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Consider a measure space $(X,\mathcal{B},\mu)$. Lots of questions have been asked in MSE regarding convergence of the $L^p$ norm of a measurable function to the $L^\infty$ norm as $p\to\infty$. I would like to restrict attention to simple functions, which it convenient as it means we have essentially no problems in justifying any computation (e.g. swapping a sum and integral).

Let $f$ be a simple function: $f=\sum_{j=1}^Jc_j1_{E_j}$ where each measurable set $E_j$ has finite measure. Of course it follows from the general case (e.g. here and here) that $\lim_{p\to\infty}\|f\|_p=\|f\|_\infty$. Is there a "simpler" proof by exploiting properties of simple functions (e.g. one can explicitly give $\|f\|_{\infty}$ which is $\max_{1\leq j\leq J}\{|c_j|\}$)?

In the case when $J=1$, say $f=c1_E$, one has $\|f\|_{p}=|c|\mu(E)^{1/p}$ and the consequence is immediate. The case when $J>0$ is unclear for me.

3 Answers3

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Without loss of generality, suppose $|c_1| > |c_j|$ for all $2\le j \le J$, so that

$$\|f \|_\infty = |c_1|. $$ We can also assume (always without loss of generality) $E_i \cap E_j = \emptyset$ for $i \neq j$. We have

$$\| f\|_p =\left( \sum_{j=1}^J |c_j|^p m(E_j)\right)^{1/p} =|c_1| \left(\sum_{j=1}^J \left|\frac{c_j}{c_1} \right|^p m(E_j)\right)^{1/p}.$$

Since $|c_1|>|c_j|$ for $2 \le j \le J$, we have

$$\sum_{j=1}^J \left|\frac{c_j}{c_1} \right|^p m(E_j) \to m(E_1) $$

as $p \to +\infty$. Therefore

$$\left(\sum_{j=1}^J \left|\frac{c_j}{c_1} \right|^p m(E_j)\right)^{1/p} \to 1 $$

and consequently $\| f \|_p \to |c_1| = \|f\|_\infty$.

Stefano
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  • I was stuck with the calculation of $\int_X|\sum_{j=1}^Jc_j1_{E_j}|^p$ while I didn't notice that one can take the summation sign out when $E_j$ are pairwise disjoint. I'm now checking the limit calculation. –  Mar 08 '17 at 01:40
  • Thank you for your answer. The way you calculate the limit looks a little mysterious to me. Would you elaborate? –  Mar 08 '17 at 01:47
  • $|c_j/c_1|^p \to 0$ if $j \neq 1$, so the only term that survives in the first limit is the one with $j=1$. The second limit is of the form $a(p)^{1/p}$, where $a(p) \to m(E_1)$ as $p \to+\infty$. Therefore

    $$a(p)^{1/p} = e^{\frac{\log a(p)}{p}} \to e^0 = 1. $$

     – Stefano Mar 08 '17 at 01:51
  • Thanks. I don't know what $a(p)$ mean in your comment though, I figured out the limit and wrote an answer for elaboration. –  Mar 08 '17 at 02:04
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If $f=\sum_{j=1}^Jc_j1_{E_j}$ then

$$\int |f|^p=\sum_{j=1}^J\int_{E_j}{|c_j|}^p=\sum_{j=1}^J{|c_j|}^p|E_j|$$

Hence, $\left(\int |f|^p\right)^{1/p}=\left(\sum_{j=1}^J{|c_j|}^p|E_j|\right)^{1/p}$. Do you think you can finish it?

Fimpellizzeri
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  • Thank you for your answer. In your calculation of the $L^p$ norm for $f$, do you need to assume that the $E_j$ are pairwise disjoint? –  Mar 08 '17 at 02:18
  • Yes, I had assumed that. Generally, when one uses simple function, this is implied. See for instance this wikipedia link. – Fimpellizzeri Mar 08 '17 at 02:45
  • Fair enough. After all, one can eventually assume the sets are pairwise disjoint. –  Mar 08 '17 at 03:10
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[This was intended for elaborating Stefano's answer.] Using homogeneity of the $L^p$ norm, one can assume that $$ \|f\|_\infty=|c_1|=1. $$ By homogeneity again, one can further assume that $\mu(E_1)=1$. By assuming that $E_j$ are mutually disjoint (and all the $|c_j|$ distinct) one would end up with calculating the limit $$ \lim_{p\to\infty} \big(1+h(p)\big)^{1/p} $$ where $h(p):=|c_2|^p{\mu(E_2)}+\cdots+ |c_J|^p{\mu(E_J)}$. But one has $$ 1\leq [1+h(p)]^{1/p}\leq 2^{1/p} $$ for large enough $p$ since $|c_j|< 1$ for $j\geq 2$; one also has $\lim_{p\to\infty}2^{1/p}=1$.

  • I don't see where the assumption that $|c_1|=1$ comes from, and it is most definitely not needed. The important things are (1) $\lim_{p\to\infty}x^p=0$ if $|x|<1$, (2) $\lim_{p\to\infty}x^{1/p}=1$ if $x>0$. – Fimpellizzeri Mar 08 '17 at 02:11
  • When $|f|\infty>0$, showing $\lim{p\to\infty}|f|p=|f|\infty$ is the same as $\lim_{p\to\infty}|\frac{f}{|f|\infty}|_p=1$. One can thus assume that $|f|\infty=1$. –  Mar 08 '17 at 02:36