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

<br /> \int |f_n| &lt; \epsilon<br />

whenever n &gt; 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

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

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

<br /> ||f||_1 = \int |f|.<br />

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.