After this question : 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 $\lim \int_0^1 f_n=\int_0^1 g$ ?

  • $\begingroup$ +1. Maybe it is worthwhile mentioning in the post that you assume $g$ to take values in $\mathbb{R}$ rather than in $[-\infty, \infty]$? $\endgroup$ – Jochen Glueck Aug 19 '18 at 18:49

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), 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, $M + \sup_n |u_n(0)| + \sup_n |u_n(1)| + \sup_x |H(x)|$).

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  • $\begingroup$ change the definition of $v_n$, the $t$ by $x$. I choose your proof because it completed, Pinelis use a corollary, without give a proof. @Nate Eldredge $\endgroup$ – Dattier Aug 20 '18 at 8:41
  • $\begingroup$ you should corriged your tipos $\endgroup$ – Dattier Aug 20 '18 at 8:47

The answer is yes. Indeed, let $u_n:=f_n-H$ and $v:=g-H$, where \begin{equation} H(x):=\int_0^x(x-t)h(t)\,dt \end{equation} for $x\in[0,1]$. Then $u_n\to v$ pointwise and $u_n$ is concave for each $n$ (see the Detail below). So, by Lemma 1 below, $u_n\to v$ in $L^1[0,1]$ and hence $\lim \int_0^1 f_n=\int_0^1 g$.

Detail: By Taylor's theorem with the integral form of the remainder and the definition of $H$, for $x\in[0,1]$,
\begin{equation} u_n(x)=f_n(x)-H(x)=f_n(0)+f'_n(0)x+\int_0^1(x-t)_+[f''_n(t)-h(t)]\,dt. \end{equation} Now the concavity of $u_n$ follows because $(x-t)_+$ is convex in $x$ and $f''_n<h$.

Lemma 1. Suppose that $f_n$ are convex real-valued functions on $[0,1]$ converging pointwise to a real-valued function $f$. Then $f_n\to f$ in $L^1[0,1]$.

Proof. The function $f$ is real-valued and convex and hence bounded from below. So, by Corollary 3, all the functions $f_n$ are uniformly bounded from below. On the other hand, all the convex functions $f_n$ are uniformly bounded from above by $\sup_n(f_n(0)\vee f_n(1))$. So, Lemma 1 follows by dominated convergence.

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  • $\begingroup$ why are, $u_n$ and $v$, monotone for the same piecewises ? @losif Pinelis $\endgroup$ – Dattier Aug 19 '18 at 19:55
  • $\begingroup$ @Dattier : see the edited version. $\endgroup$ – Iosif Pinelis Aug 19 '18 at 20:26
  • $\begingroup$ @NateEldredge : This is a good point. I'll think how to show that a sequence of concave functions converging pointwise on $[0,1]$ to a function in $L^1$ converges in $L^1$; I think this must be true. $\endgroup$ – Iosif Pinelis Aug 19 '18 at 20:49
  • $\begingroup$ @NateEldredge : Please see the latest edit. $\endgroup$ – Iosif Pinelis Aug 19 '18 at 21:14
  • $\begingroup$ @IosifPinelis: Yep, I think we basically found the same argument :-) $\endgroup$ – Nate Eldredge Aug 19 '18 at 21:16

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