Convergence and Growth Rate of Lp Norms

[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} &lt; \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.

 

[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.

 

[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.

 

[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).