A surprising (to me) consequence of Hahn-Banach is that when a sequence converges weakly then there is another sequence made of (finite) convex combination which converges in norm (to the same element).


In particular, this has a consequence in the context of $\ell^p$ spaces (the space of $p$-summable sequences, with $1 < p < \infty$). Assume there is a sequence of elements $\lbrace x\_n \rbrace$ (with $x\_n \in \ell^r$ for all $n$ and $1 < r < p$) which converges weakly (in the sense of $\ell^p$) to an element $x \in \ell^r$. Under the extra hypothesis that $\|x\_n\|\_{\ell^r} \leq K$ (for some $K \in \mathbb{R}$), the above theorem implies that there will be another sequence (given by finite convex combinations) which converges in $\ell^r$.


$\textbf{Question:}$ Suppose a sequence $\lbrace y\_n \rbrace$ converges in $\ell^p$ norm to $y$, where $y\_n$ and $y \in \ell^r$ (again with $1 < r < p < \infty$). Let $Y = \overline{ \textrm{Span} \lbrace y\_n \rbrace }^{\ell^r}$ be the closure of the space generated by the $y\_n$s in $\ell^r$. Does $y \in Y$? Since this is probably false (but I would be happy with a counter-example), are there conditions (weaker than boundedness of the sequence) which may allow to reach the same conclusion?



Weak convergence implying norm convergence
No. In $\ell\_r \oplus\_r \ell\_r$ let $y\_n = (z\_n, x\_n)$ where $x\_n$ are disjoint, $\|x\_n\|\_r=4^n$, $\|x\_n\|\_p \to 0$, $\|z\_n\|\_r = 1$, and $\|z\_n-y\|\_r \to 0$ for some non zero $y$. 


