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Let $1< p < \infty$, $\{f_n\}_n \subset L^p[0,1]$ s.t. $f_n:[0,1] \to \mathbb{R}$, $f_n \to f$ a.e., and $||f_n||_{L^p} \leq M < \infty$ for all $n$. Then, given $1 \leq q < p$, we want to show that $f_n \to f$ in $L^q$.

I think that you can show this result by using the Rellich-Kondrachov theorem (i.e. $L^p \hookrightarrow L^q$ compactly for $q < p$) to extract a convergent subsequence in $L^q$. Then you can use the fact that $f_n \to f$ a.e. to show that the whole sequence must converge to $f$ in $L^q$.

However, I was wondering if this approach might be a little overkill, and if I should be using some less overpowered tools to prove the statement (assuming the argument above is valid).

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up vote 7 down vote accepted

The idea is that since $f_n$ converges to $f$ pointwise, it converges to $f_n$ almost uniformly, meaning that for any $\epsilon > 0$ one can remove a set $A$ of measure $< \epsilon$ and then $f_n \rightarrow f$ uniformly on $[0,1] - A$. Then by the uniform convergence theorem $\lim_{n \rightarrow \infty} \int_{[0,1] - A} |f_n - f|^q = 0$.

So far we haven't used that $q < p$. But by Holder's inequality, applied to $f_n^q \times 1$ and the exponents ${p \over q}$ and ${p \over p - q}$ you have $$\int_A|f_n - f|^q \leq |A|^{p - q \over q}(\int_A|f_n - f|^p)^{q \over p}$$ $$\leq |A|^{p - q \over q}(\int_0^1|f_n - f|^p)^{q \over p}$$ $$ < C\epsilon^{p - q \over q}$$ Here $C$ depends on $M$ (and you implicitly use $||f||_p \leq M$ which you can get using Fatou's Lemma). Combining with the uniform convergence off of $A$, this implies that $$0 \leq \limsup_{n \rightarrow \infty} \int_0^1 |f_n - f|^q < C\epsilon^{p - q \over q}$$ This is true for all $\epsilon > 0$, so letting $\epsilon \rightarrow 0$ you obtain $$\lim_{n \rightarrow \infty} \int_0^1 |f_n - f|^q = 0$$

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Nice argument! When you subtract the $\liminf \int_0^1 |f_n - f|^q$ term, are you just bounding it below by zero? – user1736 Aug 5 '11 at 15:48
@user1736 yes. I will edit that line to simplify it. – Zarrax Aug 5 '11 at 16:09

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