I have a sequence $a_n$ which is bounded in a separable Hilbert space and I know that there is a subsequence $n_j$ with $$a_{n_j} \rightharpoonup a$$ and I also have for all $k$ integer that $$a_{n_j + k} \to a.$$
Does this imply that $a_n$, the original sequence, has the weak limit $a$?
Consider the integer sequences $n_j = 2^j$ and $m_j = 3 \cdot 2^j$. Further, consider the Hilbert Space $\Bbb{R}$ and the sequence $$a_n = \begin{cases} 1 & \text{if } \exists j \in \Bbb{N} : n = m_j \\0 & \text{otherwise.} \end{cases}$$ Fix $k \in \Bbb{N}$ and consider the set $\{n_j + k\}_{j=1}^\infty$. I claim that it has a finite intersection with $\{m_j\}_{j=1}^\infty$.
Why? Simply count the number of $1$s in the binary digits. The digits of $m_j$ are always two consecutive $1$s, followed by a number of $0$s. For large enough $j$ (in particular, for $j$ such that $j > \log_2 k$), the digits of $n_j + k$ will be a single $1$ digit, followed by a number of $0$s, then the binary digits for $k$. If we pick $j > \log_2 k + 1$, then this guarantees at least one $0$ between consecutive $1$ digits, meaning the resulting number cannot belong to $\{m_j\}_{j=1}^\infty$.
It therefore follows that $a_{n_j + k}$ is eventually the $0$ sequence. However, the sequence $a_n$ does not converge to $0$, even weakly, as it has a subsequence converging to a different limit.
So no, we cannot conclude that $a_n$ converges weakly from the given hypotheses.