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

• AxiomOfChoice
In summary, to show that a sequence of functions converges to 0 in the L^1 norm, one must show that the integral of the absolute value of each function is less than any given epsilon for all n greater than a certain N. It is not possible for a sequence of functions to converge uniformly to 0 and not converge to 0 in the L^1 norm. The L^1 norm is defined as the integral of the absolute value of a function, and it is not always less than the supremum norm. An example of this is a triangle function with base 2n and height 1/n, which converges uniformly to 0 but does not converge to 0 in the L^1 norm. This
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!

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

hamster143 said:
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?

Anyone? Anyone?

AxiomOfChoice said:
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.]

Landau said:
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.]

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 $$\| f \|_1 = \infty$$.

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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.

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.

## 1. What does it mean for a sequence of functions to converge to zero in the L1 norm?

Convergence to zero in the L1 norm means that the values of the functions in the sequence approach zero as the index of the sequence increases. In other words, the functions become "closer" to the zero function as the sequence progresses.

## 2. What is the significance of the L1 norm in measuring convergence?

The L1 norm, also known as the "taxicab norm", is a measure of distance or "closeness" between functions. When a sequence of functions converges to zero in the L1 norm, it means that the functions are becoming increasingly similar or "close" to the zero function.

## 3. How is convergence in the L1 norm different from other types of convergence?

Convergence in the L1 norm is specific to the measure of distance or "closeness" between functions. Other types of convergence, such as pointwise convergence or uniform convergence, focus on the behavior of individual function values or the overall behavior of a function, respectively.

## 4. What are some applications of convergence to zero in the L1 norm?

Convergence in the L1 norm is commonly used in the analysis of numerical methods and algorithms, as well as in the study of Fourier series and other mathematical concepts. It also has applications in engineering, physics, and other scientific fields.

## 5. Are there any conditions that must be met for a sequence of functions to converge to zero in the L1 norm?

Yes, there are certain conditions that must be met for a sequence of functions to converge to zero in the L1 norm. For example, the functions in the sequence must be integrable and the integral of the absolute value of each function must approach zero as the index of the sequence increases. These conditions can vary depending on the specific context in which the convergence is being studied.

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