# The Unapologetic Mathematician

## The L¹ Norm

May 28, 2010 -

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