7
$\begingroup$

I am struggling to understand what the space $L^1$ is, and what it means for a function to be $L^1$.

A friend told me that a function $f$ is $L^1$ if $\int_\mathbb{R} |f|$ is finite. It is $L^2$ if $(\int_\mathbb{R} |f|^2)^{1/2}$ Firstly is this correct? I have looked online but the definitions seem complicated. I'm not studying a Lebesgue integration course, and I just want a basic understanding of what these spaces are.

He explained the spaces as saying they contained functions that decay to zero, is this correct?

Could somebody provide a simple definition/intuitive explanation for me to view these spaces, and give me some examples of functions they contain(and better yet functions they don't contain).

I have never studied Lebesgue Integration, and am now taking a graduate Fourier Analysis course, so perhaps I should study it in some more detail, but for the moment I'd just like to understand a bit more about these spaces.

-----EDIT------- So for example are $sin$ and $cos$ $L^1$?

$\endgroup$
  • 3
    $\begingroup$ Your friend's definition misses out on some subtleties, but it's basically correct. I'm afraid a rigorous definition cannot avoid Lebesgue integration and measure theory. $\endgroup$ – Alex Becker Apr 8 '14 at 23:04
  • $\begingroup$ Yes, your friend is correct and nothing subtle is going on here once you know how to integrate. $\endgroup$ – hot_queen Apr 9 '14 at 3:14
9
$\begingroup$

Given a measurable space $X$ equipped with a measure $\mu$, a function $f : X \to \mathbb{C}$ which is defined almost everywhere (that is, up to a set of measure $0$) is said to be an element of $L^1$ if

$$\int_X |f| d\mu < \infty$$

More properly, we have to identify functions which are equal almost everywhere, so the elements of the Lebesgue space $L^1$ are really equivalence classes of functions under the relation of being almost everywhere equal - but this is a technical note.


Practically speaking, a real or complex valued measurable function on the real line with respect to Lebesgue measure is an element of $L^1$ if $$\int_{-\infty}^{\infty} |f(x)| dx < \infty$$ So a function like $\chi_{[0,1]}$ which is $1$ on $[0,1]$ and $0$ outside is in $L^1$, as is $e^{-x^2}$.

If $f$ is continuous enough, this coincides with the usual Riemann integral. Now it's fairly easy to prove that $$\int_{\mathbb{R}} |\sin x| dx$$ can't be finite, so $\sin x \notin L^1(\mathbb{R})$. In some sense, $L^1$ functions have to decay to $0$ at $\pm \infty$: In fact, one way to think of $L^1$ is that it's the completion of

$$C_C = \{\text{continuous functions supported on a compact set}\}$$

under the metric induced by integration (again, with slight technical caveats).


So in short, ignoring the technical definitions, $L^1$ functions are exactly those functions which have small enough spikes and decay fast enough that $\int |f| < \infty$.

$\endgroup$
6
$\begingroup$

An $L^1$ functional from a space $X$ to $\mathbb{R}$ is an $\mu$-measurable function such that $$ \int_{X} |f|\,d\mu < \infty. $$

$\endgroup$
  • $\begingroup$ Note that typically, we define $L^1$ to be the measurable functions $X \to \mathbb{C}$ satisfying the above condition. $\endgroup$ – Ink Apr 9 '14 at 0:52
  • 1
    $\begingroup$ With the caveat that we identify functions which are equal almost everywhere with respect to the given measure. $\endgroup$ – user61527 Apr 10 '14 at 1:02
1
$\begingroup$

We can, in fact, give a definition that doesn't require the Lebesgue measure or measure theory. We define a function $f: \mathbb{R} \to \mathbb{R}$ to be Lebesgue integrable ($f \in \mathcal{L}^1(\mathbb{R})$) iff there is a sequence $(f_n)_{n=1}^\infty$ of continuous, compactly supported functions $f_n \in C_c(\mathbb{R})$ such that the following two properties hold:

  1. $\displaystyle \sum_{n=1}^\infty \int_{\mathbb{R}} |f_n| < \infty$,
  2. If $x \in \mathbb{R}$ satisfies $\displaystyle \sum_{n=1}^\infty |f_n(x)| < \infty$, then $ \displaystyle\sum_{n=1}^\infty f_n(x) = f(x)$.

In the above, the integral $\int_{\mathbb{R}} |f_n|$ is the Riemann integral; it exists and is finite for all $f_n$ because $f_n \in C_c(\mathbb{R})$.

Here I've written $\mathcal{L}^1(\mathbb{R})$ instead of $L^1(\mathbb{R})$ because they're technically different. The former contains the Lebesgue integrable functions, and the latter treats Lebesgue integrable functions which are equal almost everywhere as equivalent.

The upshot is that in some sense, we can think of a function $f \in \mathcal{L}^1(\mathbb{R})$ as the limit of $C_c(\mathbb{R})$ functions which vanish at infinity in a way that restricts their (absolute) integrals over all of $\mathbb{R}$ to be finite, and hence $\int_\mathbb{R} |f|$ to be finite.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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