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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.

Mittens
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Johnstein
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  • I'm pretty sure you need that $f\in L^\infty$ as well, so that the statement is actually true. – K. Y Jun 30 '20 at 01:21
  • 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 '20 at 01:25
  • @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 '20 at 01:46
  • See functions in $L^p$ but not $L^\infty$ – K. Y Jun 30 '20 at 02:02
  • @K. Y I think $\infty$ also can be the limit of some sequence,So... – Johnstein Jun 30 '20 at 02:06

1 Answers1

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The following argument does not really on any finiteness assumptions on $\mu$. The only general assumptions is that $f\in L_r$ for some $r>0$.

If $\|f\|_\infty=0$ the conclusion is trivial. There are two cases other cases to consider: $0<\|f\|_\infty<\infty$ and $\|f\|_\infty=\infty$.

  1. Assume first that $0<\|f\|_\infty<\infty$ and $f\in L_r$ for 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 yields $$\frac{\|f\|_p}{\|f\|_\infty}\leq\Big(\frac{\|f\|_r}{\|f\|_\infty}\Big)^{r/p}\xrightarrow{p\rightarrow\infty}1$$ 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, $\|f\|_\infty\leq\liminf_p\|f\|_p$. The conclusion follows.

  2. Now assume that $\|f\|_p=\infty$ and $f\in L_r$ for some $r>0$. Then $0<\mu(|f|>n)\leq\frac{1}{n^r}\|f\|^r_r<\infty$ 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 $n\in\mathbb{N}$. The conclusion follows.

The assumption $f\in L_r$ for some $r>0$ may not dropped in general. Consider the measure space $(\mathbb{R},\mathscr{B}(\mathbb{R}),m)$ where $m$ Lebesgue's measure. The function $f(x)=\mathbb{1}_{(0,\infty)}(x)$ is measurable, and $\|f\|_p=\infty$ for all $p>0$, while $\|f\|_\infty=1$.

Mittens
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