If $\lim_{i \rightarrow \infty}f_i(x)=g(x)$ and $f_i'' \geq 0$, prove that $g''\geq0$ Let $(f_i)_{i=1}^\infty$ and $g$ be twice-differentiable real-valued functions on $\mathbb{R}$, with $f_i'' \geq 0$. Suppose that
$$\lim_{i \rightarrow \infty}f_i(x)=g(x)$$
for all $x\in \mathbb{R}$. Show that $g''\geq0$.
No clue here...wrote down the definition for pointwise convergence and initial thought is to invoke some sort of contradiction. Something like, assume for the sake of contradiction that $g''<0$, so then there exists some negative number $a$ such that $g''<a$. But not sure how to tie $f_i(x) \rightarrow g(x)$ to $f''_i(x) \rightarrow g''(x)$. Thoughts/Suggestions?
 A: For a twice-differentiable function $f$, $f''\ge 0$ if and only if $f$ is convex. Therefore it is sufficient to check that $g$ is convex, but that is easy: just take a limit from the inequality
$$f_n(\lambda x + (1-\lambda)y)\le \lambda f_n(x) + (1-\lambda) f_n(y)$$
for $0\le \lambda \le 1$.
A: Consider that for $x\in \mathbb{R}$, \begin{align*} g''(x) &= \lim_{h\to 0} \frac{g(x+h)+g(x-h)-2g(x)}{h^2} \\ &= \lim_{h\to 0} \lim_{n\to \infty} \frac{f_n(x+h)+f_n(x-h)-2f_n(x)}{h^2} \end{align*} As $f_n''\geq 0$, we have that $f_n$ is convex for all $n$. Therefore, $$f_n(tx_1+(1-t)x_2)\leq tf_n(x_1)+(1-t)f_n(x_2)$$ for all $x_1, x_2\in \mathbb{R}$ and $t\in [0, 1]$. If we let $x_1 = x-h$, $x_2 = x+h$, and $t = \frac{1}{2}$, we therefore get $$f_n(x)\leq \frac{1}{2}(f_n(x+h)+f_n(x-h))$$ This implies that $$\frac{f_n(x+h)+f_n(x-h)-2f_n(x)}{h^2}\geq 0$$ for all $x\in \mathbb{R}$, $h > 0$, and $n\in \mathbb{N}$. Therefore, $$\lim_{n\to \infty} \frac{f_n(x+h)+f_n(x-h)-2f_n(x)}{h^2}\geq 0$$ for all $x\in \mathbb{R}$ and $h > 0$, which implies that $$g''(x) = \lim_{h\to 0} \lim_{n\to \infty} \frac{f_n(x+h)+f_n(x-h)-2f_n(x)}{h^2}\geq 0$$ for all $x\in \mathbb{R}$.
A: First, note that it is important that we are given that $g''$ exists in the first place. Otherwise one readily finds $f_i$ that converge to $g(x)=|x|$, for example.
Assume $g''(x_0)<0$. Then $g'$ is strictly decreasing in a neighbourhood of $x_0$ and we find $x_1<x_0<x_2$ such that the straight sekant from $(x_1,g(x_1))$ to $(x_2,g(x_2))$ runs below the point $(x_0,g(x_0))$. Then for $i\gg 0$, the sekant from $(x_1,f_i(x_1))$ to $(x_2,f_i(x_2))$ runs below the point $(x_0,f_i(x_0))$, contradicting convexity of $f_i$.
