# About sequence of convex function

Let $(f_n)_{n\in\mathbb{N}}$ be convex $C^2$ functions such that $$f_n(x)\xrightarrow[n\to\infty]{}f(x)\in\mathbb{R}\ \forall x\in\mathbb{R}.$$ Do you think it's true that for Lebesgue-almost every $x\in\mathbb{R}$ the sequence $(f_n''(x))_{n\in\mathbb{N}}$ is bounded?

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No. For $n = 2^k + m$, $n,k,m$ integers with $0 \le m < 2^k$, let $g_n$ be continuous with $0 \le g_n(x) \le 1$, $g_n(x) = 1$ for $m 2^{-k} \le x \le (m+1) 2^{-k}$, $g_n(x) = 0$ for $x \le (m-1) 2^{-k}$ or $x \ge (m+1) 2^{-k}$. Let $f_n(x) = \int_0^x (x-s) g_n(s)\ ds$.
Thus $f_n'' = g_n$. Now $0 \le f_n(x) \le |x| \int_0^1 g_n(s)\ ds < 3 \cdot 2^{-k} |x| \le 6 |x|/n$, so $f_n \to 0$. But every $x \in [0,1]$ has $g_n(x) = 1$ for infinitely many $n$.
No, I don't think so. Take a sequence $h_n$ of nonnegative continuous functions on $[0,1]$ such that $h_n\to 0$ in $L^1$ norm, but for every $x$ the sequence $\{h_n(x)\}$ is unbounded. (Think of tall and thin trapezoids). Integrate twice to get $f_n$ such that $f_n''=h_n$ and $f_n(0)=f_n'(0)=0$. Since $h_n$ converge in $L^1$, you have uniform convergence of $f_n'$, and a forteriori of $f_n$.