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. 
 A: (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. 
