Is there a "simple" proof of $\lim_{p\to\infty}\|f\|_p=\|f\|_\infty$ for simple functions?

Question

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 $\max_{1\leq j\leq J}\{|c_j|\}$)?

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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.

Answer 1

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

Answer 2

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?

Answer 3

[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 $\lim_{p\to\infty}2^{1/p}=1$.