# Can weak convergence in $V$ imply strong convergence in $H$?

In the proof of the existence of strong solutions of the stationary NSE in the setting of Hilbert spaces, the following argument is made in Constantin and Foias's Navier-Stokes Equations (p60):

Let $\Omega$ be an open bounded subset of $\mathbb{R}^n$, $n=2$ or $3$. Define $V=\overline{\mathcal{V}}^{H_0^1(\Omega)^n}$ and $H=\overline{\mathcal{V}}^{L^2(\Omega)^n}$ where $$\mathcal{V}=\{f\in C_0^\infty(\Omega)^n\mid \textrm{div }f=0\}.$$ Let $u_m$ be a bounded sequence in $V$. Then there exists a subsequence $u_{m'}$ which is weakly convergent in $V$ to some $u\in V$. Passing to a subsequence if necessary, we may assume that $u_{m'}\to u$ in $H$ strongly.

Here is my question:

Why can weak convergence in $V$ imply strong convergence of a subsequence in $H$?

• Isn't $V$ compactly embedded into $H$? – Pedro Feb 28 '16 at 2:50

It is not that "weak convergence in $V$ implies strong convergence in $H$", but that a compact operator maps bounded sets to relatively compact sets.
Once one knows that $V$ is compactly embedded in $H$ (by the Rellich–Kondrachov theorem), $u_{m'}$ has a strongly convergent subsequence in $H$ since it is bounded in $V$.