# Continuously smooth functions and Lp space

1. May 14, 2012

### sdickey9480

How might I prove the following?

1) If f ∈ C(Rn) and f has compact support, then f ∈ Lp(Rn) for every 1 ≤ p ≤ ∞.

2) If f ∈ C(Rn), then f ∈ Lp_{loc}(Rn) for every 1 ≤ p < ∞.

(Where C(Rn) is the space of continuous functions on Rn)

2. May 15, 2012

### theorem4.5.9

What is special about continuous functions on compact domains? (apply this to both parts)

3. May 15, 2012

### sdickey9480

They are uniformly continuous

4. May 15, 2012

### micromass

Staff Emeritus
Yes, but what else?? Can functions on compact domains grow arbitrarly large??

5. May 15, 2012

### sdickey9480

No, b/c they are bounded.

6. May 15, 2012

### micromass

Staff Emeritus
Yes. So use that to find an estimate for the integral

$$\int_{\mathbb{R}^n} |f|^p$$

7. May 15, 2012

### sdickey9480

Not following. Could you provide a little more detail?

8. May 15, 2012

### sdickey9480

So since we are dealing with bounded continuous functions, by finding an estimate to the aforementioned integral this will in turn justify that f must also belong to Lp?

9. May 15, 2012

### micromass

Staff Emeritus
If we can show that

$$\int_{\mathbb{R}^n} |f|^p$$

is not infinite, then the function is in $L^p$. So we must find some real number C such that

$$\int_{\mathbb{R}^n} |f|^p\leq C$$

10. May 15, 2012

### Office_Shredder

Staff Emeritus
It might help to first answer the (hopefully easy) question

If f(x) is a continuous function on the reals, is $$\int_a^b |f(x)|^p dx$$ ever infinite?

If you can figure out the answer to this you can solve micromass's question

11. May 16, 2012

### sdickey9480

Won't the integral always be finite? Hence do we even need to find a particular C, M, etc.? Can't we just assume there exists one, again b/c integral is finite? Am I using this same idea for 1) and 2)?

12. May 16, 2012

### micromass

Staff Emeritus
Why do you think the integral will always be finite?? Where did you use compactness??

13. May 16, 2012

### sdickey9480

Finite because it's bounded on a compact interval?

14. May 16, 2012

### sdickey9480

Can I just prove the space of smooth continuous functions is dense in Lp, hence if a function belongs to C(R) it belongs to Lp(R). If so, what's the difference in the proof of 1) & 2)?