# Question about $\lim _{q \rightarrow \infty}\|f\|_{q}=\|f\|_{\infty}$

Let $$(X,B,\mu)$$ be a complete measure space,Show that $$\lim _{q \rightarrow \infty}\|f\|_{q}=\|f\|_{\infty}, \quad \forall f \in \bigcup_{p} \bigcap_{p \leqslant q<\infty} L^{q}$$ So,$$\lim _{q \rightarrow \infty}\|f\|_{q}$$ , $$\|f\|_{\infty}$$ are equal-norm with space $$L^{\infty} \cap\left(\bigcup_{p} \bigcap_{p \leqslant q} L^{q}\right)$$.

Case 1: $$m(X)<\infty$$.It's easy to prove that.

Case 2: $$m(X)=\infty$$. I have no idea about it,And I started to doubt the correctness of this conclusion. Can somebody give me a hint for this problem or just give an example to prove that this is a wrong conclusion when $$m(X)=\infty$$.

• I'm pretty sure you need that $f\in L^\infty$ as well, so that the statement is actually true. – K. Y Jun 30 at 1:21
• I'm not sure that you need cases for $m(X)$. One direction is simply the application that $||f||_q \leq ||f||_p^{p/q}||f||_\infty^{1-p/q}$ – K. Y Jun 30 at 1:25
• @K. Y Thanks for your answer first.Let $f \in \bigcup_{p}\bigcap_{p \leqslant q<\infty}L^{q}$,but not $L^{\infty}$,then exists $q_0 \in \mathbb{N}$,$\forall q>q_0,f \in L^q$.We define $E=\{x:|f(x)|>M\}$,Then $0<m(E)<\infty$(If $m(E)=\infty,\|f\|_{q} \geqslant\left(\int_{E}|f|^{q} d \mu\right)^{\frac{1}{q}} \geqslant M|E|^{\frac{1}{q}}$,We get $f \notin L^q$).Observing inequality $\|f\|_{q} \geqslant\left(\int_{E}|f|^{q} d \mu\right)^{\frac{1}{q}} \geqslant M|E|^{\frac{1}{q}}$,Let $q \rightarrow \infty$,Actually we prove that $\varliminf_{q \rightarrow \infty}\|f\|_{q} \geqslant\|f\|_{\infty}$. – Johnstein Jun 30 at 1:46
• – K. Y Jun 30 at 2:02
• @K. Y I think $\infty$ also can be the limit of some sequence,So... – Johnstein Jun 30 at 2:06

Assume $$0<\|f\|_\infty<\infty$$ and $$f\in L_r$$ or some $$r>0$$. Then $$|f|/\|f\|_\infty<1$$ a.s.. For $$p>r$$

$$\frac{|f|^p}{\|f\|^p_\infty}\leq \frac{|f|^r}{\|f\|^r_\infty}\in L_1$$

hence $$p\in E:=\{s: \|f\|_s<\infty\}$$. Integrating on both side leads to

$$\frac{\|f\|_p}{\|f\|_\infty}\leq\Big(\frac{\|f\|_r}{\|f\|_\infty}\Big)^{r/p}\xrightarrow{p\rightarrow\infty}1$$ That is $$\limsup_p\|f\|_p\leq \|f\|_\infty$$

By the Markov-Chebyshev inequality, for any $$0<\alpha<\|f\|_\infty$$

$$0<\alpha\big(\mu(|f|>\alpha)\big)^{1/p}\leq\|f\|_p$$

Hence $$\alpha\leq\liminf_p\|f\|_p$$ and so, $$\|f\|_\infty\leq\liminf_p\|f\|_p$$.

If $$\|f\|_p=\infty$$ and $$f\in L_r$$ for some $$r>0$$ then $$0<\mu(|f| and so

$$0n)\big)^{1/p}\leq\|f\|_p\quad\text{for}\quad p\geq r$$

This implies $$n\leq\liminf_p\|f\|p$$ for any $$n\in\mathbb{N}$$.

• Excellent Answer! – Johnstein Jun 30 at 1:57
• Notice that is enough to have $f\in\bigcup_{p>0}L_p$ for the statement to hold. – Oliver Diaz Jun 30 at 2:00
• You are right.Thanks. – Johnstein Jun 30 at 2:13
• Actually,Because $f\in L_r$ for some $r>0$,Then $\mu(|f|>n)<K(n)<\infty$. – Johnstein Jun 30 at 2:18
• @Johnstein: Here is a related problem that may be of interest which can be solve with methods similar to what we did here. math.stackexchange.com/questions/1482933/… – Oliver Diaz Jun 30 at 22:49