Is $f$ a Convex Function? Let $\Omega\subset\mathbb{R}^n$ be a open convex set and $A=\{x_1,...,x_k\}\subset\Omega$ a finite set. Suppose $f:\Omega\rightarrow\mathbb{R}$ is a $C^1$ function in $\Omega\setminus A$ and continuous in $\Omega$ such that $$(f'(u)-f'(v))(u-v)\geq 0,\ \forall\ u,v\in\Omega\setminus A$$
Can I conclude that $f$ is convex?
Edit. It was missing the hypothesis of continuity.
 A: First: to show that a function defined on an open convex set in $\mathbb R^n$ is convex, it suffices to show that its restriction to any line is convex.  Therefore we may assume $n=1$.  
So: let $\Omega$ be an open interval in $\mathbb R$, let $A \subset \Omega$ be a finite set, let $f : \Omega \to \mathbb R$ be a function.  Assume $f$ is continuous on $\Omega$ and differentiable on $\Omega \setminus A$.  Assume: for $u,v \in \Omega \setminus A$, if $u \le v$ then $f'(u) \le f'(v)$.  We must show that $f$ is convex on $\Omega$.  That is: Let $a,b,c \in \Omega$, $a<b<c$.  Then
$$
f(b) \le \frac{c-b}{c-a}f(a) + \frac{b-a}{c-a}f(c) .
\tag{$*$}
$$
Now, if we prove ($*$) whenever $f$ is differentiable at $b$, it follows for general $b$ by continuity.  So let's assume $f$ is differentiable at $b$.
Claim 1.  If $u,v \in \Omega$ and $b \le u < v$, then
$$
f'(b) \le \frac{f(u)-f(v)}{u-v} .
\tag{1}$$
Claim 2.  If $u,v \in \Omega$ and $u < v \le b$, then
$$
\frac{f(u)-f(v)}{u-v} \le f'(b) .
\tag{2}$$
Note: Claims 1 and 2 are enough to finish the proof:  Indeed, from Claim 1 we have $f(c) \ge f(b)+(c-b) f'(b)$, and from Claim 2 we have $f(a) \ge f(b) - (b-a) f'(b)$.  Multiply the first inequality by $(b-a)$, the second inequality by $(c-b)$ and add to get
$$
(b-a)f(c)+(c-b)f(a) \ge (c-a)f(b),
$$
as required.
Proof for Claim 1.  First, if $(u,v) \cap A = \varnothing$, regardless of whether $u$ or $v$ belong to $A$, then $(f(v)-f(u))/(v-u)$ is $f'$ evaluated at some point  between $u$ and $v$ by the Mean Value Theorem.  So ($1$) is true.  For the general case: Let $A \cap (u,v) = \{x_1,x_2,\dots,x_n\}$ with $x_1 < x_2 < \dots < x_n$.  Then
$$
\begin{align}
(x_1-u)f'(b) &\le f(x_1)-f(u),
\\
(x_2-x_1)f'(b) &\le f(x_2)-f(x_1),
\\
(x_3-x_2)f'(b) &\le f(x_3)-f(x_2),
\\ ... \\
(x_n-x_{n-1})f'(b) &\le f(x_n)-f(x_{n-1}),
\\
(v-x_{n})f'(b) &\le f(v)-f(x_{n}).
\end{align}
$$
Add to get
$$
(v-u)f'(b) \le f(v)-f(u)
$$
as claimed.  
The proof for Claim 2 is similar.  
