A subsequence of a weakly convergent sequence $x_n$ is strongly convergent to $x$, then the sequence $x_n$ is strongly convergent to $x$?

Question

If $x_n$ in a normed space $X$ is weakly convergent to $x$, and if there is a subsequence $y_k$ of it such that this subsequence is strongly convergent to $x$, then can we say that the sequence $x_n$ is strongly convergent to $x$?

Answer 1

Hint: Take $(x_n)_n$ and $(y_n)_n$ such that $x_n$ strongly converges to $x$ but $y_n$ only weakly converges to $y$. Then the sequence $(z_n)_n$, with $z_{2n}=x_n$, $z_{2n+1}=y_n$, does not strongly converge.

Answer 2

No: let $x_n$ be defined in the following way. Consider a sequence $\left(y_n\right)_{n\geqslant 0}$ which converges weakly to $0$, but not strongly and let $x_{2n}:=y_n$ and $x_{2n+1}:=0$.

Answer 3

Take the classical example of a Hilbert space with $e_n$ infinite countable ONB. Define a sequence $a_{2n+1} = e_n$ and $a_{2n}=0$. Then $a_{2n}\to 0$ strongly and $a_n$ only weakly.