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Let $({f_n})_{n\geq 1}$ be a sequence of functions on $[0,1]$ such that $\lim _{n\rightarrow \infty}f_{n}(x)=f(x)$ almost everywhere with respect to Lebesgue measure and $$\sup_{n\geq 1}||f_n||_{L^4}<\infty.$$

Prove that $$\lim_{n\rightarrow \infty}||f_n - f||_{L^3} = 0.$$

Since we're on a finite measure space I see how we get that $f_n \in L^3$. But I don't see how I can use a convergence theorem on $L^p$ larger than one.

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  • $\begingroup$ My approach is to note that $F_n=|f_n-f|\leq2^p(||f_n||^p+||f||^p)=g_n$ so $F_n \rightarrow 0$ and $g_n \rightarrow g = 2^p||f||$ which is integrable. Thus($F_n\leq g_n$) by the Dominated Convergence Theorem we're finished. But this assumes that $||f_n||_p \rightarrow ||f||_p$. $\endgroup$
    – Dave
    Aug 16, 2012 at 1:54
  • $\begingroup$ that should be $F_n=|f_n-f|^p....$ $\endgroup$
    – Dave
    Aug 16, 2012 at 2:01

3 Answers 3

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A sequence bounded in $L_p$, $p>1$, is uniformly integrable. Using this fact with $p=4/3$, it follows that $(f_n^3)$ is uniformly integrable. It now follows from Vitali's Convergence Theorem that $f_n$ converges to $f$ in the $L_3$-norm.

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You can show that the sequence is uniformly integrable and arrive at your conclusion.

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  • $\begingroup$ I'm not sure what you mean. $\endgroup$
    – Dave
    Aug 16, 2012 at 1:08
  • $\begingroup$ The link you sent me is blank. $\endgroup$
    – Dave
    Aug 16, 2012 at 1:31
  • $\begingroup$ Try this: en.wikipedia.org/wiki/Uniform_integrability $\endgroup$
    – ncmathsadist
    Aug 16, 2012 at 1:35
  • $\begingroup$ Yea I found that page but its quite unclear to me. $\endgroup$
    – Dave
    Aug 16, 2012 at 1:39
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Using Fatou's Lemma, $$ \|f\|_{L^4}\le\liminf_{n\to\infty}\|f_n\|_{L^4}\tag{1} $$

Since $f_n$ converges pointwise, for any $\epsilon>0$ Egorov's Theorem states that except on $E_{\large\epsilon}$, a set of measure $\epsilon$, $f_n\to f$ uniformly. Therefore, $$ \lim_{n\to\infty}\int_{\large E_{\Large\epsilon}^c}\left|f_n(x)-f(x)\right|^3\,\mathrm{d}x=0\tag{2} $$ Furthermore, by Hölder's Inequality with $p=4/3$ and $q=4$, we have $$ \limsup_{n\to\infty}\int_{\large E_{\Large\epsilon}}\left|f_n(x)-f(x)\right|^3\,\mathrm{d}x\le\epsilon^{1/4}\sup_n\|f_n-f\|_{L^4}^3\tag{3} $$ Since $\epsilon$ is arbitrary, we get $$ \lim_{n\to\infty}\int_0^1\left|f_n(x)-f(x)\right|^3\,\mathrm{d}x=0\tag{4} $$

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  • $\begingroup$ The fact that $f \in L^4$ is also immediate from Fatou's lemma. $\endgroup$
    – Nate Eldredge
    Aug 21, 2012 at 2:13
  • $\begingroup$ @NateEldredge: Duh! I looked at Fatou and completely missed that since $f$ was the limit, it was also the $\liminf$. Thanks. $\endgroup$
    – robjohn ♦
    Aug 21, 2012 at 12:53

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