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

Question

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.

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 $1$.
Then the function $f_n(x)=\exp(is_nx)$ converges pointwise to the function $f(x)=\exp(ix)$.
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.
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.
The sequence of Cesaro means converges in $B^2(\mathbb{R}^d)$ therefore it also converges in $L^2_\text{loc}(\mathbb{R}^d)$.
Any sequence that converges in $L^2_\text{loc}(\mathbb{R}^d)$ has a subsequence which converges pointwise almost everywhere.
However, the original sequence $f_n$ converges pointwise to $f$, therefore $f=g$.
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.
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. — Thangachelli Debopritama, Sep 08 '19 at 18:08
@mathworker21 I have added more information for points 3, 4, 8. — Thangachelli Debopritama, Sep 09 '19 at 06:07
Also, isn't there a quotienting issue, cause the seminorm is a seminorm, not a norm. — mathworker21, Sep 09 '19 at 07:10
@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. — Thangachelli Debopritama, Sep 09 '19 at 07:13
See my answer below. If you object to the proposed definition of the title of your question, I'll delete my answer. — mathworker21, Sep 09 '19 at 07:38

score 2 · Accepted Answer · answered Sep 09 '19 at 07:37

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.

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

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