# Rudin Lp question

1. Dec 2, 2007

### redrzewski

This is problem 3.9 (from second edition of daddy rudin). (This isn't homework).

Suppose f is Lebesgue measurable on (0,1), and f is not essentially bounded.

Is it true that for every function g on $$(0,\infty)$$ such that
$$g(p) \rightarrow \infty$$ as $$p \rightarrow \infty$$

one can find an f such that
$$\|f\|_{p} \rightarrow \infty$$ as $$p \rightarrow \infty$$ with$$\|f\|_{p} < g(p)$$ for all sufficiently large p?

Any hints are appreciated.

I know that given any function f, the set of p for which $$\|f\|_{p} < \infty$$ can be any connected subset of $$(0,\infty)$$.

Hence, given any r with 1 < r < $$\infty$$, we can find a fuction such that the $$\|f\|_{p} < \infty$$ if p < r but that the norm is infinite for p >= r.

But say we choose g(p)=p. The above mechanism fails to find a suitable f, since we can only choose some r finite, then g(p) < $$\|f\|_{p}$$ for all p > r.

On the other hand, I don't know how to prove that such a function f can't exist. But every function I can think of that meets the above criteria always has a finite inflection point where the norm goes infinite.

So I'm stumped.

Last edited: Dec 2, 2007
2. Dec 2, 2007

### Xevarion

I think this should be true. You just need to make $$f$$ have a singularity that is really 'weak'. For example, you could have the function look like $$\frac{1}{\log(x)}$$. That one is in $$L^p$$ for all large $$p$$ but the $$L_p$$ norm is growing as a function of $$p$$. Of course this isn't a full solution to the problem but it should give you some idea of what's going on.. I hope it helps.

3. Dec 3, 2007

### redrzewski

I think I'm getting closer. The hint was actually in a different problem.

I think the function i need is like:

$$\log(\frac{1}\sqrt(x))$$

I'll keep plugging away.

4. Dec 3, 2007

### Xevarion

Well, you won't get a specific function. The idea is that given any $$g$$ that specifies an asymptotic growth condition, you can find a function whose $$L^p$$ norms grow even slower than $$g$$. You really don't need to construct an explicit one (it might be extremely hard if $$g$$ is really complicated).