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Let $1\leq p<q<\infty$, deduce from each one of the following statements that $L^q[0,1]$ is a (Baire) first category subset in $L^p[0,1]$:

  1. Let $p<r<s<q$ and $\beta=s/(s-1)$, we define $g_n=n\cdot\chi_{[0,n^{-\beta}]}\; (n\in\mathbb{N})$ and $$ f(t)= \left\{ \begin{array}{ll} t^{-1/r} & 0<t\leq 1 \\ 0 & t=0. \end{array} \right.\quad(*) $$ Then: $$ \lim_{n\rightarrow\infty}\int_0^1{f(t)\,g_n(t)\;dt}=0 $$ For all $f\in L^q[0,1]$, but not for $f\in L^p[0,1]$ defined above $(*)$.

  2. The inclusion $L^q[0,1]\hookrightarrow L^p[0,1]$ is s non surjective continuous map.

My work: I have proved that both statements are true, which is not trivial. But I have real struggles to deduce from each of them that $L^q[0,1]$ is a (Baire) first category subset in $L^p[0,1]$. I think that both of them are related with some dense subset, but I am really...

Mittens
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Sergio Ferrer
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  • I transcribed it from a Spanish text, maybe there are some mistakes. Why did you think that? – Sergio Ferrer Apr 07 '23 at 11:35
  • Like I mentioned before, (1) is to show that $L_q\neq L_p$. One already knows that $L_q\subset L_p$ from Holder's inequality which is independent from (2). You can now prove that the closed balls in $L_q$ are closed in $L_p$ by Fatou's lemma for example (again independent from (2)). The rest follows the same lines as in my posting: $L_q=\bigcup_n\overline{B}_q(0;n)$, ... – Mittens Apr 07 '23 at 13:55

1 Answers1

1

Here is a proof of the statement starting from (2):

That fact that $\iota: L_q=L_q[0,1] \hookrightarrow L_p=L_p[0,1]$ is non surjective but continuous map implies that

  • $L_q:=L_q[0,1]\varsubsetneq L_p:=L_p[0,1]$: $\|\iota(f)\|_p=\|f\|_p\leq C\|f\|_q$ for all $f\in L_q$ and some constant $C>0$ ($C=1$ by Holder's)
  • The closed ball $\overline{B}_q(0;c):=\{f\in L_q[0,1]: \|f\|_q\leq c\}$ in $L_q$ is also closed in $L_p$: $\iota^{-1}(\overline{B}_q(0;c))=\overline{B}_q(0;c)$ and $\overline{B}_q(0;c)$ is closed in $L_q$.

To complete the proof of the statement from (2), notice that $$L_q=\bigcup_n\overline{B}_q(0;n)$$ Suppose there is $m$ such that $\overline{B}_q(0;m)$ has nonempty interior in $L_p$. Then, there is $f\in \overline{B}_q(0;m)$ and $r>0$ such that $B_p(f; r)=\{g\in L_p: \|f-g\|_p<r\}\subset \overline{B}_q(0;m)$. For any $h\in L_p$, choose $\varepsilon>0$ small enough so that $\varepsilon\|h-f\|_p<r$. Then $$f+\varepsilon(h-f)\in B_p(f; r)\subset \overline{B}_q(0;m)\subset L_q$$ and so, $h\in L_q$. It follows that $L_p= L_q$ which is a contradiction. Consequently, each $\overline{B}_q(0;n)$ is nowhere dense whence we conclude that $L_q$ is of first category in $L_p$.

Here is a proof of (2): If $f\in L_q$, the $|f|^p\in L_{q/p}$. By Holder's inequality $$\|f\|^p_p=\int_{(0,1]}|f|^p\leq \big(\int|f|^q\big)^{p/q}=\|f\|^p_q$$

To show that $L_q\neq L_p$, consider any function of the form $g(t)=t^{-\alpha}$ with $\frac1q\leq \alpha<\frac1p$.

(1) implies that $L_q\neq L_p$. It follows from Holder's inequality that $L_q\subset L_p$. Now, one can prove directly that $\overline{B}_q(0;c)$ is closed in $L_p$: Suppose $(f_n:n\in\mathbb{N})\subset \overline{B}_q(0;c)$ converges in $L_p$ to some $f\in L_p$. Then $f_n$ converges to $f$ pointwise almost surely along some subsequence $n'$. By Fatou's Lemma $$\int|f|^q\,dm\leq \liminf_{n'}\int |f_{n'}|^q\,dm\leq c^q$$ Notice that the arguments here are independent of the statement in (2).

To prove that $L_q$ is of first category is now as above.

Mittens
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  • The exercise is longer and I had to prove before that $L^q[0,1]\subsetneq L^p[0,1]$ and also, $B_q(0,n)$ are closed empty interior in $L^q[0,1]$, that implies $Lq$ is of first category in $Lp$. My problem is to proved that just using (1) or (2), not together. – Sergio Ferrer Apr 07 '23 at 11:40
  • @SergioFerrer: I added a proof starting from (1). Let me know if you have further questions regarding this posting. – Mittens Apr 07 '23 at 18:29