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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 $p$?

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    @Yuval, Thanks. I have corrected. For the fast growth part, we indeed need to add the condition.2011-10-28

2 Answers 2

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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 $[a,\infty)$).

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    The algorithm I had in mind was: (i) Fix your sequence $\{c_n\}$; (ii) maximize $g_n(p)=(cn/\Phi(p))^p$ in $p$ for each $n$, say g_n(p) for all $p$; (iii) take m(En)<1/(2^nM_n). By fixing a clever choice of $c_n$, you've essentially managed to do (ii) and (iii) in general. So you give up a little bit of choice and gain some efficiency. The reason I decided not to use a specific $\{c_n\}$ in writing my answer was that I wanted to make transparent the properties I was using, namely, that $c_n\to\infty$ with $n$ and c_n>0. (Had I thought of your choices I might have changed my mind. :))2011-10-28
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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).

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    Neat argument, Thanks!2011-10-28