Suppose $f_n:X\subseteq\mathbb{R}^d\to\mathbb{R}$ is a convex function where $X$ is an open convex set. If $f_n$ converges to $f$ uniformly on compact sets, is $f$ a convex function?
I know that $f$ is a convex function on all compact subsets of $X$ but I'm not sure if it is on all $X$. So here's an attempt. Suppose, by contradiction, that there exists $x_1,x_2\in X$ and $t\in [0,1]$ such that $$ f(tx_1 + (1-t)x_2) > tf(x_1) + (1-t)f(x_2). $$ However, there is a compact set containing $x_1,x_2$ and so the above inequality contradicts the fact that $f$ is convex on all compact sets. Hence, there cannot exist such an $x_1,x_2$ $\implies f$ is convex on $X$. Are there any issues with my proof?
As already remarked by max_zorn and copper.hat, pointwise convergence is sufficient: if $(f_n)_n$ is a sequence of functions which are all convex in $X$ then also the pointwise limit $f$ is convex in $X$. In fact, for all $n\in\mathbb{N}$, for all $x_1,x_2\in X$, and for all $t\in [0,1]$ we have that $$ f_n(tx_1 + (1-t)x_2)\leq tf_n(x_1) + (1-t)f_n(x_2)$$ and after taking the limit as $n$ goes to infinity we obtain $$f(tx_1 + (1-t)x_2)\leq tf(x_1) + (1-t)f(x_2).$$ Notice that strict convexity is not preserved even for uniform convergence. Take for example the sequence $(x^n)_{n\geq 2}$ which are strict convex in $X=[0,1/2]$, the uniform limit in $X$ is the constant function $0$ which is not strict convex.