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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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    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 $||f_n||_p \rightarrow ||f||_p$.2012-08-16
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    that should be $F_n=|f_n-f|^p....$2012-08-16

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