{b_n\}$ in $\mathbb{R}^2$ both converging to $(0,0)$ such that $f(a_n)\rightarrow a$ and $f(b_n)\rightarrow b$ with $b<a$. Then you show that $f(-1,0)=\infty$ (which I'm still trying to understand), which is a contradiction. – luftbahnfahrer Apr 5 '17 at 16:18
• I add explanation about other cases. – HK Lee Apr 5 '17 at 21:13