Convergence of a sequence of functions to zero in the L1 norm?

[AxiomOfChoice]

I just want to make sure I'm straight on the definition.

Am I correct in assuming that, if I want to show that a sequence $\langle f_n \rangle$ of functions converges to 0 in the $L^1$ norm, I have to show that, for every $\epsilon > 0$, there exists $N \in \mathbb N$ such that

$$\int |f_n| < \epsilon$$

whenever $n > N$?

Also, is it possible for a sequence of functions to converge uniformly to 0 and yet *not* converge to 0 in the $L^1$ norm? (I'm pretty sure I have an example of this if the above definition is correct.)

Thanks!

 

[hamster143]

Yes, if by $$L^1$$ you mean $$L^1(-\infty;+\infty)$$.

 

[AxiomOfChoice]

Yes, if by $$L^1$$ you mean $$L^1(-\infty;+\infty)$$.

Thanks! Yes, I'm only asking about $L^1(\mathbb R)$.

And how about my second question concerning uniform convergence? Is your answer to that "yes" as well?

 

[AxiomOfChoice]

Anyone? Anyone?

 

[Landau]

Also, is it possible for a sequence of functions to converge uniformly to 0 and yet *not* converge to 0 in the $L^1$ norm?

I don't think so. We have $$\|...\|_1\leq \|...\|_\infty$$, so every sequence of functions converging in de supremum-norm converges in the 1-norm, with the same limit. [The converse is not true.]

So, what's your example?

 

[AxiomOfChoice]

I don't think so. We have $$\|...\|_1\leq \|...\|_\infty$$, so every sequence of functions converging in de supremum-norm converges in the 1-norm, with the same limit. [The converse is not true.]

So, what's your example?

Well, just consider a function that's a triangle with base $2n$ (centered at the origin) and height $1/n$. Then the integral of all functions in the sequence is always 1, so it can't possibly converge to zero in the $L^1$ norm, but it converges to 0 uniformly (right?) on $\mathbb R$ since

$$\sup_{x\in \mathbb R} |f_n| = 1/n \to 0 \text{ as } n\to \infty.$$

I may have gotten a definition wrong somewhere here, so please feel free to correct me if I have. To make sure we're on the same page, note that by the $L^1$ norm of $f$, I mean

$$||f||_1 = \int |f|.$$

Also, I'm not sure about the veracity of your claim that $$\|...\|_1\leq \|...\|_\infty$$. What about the constant function $$f = 1$$? We have $$\| f \|_\infty = 1$$, but $$\| f \|_1 = \infty$$.

 

[AxiomOfChoice]

I think the reason I'm confused about this is that I'm thinking about the fact from second-semester real analysis that if a sequence of functions converges uniformly to a function, the integrals converge to the integral. So I keep thinking the sequence of functions in my example should converge to 0 in the $$L^1$$ norm. But this is only on a bounded interval $$[a,b]$$, correct? My example is crap if we restrict ourselves to any bounded interval, but I think it holds water if we're allowed to consider the whole real line.

 

[Landau]

Hah, I was indeed working with $$\|f\|_1:=\int_0^1|f(t)dt$$. I should have read the first reply, I apologize.
So it looks like you understand it perfectly, you last statement about compact intervals is also correct.