How to prove that weakly convergence in $L^{p_2}$ implies weakly convergence in $L^{p_1}$ where $1\leq p_1\leq p_2<\infty$?

Question

As the title suggested, suppose that $1\leq p_1\leq p_2<\infty$. Let $\{f_n\}$ be a sequence of functions in $L^{p_2}([0,1])$ and $f\in L^{p_2}([0,1])$. Show that $f_n \rightharpoonup f$ in $L^{p_2}([0,1])$ implies $f_n \rightharpoonup f$ in $L^{p_1}([0,1])$. And I assume that the converse is false. Is there any counterexample? If not, how to prove it?

Answer 1

Suppose $1/q_i+1/p_i=1$, so $q_1 \ge q_2$.

(a) By the Riesz representation theorem, the assumption that $f_n \rightharpoonup f$ weakly in $L^{p_2}([0,1])$ means that $$\int_0^1 f_n g \,dx \to \int_0^1 f_n g \,dx \quad (*) $$ for all $g \in L^{q_2}([0,1])$. Since $L^{q_1}([0,1]) \subset L^{q_2}([0,1])$, $(*)$ holds for all $g \in L^{q_1}([0,1])$, so $f_n \rightharpoonup f$ weakly in $L^{p_1}([0,1])$ as well.

(b) On the other hand, suppose that $1 \le p_1<p_2$. Fix $p \in (p_1,p_2)$, which implies that $q:=p/(p-1)$ is in the interval $(q_2,q_1).$ Let $$g(x):=x^{-1/q} \quad \text{and note that} \quad g \in L^{q_2}([0,1])\,.$$ The functions $\{f_n\}_{n \ge 1}$, defined by $$f_n(x)=x^{-1/p}{\bf 1}_{[e^{-n},e^{1-n}]}(x)$$ are all in $L^{p_2}([0,1])$. Moreover, the sequence $\{f_n\}_{n \ge 1}$ converges strongly to $0$ in $L^{p_1}([0,1])$, since $$\|f_n\|_{p_1}^{p_1} \le e^{np_1/p-1} \to 0 \,$$ as $n \to \infty$. However, $$\int_0^1 f_n g\, dx =\int_{e^{-n}}^{e^{1-n}} \frac{dx}{x} =1 \,,$$ so the sequence $\{f_n\}_{n \ge 1}$ does not converge weakly to $0$ in $L^{p_2}([0,1])$. It cannot converge weakly to another limit in $L^{p_2}([0,1])$ by part (a).