TITLE: How slow/fast can $L^p$ norm grow?
QUESTION [34 upvotes]: 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 $p$?

REPLY [9 votes]: Here's a similar solution for the case where we want the norm to grow arbitrarily fast. As mentioned in my comment, we must assume that $\Phi$ obtains a finite maximum on finite intervals (since the $p-$th norm is an increasing function in $p$).
Let $$C_n = \max(2^n, \max_{p \in [n,n+1]} \Phi(p)).$$ We construct a function $f$ such that $C_n \leq \|f\|_n < \infty$ for all natural $n$. The construction is of the general form $$ f = \sum_{n \geq 1} a_i \chi_{B_i}, \quad \mu(B_i) \triangleq \mu_i, \quad \sum_{i \geq 1} \mu_i < \infty. $$ We define $$ a_n = C_n^{n+1}, \mu_n = C_n^{-n^2}. $$ Since $C_n \geq 2^n$, the series $\sum_{i \geq 1} \mu_i$ clearly converges.
On the one hand, for every $n$ we have
$$ \| f \|_n^n \geq a_n^n \mu_n = C_n^n. $$
On the other hand, for every $n$ we have
$$ \| f \|_n^n = \sum_{i = 1}^n a_n^n \mu_n + \sum_{i \geq n+1} a_i^n \mu_i. $$
In order to show that the norm exists, it is enough to show that the latter series converges:
$$ \sum_{i \geq n+1} a_i^n \mu_i = \sum_{i \geq n+1} C_i^{n(i+1) - i^2} \leq \sum_{i \geq n+1} C_i^{-1} < \infty. $$
We used $C_i \geq 1$ and $n(i+1) - i^2 \leq -1$, which follows from $n(n+2) - (n+1)^2 = -1$ and the fact that $n(i+1) - i^2$ is decreasing for $i \geq n/2$ (calculus).