How slow/fast can $L^p$ norm grow? 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 $p$?
 A: 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).
A: 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 $[a,\infty)$).
