Weak convergence in $L^q([0,1]) \subset L^p([0,1])$.

Question

Suppose that $1 \leq p < q < \infty$ so that $L^q([0,1]) \subset L^p([0,1])$.

I want to show that for $f_n \in L^q([0,1])$, $f_n \to f$ weakly in $L^q \iff f_n \to f$ weakly in $L^p$.

The forward direction is clear from the reverse-inclusion of the dual space (e.g. see here).

I'd appreciate any help showing that the reverse direction is true.

Answer 1

Assume that $f_n\to f$ weakly in $\mathbb L^p$. We would like to find a condition which ensures that $f_n\to f$ weakly in $\mathbb L^q$. First of all, we should have that $f_n\in\mathbb L^q$, which is not automatic. Second, a weakly convergent sequence in a Banach space is bounded, hence we should also assume that $\sup_n\left\lVert f_n\right\rVert_q$ is finite, as Giuseppe Negro suggest.

So a not hopeless statement could be:

Let $1\lt p<q\lt +\infty$ and let $\left(f_n\right)_{n\geqslant 1}$ be a sequence which converges weakly to $f$ in $\mathbb L^p$ and such that $\sup_n\left\lVert f_n\right\rVert_q<+\infty$. Then $f_n\to f$ in $\mathbb L^q$.

Under these assumptions, a subsequence $\left(f_{n_k}\right)_{k\geqslant 1}$ converges in $\mathbb L^q$ to some $g$. By the implication proved in the opening post, we have that $f_{n_k}\to g$ weakly in $\mathbb L^p$ hence $f=g$ and we deduce that $f\in\mathbb L^q$. For the moment, we have proven that a subsequence of $\left(f_n\right)_{n\geqslant 1}$ converges to $f$ weakly in $\mathbb L^q$. Nothing prevents us from applying to a subsequence of $\left(f_n\right)_{n\geqslant 1}$. Therefore, we get that for each subsequence of $\left(f_n\right)_{n\geqslant 1}$, we can extract a further subsequence which converges weakly to $f$ in $\mathbb L^q$.