The Unapologetic Mathematician

Mathematics for the interested outsider

The L¹ Norm

We can now introduce a norm to our space of integrable simple functions, making it into a normed vector space. We define

\displaystyle\lVert f\rVert_1=\int\lvert f\rvert\,d\mu

Don’t worry about that little 1 dangling off of the norm, or why we call this the “L^1 norm”. That will become clear later when we generalize.

We can easily verify that \lVert cf\rVert_1=\lvert c\rvert\lVert f\rVert_1 and that \lVert f+g\rVert_1\leq\lVert f\rVert_1+\lVert g\rVert_1, using our properties of integrals. The catch is that \lVert f\rVert_1=0 doesn’t imply that f is identically zero, but only that f=0 almost everywhere. But really throughout our treatment of integration we’re considering two functions that are equal a.e. to be equivalent, and so this isn’t really a problem — \lVert f\rVert_1=0 implies that f is equivalent to the constant zero function for our purposes.

Of course, a norm gives rise to a metric:

\displaystyle\rho(f,g)=\lVert g-f\rVert_1=\int\lvert g-f\rvert\,d\mu

and this gives us a topology on the space of integrable simple functions. And with a topology comes a notion of convergence!

We say that a sequence \{f_n\} of integrable functions is “Cauchy in the mean” or is “mean Cauchy” if \lVert f_n-f_m\rVert_1\to0 as m and n get arbitrarily large. We won’t talk quite yet about convergence because our situation is sort of like the one with rational numbers; we have a sense of when functions are getting close to each other, but most of these mean Cauchy sequences actually don’t converge within our space. That is, the normed vector space is not a Banach space.

However we can say some things about this notion of convergence. For one, a sequence \{f_n\} that is Cauchy in the mean is Cauchy in measure as well. Indeed, for any \epsilon>0 we can define the sets

\displaystyle E_{mn}=\left\{x\in X\big\vert\lvert f_n(x)-f_m(x)\rvert\geq\epsilon\right\}

And then we find that

\displaystyle\lVert f_n-f_m\rVert=\int\lvert f_n-f_m\rvert\,d\mu\geq\int\limits_{E_mn}\lvert f_n-f_m\rvert\,d\mu\geq\epsilon\mu(E_mn)

As m and n get arbitrarily large, the fact that the sequence is mean Cauchy tells us that the left hand side of this inequality gets pushed down to zero, and so the right hand side must as well.

This notion of convergence will play a major role in our study of integration.

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May 28, 2010 - Posted by | Analysis, Functional Analysis, Measure Theory

11 Comments »

  1. ### That is, the normed vector space is not a Banach space. ###

    But L^1 is certainly a Banach space…

    Comment by Cristi | May 29, 2010 | Reply

  2. Be careful, Cristi: like I said, the situation is sort of like with the rational numbers, which are not complete. The space we have ahold of now is not the full space L^1(X), but only the space of integrable simple functions.

    It’s easy to construct a sequence of integrable simple functions which is mean Cauchy, but which do not converge in the mean to an integrable simple function.

    Comment by John Armstrong | May 29, 2010 | Reply

  3. You’re right, of course. But why not define the norm for all good functions (like i first thought you did)?

    Comment by Cristi | May 31, 2010 | Reply

  4. That’s exactly the point of Lebesgue’s approach to integration, Cristi: first you integrate simple functions, and then you approach other functions as limits of simple functions. You can’t define the L^1 norm on “integrable functions” until you know what an “integrable function” is.

    Comment by John Armstrong | May 31, 2010 | Reply

  5. Yes, but you can define the integral of simple functions, then the integral of positive/integrable functions, and derive the properties of all integrals directly, instead of spending too much time on simple functions only. This is just my taste (based on Rudin), maybe you want to take a more pedagogical approach.

    Comment by Cristi | June 1, 2010 | Reply

  6. Okay, Cristi, so sketch it out: how do you go from simple integrable to positive integrable functions? Make sure that you know the integral of a positive integrable function doesn’t depend on which sequence of simple integrable functions you pick to converge to it.

    Comment by John Armstrong | June 1, 2010 | Reply

  7. You define the integral of a positive measurable function f to be the supremum of the integrals of those simple functions that are at most f. Then you prove the Monotone Convergence Theorem. This is how Rudin does it in the first chapter of R&C and it’s very quick. But not everybody likes his style.

    Comment by Cristi | June 1, 2010 | Reply

  8. One thing I’ve wondered is why we do not define the measure of any non-negative f as the measure of { (x,y) : y <= f(x) } in the R^2 product measure.

    Comment by Tom | June 19, 2010 | Reply

  9. Well, Tom, for one thing we haven’t talked about multiple Lebesgue integrals and the analogue of Fubini’s theorem. In practice, once we do it should work out to be just the measure you refer to.

    Comment by John Armstrong | June 19, 2010 | Reply

  10. Hi John, I understand that you want to take a particular order, but is there a big drawback to my suggestion which is why all authors (that I know of) avoid it?

    Comment by Tom Ellis | June 19, 2010 | Reply

  11. It’s possible to define the L^1 norm like that, yes, but then you basically have to do integrals anyway. It doesn’t really simplify anything.

    Comment by John Armstrong | June 19, 2010 | Reply


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