# Does weak convergence in $W^{1,p}$ imply strong convergence in $L^q$?

Does weak convergence in $W^{1,p}(\Omega)$ imply strong convergence in $L^q(\Omega)$ when $\Omega$ is bounded?

If $f_j$ converges weakly to $f$ in $W^{1,p}$, what can we say about the $L^q$ convergence? or at least, are the $L^p$ norms of the $f_j$ bounded?

#### Solutions Collecting From Web of "Does weak convergence in $W^{1,p}$ imply strong convergence in $L^q$?"

In the case $q<p^*$ the Rellichâ€“Kondrachov theorem applies, if your domain $\Omega$ has the required properties.

If $q=p^*$, the $L^q$ norm is still controlled by the $W^{1,p}$ norm; and since a weakly convergent sequence is norm-bounded, $L^q$ norm is bounded. Moreover, you will have weak convergence in $L^q$, since linear maps preserve weak convergence. But it will not be strong in general. For example, pick $1<p<n$ and consider the sequence $f_k(x) = k^{\frac{n}{p}-1} (1-k|x|)^+$. This is a bounded sequence in $W^{1,1}(B^n)$, where $B^n$ is the $n$-dimensional unit ball. It converges weakly in $W^{1,p}$, specifically to zero. (Indeed, it converges to $0$ in the sense of distributions; on the other hand $W^{1,p}$ is reflexive, so a weakly convergent subsequence can be extracted; the weak limit has to be the same $0$, so the whole sequence converges weakly to $0$.) On the other hand, the $L^q$ norm for $q=p^*=np/(n-p)$ stays the same for all $k$.