Let us consider the sequence space $c\_0$ with the equivalent norm
$$\Vert x \Vert^2 = \max\_{i\ge1} \vert x^i \vert^2 + \sum\_{i=2}^{\infty} 2^{-i+1} \vert x^i \vert^2 $$
for $x=(x^1,x^2,\ldots)\in c\_0$.


Let us take two sequences $(x\_{2n})$ and $(x\_{2n+1})$ in $c\_0$ such that $x\_{2n} \xrightarrow{w} x$ and $x\_{2n+1} \xrightarrow{w} y$ and $I\_n:=2\Vert x\_{2n} \Vert^2 + 2 \Vert x\_{2n+1} \Vert^2 - \Vert x\_{2n} + x\_{2n+1} \Vert^2 =0$ for all $n$. Does it then follow that $x=y$?


This question can be restated simply as follows: Take any $x$ and $y$ in $c\_0$ such that $I(x,y):=2\Vert x \Vert^2 + 2 \Vert y \Vert^2 - \Vert x+y \Vert^2 =0$. Does it then follow that $x=y$?


Or even more simply: Take any $x$ and $y$ in $c\_0$ such that $\Vert x+y \Vert=\|x\|+\|y\|$. Does it then follow that $x$ and $y$ are linearly dependent?



Renorming on a separable Banach space
The answer is no. E.g., let $x\_{2n}=x:=(\frac12,\frac12,\frac13,\frac14,\ldots)$ and $x\_{2n+1}=y:=(0,\frac12,\frac13,\frac14,\ldots)$ (for all $n$). Then (i) $x\_{2n}=x\to x$ and $x\_{2n+1}=y\to y$ in any sense and (ii) $\|x\_{2n}+x\_{2n+1}\|=2\|x\_{2n}\|=2\|x\_{2n+1}\|$ and hence $I\_n=0$ for all $n$. However, $x\ne y$. $\quad\Box$


