Let $\Omega \subset \mathbb{R}$ be a bounded smooth domain, $V \subset L^2(\Omega)$ a Hilbertspace and $\{V_n\}$ a sequence of subspaces such that \begin{align*} V_1 \subset V_2 \subset \dots \quad \text{and} \quad \overline{\bigcup_{n \in \mathbb{N}} V_n} = V \, (\text{w.r.t. } V\text{-norm} ). \end{align*} Furthermore, there is a sequence $\phi_k \in L^\infty(\Omega)$ (uniformly bounded) and a limit $\phi \in L^\infty(\Omega)$ such that \begin{align*} \sup_{v_k \in V_k, \|v_k\| = 1} \int_\Omega (\phi_k - \phi) v_k \ dx \leq \lambda_k \to 0 \text{ as } k \to \infty \end{align*} Can we somehow conclude the convergence \begin{align*} \sup_{v \in V, \|v\| = 1} \int_\Omega (\phi_k - \phi) v \ dx \to 0 \text{ as } k \to\infty \end{align*} and if yes, how can we prove this result? Cf. also this question.

  • $\begingroup$ Does $\Omega$ have finite measure? Without it I'm not sure your integrals are well defined, since you are using the $L^2$-Norm for $V$, right? $\endgroup$ – humanStampedist Jul 4 '18 at 8:52
  • $\begingroup$ Yes. I forgot to mention this. $\endgroup$ – FredTheBread Jul 4 '18 at 9:27
  • $\begingroup$ Do you assume $\phi_k\to \phi$ in $L^\infty(\Omega)$? I could not tell if "a limit" refers to convergence in $L^\infty$ or to convergence in the sense described in the question; I assumed the latter. $\endgroup$ – user357151 Jul 4 '18 at 18:42

No. Consider a one-dimensional domain $\Omega = (0, 1)$ for simplicity. Let $\phi_k(x) = \exp(2\pi i k x)$ and $\phi(x)=0$. Let $V_n$ be the span of $\{\phi_k : k < n\}$. Then
$$\sup_{v_k \in V_k, \|v_k\| = 1} \int_\Omega (\phi_k - \phi) v_k \ dx = \sup_{v_k \in V_k, \|v_k\| = 1} \int_\Omega \phi_k v_k \ dx = 0 $$ by orthogonality. But if the same supremum is taken over $v\in V$ (where $V$ is $\overline{\bigcup V_n}$), then by virtue of $\phi_k\in V$ we have $$\sup_{v \in V, \|v\| = 1} \int_\Omega \phi_k v \ dx \ge \int_\Omega \phi_k \phi_k \ dx = 1 $$


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