my question is the following.

Given a sequence $(f_k)_k$ in $W^{1,q}(\Omega)$ with $q \in (1,\infty)$ and $\Omega \subseteq \mathbb{R}^n$ open and bounded. If I want to show $f_k$ converges in $W^{1,q}(\Omega)$ weakly, is it then sufficient to show that the corresponding sequence of regular distributions $([f_k])_k$ converges in $D'(\Omega)$, the locally convex space of distributions? I tend to say yes since the testfunctions $D(\Omega)$ are dense in $W^{1,q}(\Omega)$ and by that $W^{1,q}(\Omega)'$ is dense in $D'(\Omega)$. But I'm not quite sure. Is anyone familiar with stuff like this?

  • $\begingroup$ If the sequence $\{f_k\}$ is additionally bounded in $W^{1,q}(\Omega)$, the density argument is ok. If it is not bounded, it does not converge weakly. $\endgroup$ – gerw May 27 '16 at 17:26
  • $\begingroup$ oh sorry I forgot to write that I want to apply this to bounded sequences as you said. $\endgroup$ – CandyOwl May 27 '16 at 17:28
  • $\begingroup$ Note that $D(\Omega)$ is not dense in $W^{1,q}(\Omega)$ but merely in $W_0^{1,q}(\Omega)$. $\endgroup$ – gerw May 27 '16 at 17:43
  • $\begingroup$ sure many thanks for reminding me! $\endgroup$ – CandyOwl May 27 '16 at 17:53

Let us assume that $f_k$ converges to $F \in D'(\Omega)$ in the sense that $$\int_\Omega f_k \, v \, \mathrm{d}x \to F(v) \qquad\forall v \in D(\Omega).$$

Now, since $W^{1,q}(\Omega)$ is reflexive, you get $f \in W^{1,q}(\Omega)$ and a subsequence such that $f_{n_k} \rightharpoonup f$ in $W^{1,q}(\Omega)$. In particular, $f_{n_k} \rightharpoonup f$ in $L^1(\Omega)$ and this shows $$F(v) = \int_\Omega f \, v \, \mathrm{d}x \qquad\forall v \in D(\Omega).$$

It remains to show the convergence of the whole sequence. Suppose that $f_k \not\rightharpoonup f$. Then, there is $\varepsilon > 0$, $\Phi \in W^{1,q}(\Omega)'$ and a subsequence with $|\Phi(f_{m_k} - f) | > \varepsilon$ for all $k$. However, since $\{f_{m_k}\}$ is bounded, it has a weakly convergent subsequence and its weak limit has to be $f$ (same argument as above). This is a contradiction. Hence, the whole sequence converges.


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