5
$\begingroup$

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 $\lim \int_0^1 f_n=\int_0^1 g$ ?

$\endgroup$
  • $\begingroup$ +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
3
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.