# $L^p$-space convergence

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.

• 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 $||f_n||_p \rightarrow ||f||_p$.
– Dave
Aug 16, 2012 at 1:54
• that should be $F_n=|f_n-f|^p....$
– Dave
Aug 16, 2012 at 2:01

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.

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

• I'm not sure what you mean.
– Dave
Aug 16, 2012 at 1:08
• The link you sent me is blank.
– Dave
Aug 16, 2012 at 1:31
• Aug 16, 2012 at 1:35
• Yea I found that page but its quite unclear to me.
– Dave
Aug 16, 2012 at 1:39

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 $$\lim_{n\to\infty}\int_0^1\left|f_n(x)-f(x)\right|^3\,\mathrm{d}x=0\tag{4}$$

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