# Pointwise a.e. convergence implies strong convergence?

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 $||f_n||_{p_2} \leq 1$ ?

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In general strong convergence in $L^{p_2}$ will not hold, so that route won't work. –  Nate Eldredge Jan 22 '13 at 1:51
You could mimic the proof in the first answer here. In place of Cauchy-Schwarz, use (or prove) $(\int_A |g|^{p_1})^{1/p_1}\le (\int_A |g|^{p_2})^{1/p_2}\cdot\bigl(\mu(A)\bigr)^{1/p_1-1/q_1}$ for $g\in L_{p_2}$. –  David Mitra Jan 22 '13 at 2:01

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 $L^1$.