A counterexample to convergence in $B^2$ implies convergence in $L^2_\text{loc}$

In regards to this question, I feel I can produce a complicated counterexample as follows. I wonder if I have made a mistake in this argument.

1. We know that $$m+n\sqrt{2}$$ is dense in $$\mathbb{R}$$ as $$m,n$$ vary over the integers. Therefore, let $$s_n=(a_n+b_n\sqrt{2})$$ be a sequence of numbers which converges to $$1$$.

2. Then the function $$f_n(x)=\exp(is_nx)$$ converges pointwise to the function $$f(x)=\exp(ix)$$.

3. Now, there is a subsequence of $$f_n$$ which converges weakly in the topology of $$B^2(\mathbb{R}^d)$$. Let us call this limit as $$g$$. This is due to the fact that $$B^2(\mathbb{R}^d)$$ is a Hilbert space, therefore, it is reflexive. As a result, any bounded sequence has a weakly convergent subsequence. The non-separability of $$B^2(\mathbb{R}^d)$$ is not an issue since we can always choose to work with the closed subspace of $$B^2(\mathbb{R}^d)$$ generated by the terms of the sequence. A proof is given here.

4. Then there is a further subsequence of $$f_n$$ whose Cesaro means converge to $$g$$ strongly in the norm of $$B^2(\mathbb{R}^d)$$. This too is a general property of Hilbert spaces, given a weakly convergent sequence, one can find a subsequence whose Cesaro means converge in the norm. A proof is given here.

5. The sequence of Cesaro means converges in $$B^2(\mathbb{R}^d)$$ therefore it also converges in $$L^2_\text{loc}(\mathbb{R}^d)$$.

6. Any sequence that converges in $$L^2_\text{loc}(\mathbb{R}^d)$$ has a subsequence which converges pointwise almost everywhere.

7. However, the original sequence $$f_n$$ converges pointwise to $$f$$, therefore $$f=g$$.

8. The above line of argument shows that the entire sequence $$f_n$$ converges weakly to $$f$$ in $$B^2(\mathbb{R}^d)$$. The argument is the following: We started with the sequence $$f_n$$ and showed that every weakly convergent subsequence has a further subsequence that weakly converges to $$f$$. Here is a proof.

9. But, the norms of $$f_n$$, $$||f_n||_{B^2}=1$$ converges to the norm of $$f$$, $$||f||_{B^2}=1$$. As a consequence, the sequence $$f_n$$ converges strongly to $$f$$ in the norm of $$B^2(\mathbb{R}^d)$$. See here.

But this is clearly impossible because the sequence $$f_n$$ has no convergent subsequences on account of the distance between each of them being positive.

I feel that my argument is correct but I am not able to find the flaw. I would be thankful for any help.

• @mathworker21 Yes, the distance between any distinct two of them is $\sqrt{2}$ due to orthogonality. Sep 8 '19 at 18:08
• @mathworker21 I have added more information for points 3, 4, 8. Sep 9 '19 at 6:07
• Also, isn't there a quotienting issue, cause the seminorm is a seminorm, not a norm. Sep 9 '19 at 7:10
• @mathworker21 Yes, there is a quotienting issue, but I have not thought about it much, because I felt that it might be ok to work with a representative element. I will think about it. Sep 9 '19 at 7:13
• See my answer below. If you object to the proposed definition of the title of your question, I'll delete my answer. Sep 9 '19 at 7:38

I think the quotienting is the issue. You first have to clarify what you mean by "convergence in $$B^2$$ implies convergence in $$L^2_{loc}$$". Does that mean "if $$[f_n] \to [f]$$ in $$B^2$$, then there are $$\phi_n,\phi \in L^2_{loc}$$ with $$||\phi_n||_{B^2} = ||\phi||_{B^2} = 0$$ and $$f_n+\phi_n \to f+\phi$$ in $$L^2_{loc}$$", where $$[f]$$ denotes the equivalence class of $$f$$ in the quotiented $$B^2$$? If so, then when you go from Cesaro convergence to $$g$$ in $$B^2$$ to Cesaro convergence to $$g$$ in $$L^2_{loc}$$, you are no longer looking at the Cesaro average of $$f_{n_j}$$, but rather of $$f_{n_j}+\phi_{n_j}$$ for some $$(\phi_{n_j})_j$$ and thus can no longer claim that $$g+\phi$$ is equal to $$f$$, since we don't know what the a.e. pointwise Cesaro average of $$f_{n_j}+\phi_{n_j}$$ is.