During my research, I came across the following question.


Let $(f\_n)\_n$ be a sequence in $C^2([0,1])$ converging pointwise to $g \in L^1([0,1])$. Assume that:


$\forall n\in\mathbb N, f\_n''<h$, where $h$ is locally integrable on $]0,1[$. 


Is it true that $\lim \int\_0^1 f\_n=\int\_0^1 g$ ?



Dominated convergence 2.0?
**Counterexample.** Let $f: \mathbb{R} \to \mathbb{R}$ denote your favourite test function with support in $(0,1)$ and with integral $1$. We define $f\_n(x) := n f(nx)$ for all $n \in \mathbb{N}$ and all $x \in [0,1]$.


Then $f\_n(x) \to 0$ as $n \to \infty$ for all $x \in [0,1]$ and $\int\_0^1 f\_n = 1$ for all $n$. If we define $M := \sup\_{x \in [0,1]} \lvert f''(x) \rvert$, then $\lvert f\_n''(x)\rvert \le n^3 M$ for each $x \in [0,1]$ and each index $n$. Since $f\_n$ is supported on $(0,1/n)$, it follows that
\begin{align\*}
 f\_n''(x) \le \lvert f\_n''(x) \rvert < M/x^3
\end{align\*}
for all $n \in \mathbb{N}$ and all $x \in (0,1)$.


**Remark.** I'd say this counterexample can be considered as an abstract version of the construction skechted by user37208 in the comments.


