Let $ 1 \leq p_1 < p_2 < \infty$, and suppose that $f_n$ is a sequence of functions in $L^{p_1}[a,b]$ such that $f_n \to f$ pointwise a.e. on $[a,b]$. Suppose in addition that $ ||f_n||_{p_2} \leq 1$ for every $n$, where $|| \cdot ||_{p_2}$ denotes the $L^{p_2}$ norm, then how can we show that $f_n \to f$ strongly in $L^{p_1}$ ?
What I have tried:
Since $ 1 < p_2 < \infty$ and $\{f_n\}$ has bounded $L^{p_2}$ norm, it follows that $f_n \to f$ weakly in $L^{p_2}$. Now since $[a,b]$ has finite measure, if $f_n \to f$ strongly in $L^{p_2}$, then $f_n \to f$ strongly in $L^{p_1}$ also, so this is what I am trying to show, although I do not know whether we actually do have strong convergence in $L^{p_2}$.
Given $f_n \to f$ weakly in $L^{p_2}$, we have strong convergence in $L^{p_2}$ if and only if the norms converge, i.e. $||f_n||_{p_2} \to ||f||_{p_2}$. But how can we argue that we do have convergence of the norms in this case? Maybe using the condition that $ ||f_n||_{p_2} \leq 1$ ?
Best Answer
For simplicity, consider the case $p_1 = 1$. (The general case follows, by replacing $f_n$ by $|f_n|^{p_1}$ and $p_2$ by $p_2/p_1$.)
Show that the condition $\|f_n\|^{p_2} \le 1$ implies that $\{f_n\}$ is uniformly integrable. Then use Vitali's convergence theorem to conclude that $f_n$ converges in $L^1$.