# Absolutely Continuous Functions

1. Aug 27, 2010

### Matthollyw00d

Is the space of all absolutely continuous functions complete?
I've never learned about absolutely continuous functions, and so I'm unsure of their properties when working with them. I'm fairly certain it is, but would like some verification.

Much appreciated.

2. Aug 27, 2010

### mathman

It depends on what you use for the norm. For example (for simplicity I'll assume a finite domain) - if ||f||=max|f|, the answer would be no. On the other hand if ||f||=max|f|+max|f'|, the answer would be yes.

3. Aug 28, 2010

### Matthollyw00d

I was thinking with the induced norm from $$<f,g>=\Sigma_{k=1}^n \int^1_0 f^{(k)}(t)\overline{g^{(k)}(t)}dt$$

4. Aug 28, 2010

### mathman

I'm confused? How are f(k) related to f?

5. Aug 29, 2010

### Matthollyw00d

The kth derivatives.
I'm dealing with a subset of absolute continuous functions and trying to show completeness. It seemed easier to just show it was closed if I knew the space of all absolutely continuous functions were complete w.r.t the induced metric from the above inner product. But now I don't think we can talk about that space w.r.t the inner product, since absolute continuity gives nothing about the kth derivatives existing.

I guess I'm left with showing that an arbitrary Cauchy sequence converges.

6. Aug 29, 2010

### Landau

What is n? And indeed, absolutely continuous functions need not have k-th derivatives (for any k), so this inner product does not make any sense.

So back to start: what norm do you want to use (you can't talk about completeness without talking about a norm)?

7. Aug 29, 2010

### Matthollyw00d

Well I was dealing with a subset where it did make sense.

Let me just ask a different question:
If $$\{h_n\}$$ is a sequence in H and $$\Sigma^\infty_{n=1}||h_n||<\infty$$ and I show that $$\Sigma^\infty_{n=1}h_n$$ converges in H; does that imply that H is complete?

8. Aug 29, 2010

### Landau

Assuming that (h_n) is an arbitrary sequence: yes, one of the equivalent formulations of completeness of a normed space is "every absolutely convergent series converges".

\\edit: for a proof of this, see e.g. here.

Last edited: Aug 29, 2010