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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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    $\begingroup$ In general strong convergence in $L^{p_2}$ will not hold, so that route won't work. $\endgroup$ – Nate Eldredge Jan 22 '13 at 1:51
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    $\begingroup$ 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 $g\in L_{p_2}$. $\endgroup$ – David Mitra Jan 22 '13 at 2:01
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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 $L^1$.

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  • $\begingroup$ 1) Does this argument work for $L^1(\Omega)$, $\Omega \subset \mathbb{R}^n$ either bounded or unbounded? 2) If $p_2=p_1$ (=1) then can we claim $f_n$ converges WEAKLY in $L^1$ to $f$? $\endgroup$ – Behnam Esmayli Jul 24 '18 at 21:39
  • $\begingroup$ @BehnamEsmayli: (1) It works if $\Omega$ has finite measure (and in particular if $\Omega$ is bounded). Otherwise you can pretty easily find counterexamples. (2) No, consider for example $[a,b] = [0,1]$, $f_n = n 1_{(0, 1/n)}$, $f = 0$. $\endgroup$ – Nate Eldredge Jul 28 '18 at 20:13
  • $\begingroup$ But doesn't that seq in (2) converge weakly to $f$? $\endgroup$ – Behnam Esmayli Jul 31 '18 at 2:33
  • $\begingroup$ @BehnamEsmayli: If it did, then $\int f_n g$ should converge to $\int fg$ for every $g \in L^\infty([0,1])$, since integration against $g$ is a continuous linear functional on $L^1$. Try it with $g=1$. $\endgroup$ – Nate Eldredge Jul 31 '18 at 2:45

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