7
$\begingroup$

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

Improve this question
$\endgroup$
16
  • $\begingroup$ Did you not mean $|f_n''|<h$? $\endgroup$
    – GH from MO
    Commented Aug 18, 2018 at 16:32
  • 1
    $\begingroup$ No, without absolute value $\endgroup$
    – Dattier
    Commented Aug 18, 2018 at 16:32
  • $\begingroup$ Do you mind to share what you have tried? +1 for the question.. I would really love to see what you have tried.. $\endgroup$
    – Praphulla Koushik
    Commented Aug 18, 2018 at 16:46
  • 2
    $\begingroup$ Letting $h$ be only locally integrable seems like not enough to rule out garden-variety counterexamples, such as smoothed-out triangles of base $1/n$ and height $n$ centered at $x=1/n$. $\sup f_n''$ will blow up like $n^3$, but you can still bound $f_n''$ by an $h(x)$ proportional to $1/x^3$ which is locally integrable on $(0,1)$ but obviously not integrable. $\endgroup$
    – user101142
    Commented Aug 18, 2018 at 17:36
  • 1
    $\begingroup$ @PraphullaKoushik: Every question should have a high-level tag like ca, nt, etc. (the available tags are the same as in the arXiv). For this question, the best high-level tag is ca. It is a question belonging to analysis. $\endgroup$
    – GH from MO
    Commented Aug 18, 2018 at 18:13

1 Answer 1

Reset to default
8
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ $f_n(0)=n\times f(0) $ and $f(0)>0$ so $\lim f_n(0)=+\infty$, no ? $\endgroup$
    – Dattier
    Commented Aug 18, 2018 at 23:10
  • $\begingroup$ @Dattier: We have $f(0) = 0$ since $f$ is supported in the open interval $(0,1)$. $\endgroup$
    – Jochen Glueck
    Commented Aug 18, 2018 at 23:11
  • $\begingroup$ what about the case where $h \in L^1([0,1])$ : @Jochen Glueck $\endgroup$
    – Dattier
    Commented Aug 19, 2018 at 14:43
  • $\begingroup$ @Dattier: I'm not sure about this case, unfortunately. Under the stronger assumptions that $\lvert f_n''\rvert \le h$ for $h \in L^1$ and that $(f_n'(0))$ be bounded, we can apply the dominated convergence theorem. If, however, only $f_n'' \le h$ for $h \in L^1$ (or if nothing is assumed on $(f_n'(0))$), then I don't know what happens. Maybe this would be appropriate for a new question? $\endgroup$
    – Jochen Glueck
    Commented Aug 19, 2018 at 17:04
  • 1
    $\begingroup$ here, the new question : mathoverflow.net/questions/308686/dominated-convergence-2-1 $\endgroup$
    – Dattier
    Commented Aug 19, 2018 at 18:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .