Suppose $U$ is bounded and $p>q$. Does $f_n\rightharpoonup f$ in $L^p(U)$ implies $f_n\to f$ in $L^q(U)$?

Question

Suppose $U$ is bounded and $p>q$. We know $L^p(U)$ is stronger then $L^q(U)$ in the sense that $f_n\to f$ in $L^p(U)$ implies $f_n\to f$ in $L^q(U)$. Can we relax the convergence on the stronger space and still retain the result? So is it the case that $f_n\rightharpoonup f$ in $L^p(U)$ implies $f_n\to f$ in $L^q(U)$?

Answer 1

At least in the case $\infty > p > 1$ this is usually not the case, since otherwise the embedding $L^p \hookrightarrow L^q$ would be compact.

To see this, let $(f_n)_n \subset L^p$ be a bounded sequence. Since $L^p$ is reflextive (this uses that $1 < p < \infty$), there is a subsequence $(f_{n_k})_k$ which weakly converges in $L^p$ to some $f \in L^p$. If your property would hold, this would imply that $f_{n_k} \to f$ strongly in $L^q$. Thus, if your property is true, then $L^p \hookrightarrow L^q$ was a compact embedding.

This compactness is not satisfied if (as I presume) you consider the Lebesgue measure on a set with positive measure.

A direct counterexample goes as follows, at least for $U \subset \Bbb{R}^d$ open and bounded: Consider $f_n(x) = \cos(n x_1)$. It is not hard to see (this is essentially the Riemann Lebesgue lemma) that if $g \in C_c^\infty(U)$, then $\int f_n (x) g(x) \, dx \to 0$. Since $C_c^\infty(U) \subset L^{p'}$ is dense (here, we use that $1 < p < \infty$), this implies $f_n \to 0$ with weak convergence in $L^p$. But $f_n \not\to 0$ in $L^q$. This latter point is slightly more clear if you consider $f_n(x) = e^{i n x_1}$, if you are willing to consider complex-valued functions.