5
$\begingroup$

For practice, I'm working through some of the exercises in Folland's "Real Analysis: Modern Techinques and Their Applications."

In Chapter 2, Exercise 19, Folland asks for sequences of functions $f_n \in L^1(\mathbb{R})$ with $f_n \to f$ uniformly, but such that one of the conclusions $f \in L^1(\mathbb{R})$ or $\int f_n \to \int f$ fails. I can find examples of each conclusion failing, but cannot seem to find a single example where both conclusions fail in the following way:

What is an example of a sequence of functions $f_n \in L^1(\mathbb{R})$ with $f_n \to f$ uniformly, but such that $\int f = \pm \infty$ (by which I mean that exactly one of $\int f^+$ or $\int f^-$ is $+\infty$) and also $\int f_n \not \to \int f$?

My examples:

(1) $f_n(x) = \frac{1}{x}\chi_{(1,n)}(x)$. The uniform limit $f(x) = \frac{1}{x}\chi_{(1,\infty)}(x)$ has $\int f = +\infty$. However, we also have $\int f_n = \log(n) \to \infty = \int f$.

(2) $f_n(x) = \frac{1}{n}\chi_{(0,n)}(x)$. We have $\int f_n = 1 \not \to \int f = 0$, but now $\int f = 0 \neq \pm \infty$.

I feel like I'm missing something very obvious here. Thanks for your help.

  • 0
    I don't think you can have it like in the edit: If $\int f_n$ do not tend to $\infty$ (say), it means that the negative parts of the $f_n$ are compensating for the positive parts, which would be in direct conflict with the condition that \int f^-<\infty. I can't make this a proof at the moment, but I believe David Mitra's example is the best you can hope for.2012-12-09

1 Answers 1

7

You just need to modify your first example:

Take $f(x)={1\over x}\cdot\chi_{[1,\infty)}$ and for $n$ a positive integer, define $f_n(x)= {1\over x}\cdot\chi_{[1,n]} + {-1\over n}\cdot\chi_{(n, n+ n\ln n )}$.

Then

$\ \ \ 1)\ \int_{\Bbb R} f=\infty$,

$\ \ \ 2)\ (f_n)$ converges to $f$ uniformly, since $\Vert f_n-f\Vert_\infty=2/n$,

and

$\ \ \ 3)\ $for each $n$ we have $\int_{\Bbb R} f_n=0$.