This is actually an exercise in Rudin’s

Real and Complex Analysis, L^p spaces chapter. Could anyone help me out? Thanks in advance.

Motivation:It’s well known that if we have a function f which belongs to L^p(0,1) for all p\ge 1. Then \lim_{p\rightarrow \infty}\|f\|_p=\|f\|_{\infty} (moreover, \|f\|_p is increasing in p). This is true even if \|f\|_{\infty}=\infty.

Question:How slow (fast) can \|f\|_p grow when \|f\|_{\infty}=\infty? More precisely, given any positive increasing function \Phi with \lim_{p\rightarrow \infty}\Phi(p)=\infty, can we always find a function f which belongs to L^p(0,1) for all p\ge 1, and \|f\|_{\infty}=\infty, such that \|f\|_p\le (\ge)\Phi(p) for large p?

**Answer**

I’ll answer the question of whether \|f\|_p can grow arbitrarily slowly; the answer is yes. I’m fairly certain that it can grow arbitrarily quickly as well, and, though I haven’t given it much thought, I suspect a similar argument can be concocted. The problem doesn’t seem to rely on the interval (0,1), so what I write below doesn’t either. If you really want to put everything in (0,1), it is not hard to do so.

Let \Phi be as you say. Here’s the idea: We choose disjoint sets E_n with positive (but as yet undetermined) measure, and we require that f=\sum_{n=1}^{\infty} c_n\chi_{E_n}, where \{c_n\} is some sequence of positive numbers increasing to infinity, and \chi_{E_n} is the characteristic function of E_n. This ensures that f\notin L^{\infty}. Assuming (as we may) that \Phi(p) is bounded away from 0, we see that the quotient c_n^p/\Phi(p)^p is bounded in p for each fixed n. So we are free to choose our sets E_n so small in measure that (c_n/\Phi(p))^pm(E_n)<2^{-n}, independently of p. To conclude, we simply observe that

\begin{align*}

\frac{\|f\|_p^p}{\Phi(p)^p}

&=

\frac{1}{\Phi(p)^p}\int\sum_{n=1}^{\infty}c_n^p\chi_{E_n}\\

&=

\sum_{n=1}^{\infty}\frac{c_n^p}{\Phi(p)^p}m(E_n)\\

&<

\sum_{n=1}^{\infty}2^{-n}=1.

\end{align*}

This means \|f\|_p<\Phi(p) for all p (or, if you like, for all p\in[a,\infty), where \Phi(p) is bounded away from 0 on [a,\infty)).

**Attribution***Source : Link , Question Author : Syang Chen , Answer Author : Nick Strehlke*