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Let $\{n_i\}_i$ and $\{m_i\}_i$ be a sequence of numbers.

Let $X$ be a Hilbert space. I know that there are 2 sequences $x_{n_i} \to x$ and $y_{m_i} \rightharpoonup y$.

Does it follow that there is a sequence of numbers of $\{l_i\}_i$ such that $x_{l_i} \to x$ and $y_{l_i} \rightharpoonup y$?

I know this sounds obvious but I don't know how to prove it. Maybe: every subsequence of a convergent sequence converges so let us just choose some common numbers in the sets $\{n_i\}$ and $\{m_i\}$ to create the set $\{l_i\}$. But does every subsequence in a weak-convergent sequence also converge? Is this right?

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It's false: $x_n=(-1)^n=y_n$, $n_i=2i$, $m_i=2i+1$, then $x_{n_i}=1$ and $y_{m_i}=-1$. If you take $\{l_i\}_i$ such that $x_{l_i}\rightarrow 1$ then exists $i_0$ such that $x_{l_i}=1\quad\forall i\ge i_0\Rightarrow l_i$ is even $\forall i\ge i_0\Rightarrow y_{l_i}=1\quad\forall i\ge i_0\Rightarrow y_{l_i}\rightarrow1$.

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  • $\begingroup$ Thanks. Would the result be true if $x_n$ and/or $y_n$ themselves converged (weakly and/or strongly)? $\endgroup$ – sequitor Mar 12 '14 at 12:14
  • $\begingroup$ Yes: if $x_n$ converges, you take $l_i=m_i$. $\endgroup$ – blues66 Mar 12 '14 at 12:24

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