Limit of continuous convex functions. Let $\left( \, f_n(x)\, \right)$, $x \in \mathbb{R}$, be convex continuous increasing functions. Let
$$f(x) := \lim\limits_{n \rightarrow \infty} f_n(x)$$
and assume that the limit exists for every $x$.
Is $f(x)$ convex, continuous and non-decreasing?
 A: If $(a_n)$ is a sequence of non negative reals converging to $a$, $a$ is non negative.
Regarding monotonicity
Hence if $(f_n)$ is a sequence of non decreasing functions, the limit $f$ is non decreasing as you have for $x < y$ $f_n(y)-f_n(x) \ge 0$ and therefore $f(y) -f(x) \ge 0$. However a limit of strictly increasing functions might be only non decreasing. Consider for example $f_n(x)=\frac{x}{n}$.
Regarding convexity
Hence if $(f_n)$ is a sequence of convex functions, the limit $f$ is also as you have for $x < y$ and $\lambda>0$, $(1-\lambda)f_n(x)+\lambda f_n(x) -f_n((1-\lambda)x+\lambda y) \ge 0$ and therefore $(1-\lambda)f(x)+\lambda f(x) -f((1-\lambda)x+\lambda y) \ge 0$. However a limit of strictly convex functions might not be stricly convex. Consider for example $f_n(x)=\frac{x^2}{n}$.
Regarding continuity
The sequence of functions $f_n(x) = x^n$ is a sequence of continuous functions from $[0,1]$ to $\mathbb R$, whose pointwise limit $f$ is not continuous. You have $f(x)=0$ for $x \in [0,1)$ and $f(1)=1$. You need uniform convergence to ensure the continuity of the limit.
