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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 $.

Thanks in advance.

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  • $\begingroup$ I'm pretty sure you need that $f\in L^\infty$ as well, so that the statement is actually true. $\endgroup$ – K. Y Jun 30 at 1:21
  • $\begingroup$ I'm not sure that you need cases for $m(X)$. One direction is simply the application that $||f||_q \leq ||f||_p^{p/q}||f||_\infty^{1-p/q}$ $\endgroup$ – K. Y Jun 30 at 1:25
  • $\begingroup$ @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 $\varliminf_{q \rightarrow \infty}\|f\|_{q} \geqslant\|f\|_{\infty}$. $\endgroup$ – Johnstein Jun 30 at 1:46
  • $\begingroup$ See functions in $L^p$ but not $L^\infty$ $\endgroup$ – K. Y Jun 30 at 2:02
  • $\begingroup$ @K. Y I think $\infty$ also can be the limit of some sequence,So... $\endgroup$ – Johnstein Jun 30 at 2:06
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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 $0<\mu(|f|<n)\leq\frac{1}{n^r}\|f\|_r<\infty$ and so

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

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

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  • $\begingroup$ Excellent Answer! $\endgroup$ – Johnstein Jun 30 at 1:57
  • $\begingroup$ Notice that is enough to have $f\in\bigcup_{p>0}L_p$ for the statement to hold. $\endgroup$ – Oliver Diaz Jun 30 at 2:00
  • $\begingroup$ You are right.Thanks. $\endgroup$ – Johnstein Jun 30 at 2:13
  • $\begingroup$ Actually,Because $f\in L_r$ for some $r>0$,Then $\mu(|f|>n)<K(n)<\infty$. $\endgroup$ – Johnstein Jun 30 at 2:18
  • $\begingroup$ @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/… $\endgroup$ – Oliver Diaz Jun 30 at 22:49

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