After this question : [Dominated convergence 2.0?](https://mathoverflow.net/questions/308605/dominated-convergence-2-0)


I want to know, what about the case when $h\in L^1([0,1])$.


The completed question :


Let $(f\_n)\_n$ be a sequence in $C^2([0,1])$ converging pointwise to $g \in L^1([0,1])$ and $\forall x \in [0,1], g(x)\in \mathbb R$. 


Assume that:


$\forall n\in\mathbb N, f\_n''<h$, where $h \in L^1([0,1])$. 


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



Dominated convergence 2.1?
I claim that under these assumptions, the functions $f\_n$ are uniformly bounded. Then the conclusion follows from the dominated convergence theorem.


First set $H(x) = \int\_0^x \int\_0^t h(s)\,ds$, which is $C^1$. Letting $u\_n = f\_n-H$, we have that $u\_n$ is concave (i.e. $-u\_n$ is [convex](https://en.wikipedia.org/wiki/Convex_function)), continuous, and still converges pointwise.


Now let $v\_n(x) = u\_n(x) - (x u\_n(1) + (1-x) u\_n(0))$. Now $v\_n$ is again concave and continuous, $v\_n(0)= v\_n(1)=0$, and $v\_n$ still converges pointwise (note that $u\_n(0), u\_n(1)$ both converge to finite limits). In particular we have $v\_n \ge 0$ everywhere.


Let $M\_n$ be the maximum value of $v\_n$, and let $x\_n$ be the point where it is attained. Suppose first that $x\_n \ge 1/2$. By concavity we have $v\_n(x) \ge \frac{x}{x\_n} v(x\_n) \ge x v(x\_n) = x M\_n$ for all $0 \le x \le x\_n$. In particular, we have $v\_n(1/2) \ge \frac{1}{2} M\_n$. If $x\_n \le 1/2$, we can get the same result by a similar argument (or by replacing $v\_n(x)$ with $v\_n(1-x)$).


So $M := \sup\_n M\_n \le 2 \sup\_n v\_n(1/2)$ which is finite because $v\_n$ converges pointwise. So we have $0 \le v\_n(x) \le M$ for all $x,n$. It follows easily that $f\_n$ is uniformly bounded as well (by, say, $M + \sup\_n |u\_n(0)| + \sup\_n |u\_n(1)| + \sup\_x |H(x)|$). 


