Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What does it mean for a sequence $f_n$ to converge to some function, say, $f$ in the $L^1$ norm?

Is it enough to show that $\int|f_n -f| \to 0$ or must one show as well that $f\in L^1$?

I am getting confused because I've encountered questions which asked to show that $f\in L^1$ and $\lim_{n\to \infty} \int |f_n-f| =0$. Does the latter imply the former?

share|improve this question

2 Answers 2

up vote 6 down vote accepted

Not unless you add the condition that the sequence $f_n$ is in $L^1.$ If so, then for all $n\in \mathbb{N}$

$$ \int |f| d\mu = \int |f-f_n + f_n| d\mu \leq \int |f-f_n| d\mu + \int |f_n| d\mu$$

which is finite.

If you do not require $f_n$ to be in $L^1$ then we can easily find a sequence such that $ \int |f-f_n| d\mu \to 0 $ but $f \notin L^1.$ Just pick your favorite non-integrable function $g$ and make $f_n = f = g.$

share|improve this answer
    
So if $f_n$'s are integrable, then it is enough to show that $\int|f_n-f| \to 0$?. What if the $f_n$'s are uniformly integrable? –  Josh May 7 '12 at 6:29
    
@Josh Indeed, if the $f_n$ are integrable, then the inequality I wrote above shows that if $ \int |f-f_n| d\mu \to 0$ then $f \in L^1$ as well. –  Ragib Zaman May 7 '12 at 6:31
    
This is kind of off topic. But does uniform integrability imply integrablity in the usual sense? –  Josh May 7 '12 at 6:39
    
If $\mathcal{H}$ is uniformly integrable then $\mathcal{H}\subseteq L^1$ if your measure $\mu$ is finite. I do not think it is the case when $\mu$ is not finite. –  Stefan Hansen May 7 '12 at 7:51
    
In every definition I am aware of, a uniformly integrable family is made of integrable functions, whether the measure is finite or not. –  Did May 7 '12 at 11:48

The answer of Ragib Zaman is correct. However, I think that a sentence like "$\{f_n\}_n$ converges to $f$ in $L^1$" means:

  1. $f_n \in L^1$ for every $n \in \mathbb{N}$;
  2. $\lim_{n \to +\infty} \int |f_n-f| =0$.

This is a reasonable approach, since, in any (say) metric space $X$, $x_n \to x$ is rather immaterial if $x_n \notin X$. The case of a sequence $u_n = f_n -f \in L^1$ with $f_n \notin L^1$ is really a trap for homework :-)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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