Let $1 \le p_1 < p_2 < \infty$, $\{f_n\}$ be a sequence in $L^{P_2}([0, 1])$ and $f$ belong to $L^{P_2}([0, 1])$. Prove that $f_n \rightharpoonup f$ weakly in $L^{P_2}([0, 1])$ implies $f_n \rightharpoonup f$ weakly in $L^{P_1}([0, 1])$.
is the following correct?
since $1 \le p_1 < p_2 < \infty$ then $L^{P_2}([0, 1]) \subset L^{p_1}([0, 1])$. Let $g_1 \in L^{q_1}$ and $g_2 \in L^{q_2}$ where $1=\frac{1}{p_1}+\frac{1}{q_1}$ and $1=\frac{1}{p_2}+\frac{1}{q_2}$ \begin{align} \Big|\int_0^1 f_n g_1 - \int_0^1 f g_1 \Big| & \le \int_0^1 \Big|f_n g_1 - f g_1 \Big| \\ & = \int_0^1 \Big|(f_n -f)(g_1 - g_2)\Big| + \Big|\int_0^1 f_n g_2 - \int_0^1 f g_2\Big|\\ \end{align} Pick $n$ such that $\forall n\ge N ,\Big|\int_0^1 f_n g_2 - \int_0^1 f g_2\Big| \le\frac{\epsilon}{2}$
Also since $p_1 < p_2 \implies L^{q_1} \subset L^{q_2}$. Since by weak convergence in $L^{p_2}$ , $\{f_n\}$ is bounded and $L^{q_2}$ is dense in $L^{q_1}$ $\to\textbf{Can I claim it?}$ , Then $\exists g_2$ such that $\int_0^1 \Big|(f_n -f)(g_1 - g_2)\Big|\le \frac{\epsilon}{2}$ , $\forall n$. Hence $$\Big|\int_0^1 f_n g_1 - \int_0^1 f g_1 \Big| \le \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$$
I don't think you can claim such a statement.
There is an easier way to tackle this question. Try to conclude that $q_{1}>q_{2}$, for then $g\in L^{q_{1}}\subseteq L^{q_{2}}$, so $g$ can be a test function for the weak convergence of $f_{n}\rightarrow f$ in $L^{p_{2}}$.