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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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It's a very interesting question. I don't have the answer, but if we consider $f(x)=|\ln x|^{\alpha}$ where $\alpha>0$, we cover the case $\phi(p)=p^{\alpha}$. –  Davide Giraudo Oct 26 '11 at 17:06
    
Nice observation. It took me a while to verify it. Is there a good way to see this? –  Syang Chen Oct 27 '11 at 0:41
    
@Davide Giruado According to WolframAlpha, $\|f\|_p = \Gamma(p\alpha+1)^{1/p}$ for $f(x)=|\ln x|^\alpha$. How do you get $\phi(p)=p^\alpha$ from this? –  Jeff Oct 27 '11 at 4:30
    
@Jeff, use Stirling's formula for Gamma function you will see this is true with "$=$" replaced by "$\approx$". –  Syang Chen Oct 27 '11 at 6:39
    
You should add the condition that $\Phi$ is non-decreasing (or continuous), otherwise it's easy to construct $\Phi$ for which the condition trivially fails. –  Yuval Filmus Oct 27 '11 at 14:11

2 Answers 2

up vote 11 down vote accepted

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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how can you choose $c_n$ such that $c_n^p/\Phi(p)^p$ is bounded in p? What are the $c_n$'s? –  Syang Chen Oct 27 '11 at 14:07
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@XianghongChen The sequence $\{c_n\}$ is an arbitrary sequence of positive numbers that increases to $\infty$ as $n\to\infty$. The fact that $c_n^p/\Phi(p)^p$ is bounded in $p$ for fixed $n$ has nothing to do with the choice of the sequence $\{c_n\}$ other than the fact that $c_n>0$. For any $r>0$, the ratio $r^p/\Phi(p)^p$ is bounded in $p$, simply because $\Phi(p)\to\infty$ with $p$. This is because, for sufficiently large $p$, you have $r/\Phi(p)<1$. For smaller $p$, the function $\Phi$ is bounded away from zero, so $r/\Phi(p)\leq r/\delta$ for some $\delta>0$. –  Nick Strehlke Oct 27 '11 at 15:51
    
Here is how I understand it. Assume that $\Phi$ is increasing. Let $c_n=\Phi(n), m(E_n)=(2\Phi(n))^{−n}$. Is it correct? –  Syang Chen Oct 28 '11 at 1:26
    
@Xianghong Those choices will certainly work. You can see why (as you probably did) by looking at $$ \left(\frac{c_n}{\Phi(p)}\right)^pm(E_n) = 2^{-n}\left(\frac{\Phi(n)^{p-n}}{\Phi(p)^{p}}\right). $$ Whenever $\Phi(p)>1$, the above quantity is $<2^{-n}$. That's because $\Phi(n)^{p-n}<1$ if $p<n$ and $\Phi(p)^p>\Phi(n)^{p-n}$ if $p>n$, because $\Phi$ is increasing. Of course, other choices will work too. –  Nick Strehlke Oct 28 '11 at 4:04
    
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 $g_n(p)<M_n$ 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. :)) –  Nick Strehlke Oct 28 '11 at 4:26

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! –  Syang Chen Oct 28 '11 at 1:53

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