1

Let $(X,\mu)$ be a finite measure space ($\mu(X)<\infty$) and $f_n, f$ measurable functions such that $f_n\to f$ in measure, plus $||f_n||_{L^p}<C$ for all $n$, for a fixed $p>1$, and for a fixed $C>0$. How can we conclude that $f_n\to f$ in $L^1$?

Juan Carlos Ortiz
  • 1,060
  • 1
    This is probably best answered with uniform integrability. Are you familiar with the concept? – Jose27 Mar 31 '20 at 19:20
  • I know that if we prove ${ |f_n|}$ is uniformly integrable then we can finish, but how could we prove this? – Juan Carlos Ortiz Mar 31 '20 at 19:22
  • 1
    Key is convexity, and in this case you can give a direct argument: Fix $\alpha>0$ and set $\mu_\alpha:=\mu({ X>\alpha})$, then by Jensen's inequality $$\int_{|X|>\alpha} |X|d\mu \leq \mu_\alpha^{1-1/p} \left( \int_{|X|>\alpha} |X|^pd\mu \right)^{1/p},$$ so that taking $X=f_n-f$ the result follows. – Jose27 Mar 31 '20 at 19:30
  • 1
    See this question for a direct proof (which does not use uniform integrability) – saz Mar 31 '20 at 20:11
  • @Jose27: I see how to finish with your method! I posted a solution below. You still have to prove that $f\in L^p(X)$ to finish, right? Or is there a faster way to finish without proving $f\in L^p$? – Juan Carlos Ortiz Mar 31 '20 at 20:48
  • @saz: I see, I posted a solution below using that method! Thank you :-) – Juan Carlos Ortiz Mar 31 '20 at 20:48
  • 1
    You're correct, we still have to show $f\in L^p$. – Jose27 Mar 31 '20 at 22:00

1 Answers1

1

Solution 1: (Thanks to saz in the comments, and Matthew Tyler). Assume au contraire, then after passing to a subsequence we have $||f_n-f||_{L^1}>\epsilon\forall n$. Since $f_n\to f$ in measure, after passing to a (sub)subsequence we have $f_n\to f$ a.e. Then $|f_n|^p\to |f|^p$ a.e. so by Fatou's Lemma, $$ \int_X |f|^p \le \liminf_{n\to\infty} \int_X |f|^p\le C^p \hspace{.5cm}\Rightarrow \hspace{.5cm}||f||_{L^p}\le C,$$ so $f\in L^p$, and $||f_n-f||_{L^p}\le ||f_n||_{L^p}+||f||_{L^p}\le 2C$ are also uniformly bounded. Replacing the $f_n$ with $f_n-f$, we have $||f_n||_{L^1}>\epsilon\forall n$ but $f_n\to 0$ a.e. (and in measure too), plus $\{||f_n||_{L^p}\}$ is uniformly bounded, say by $D$. Then for any $\delta>0$ we have by Holder, where $\frac{1}{p}+\frac{1}{q}=1$: $$ \epsilon \le \int_X |f_n| = \int_{\{|f_n|\le\delta\}}|f_n| + \int_{\{|f_n|>\delta\}}|f_n| \le \delta\mu(X) | ||f_n\chi_{\{|f_n|>\delta\}}||_{L^1(X)}\\ \le \delta\mu(X) + ||f_n||_{L^p(X)} ||\chi_{\{|f_n|>\delta\}}||_{L^q(X)} \le\delta\mu(X) + D\mu(\{|f_n|>\delta\})^{1/q}.$$ Convergence in measure implies $\mu(\{|f_n|>\delta\})\to 0$ as $n\to\infty$, and so the above inequality gives $\epsilon\le\delta\mu(X)$ for any $\delta>0$, absurd!

Solution 2: (by Jose27 in the comments). Notice $f\in L^p$ as above. Set $g=f_n-f, \alpha>0$ arbitrary and $\mu_a:=\mu(\{|g|>\alpha\})$. Jensen's Inequality gives

$$ \bigg|\frac{1}{\mu_\alpha}\int_{\{|g|>\alpha\}} |g|\bigg|^p \le \frac{1}{\mu_\alpha}\int_{\{|g|>\alpha\}} |g|^p \hspace{.5cm}\Rightarrow\hspace{.5cm} \int_{|g|>\alpha} |g| \le \mu_\alpha^{1-1/p}\left(\int_{\{|g|>\alpha\}} |g|^p\right)^{1/p} =\mu_\alpha^{\frac{p-1}{p}}||g||_{L^p(\{|g|>\alpha\})}.$$ Notice fixing $\alpha$, $\mu_\alpha\to 0$ as $n\to\infty$, by convergence in measure. Hence, $$||f_n-f||_{L^1} \le \int_{\{|g|\le\alpha\}} |g| + \int_{\{|g|>\alpha\}}|g| \le \alpha\mu(X) +\mu_\alpha^{\frac{p-1}{p}}||g||_{L^p(\{|g|>\alpha\})}\\ \le \alpha\mu(X) +\mu_\alpha^{\frac{p-1}{p}}(||f_n||_{L^p(X)} + ||f||_{L^p(X)}) \le \alpha\mu(X) +\mu_\alpha^{\frac{p-1}{p}}(C + ||f||_{L^p(X)}),$$ so letting $n\to\infty$ we see $\limsup_{n\to\infty} ||f_n-f||_{L^1(X)} \le \alpha\mu(X)$. Since this holds for any $\alpha>0$ we see the limsup is $0$, as desired.

Juan Carlos Ortiz
  • 1,060