Showing that a sequence converges in norm I am trying to prove the following:
Let $X$ be a normed linear space satisfying the property:  $\forall \left\{x_n\right\},  \left\{y_n\right\} \subseteq X $, we have 
$\|x_n\|=\|y_n\|=1, \|x_n+y_n\|\rightarrow 2 \Rightarrow \|x_n-y_n\|\rightarrow 0.$
If $\left\{z_n\right\} \subseteq X$ converges to $z\in X$ weakly (meaning $\displaystyle \lim_{n\rightarrow \infty} f(z_n)=z$ for all $f\in X^*$) and $\|z_n\| \rightarrow \|z\|$, then $\|z_n-z\|\rightarrow 0$.
Here is what I am trying to do:
I can consider $\left\{z \right\}$ as a sequence in $X$. I want to show that $\|z_n+z\|\rightarrow 2$. Well, since $\|z_n\| \rightarrow \|z\|=1$, then since $\|z_n+z\|\leq\|z_n\|+\|z\|$, then $\displaystyle \lim_{n\rightarrow \infty} \|z_n+z\| \leq 2\|z\|=2$. 
I can't figure out how to possibly show that $\displaystyle \lim_{n\rightarrow \infty} \|z_n+z\| \geq 2$. How would I even incorporate the weak convergence assumption? Any help would be greatly appreciated! Thank you.
 A: Assume without loss of generality that $\|z\|=1$. You have
$$
\|z_n+z\|=\sup\{|f(z_n+z)|:\ f\in X^*,\ \|f\|=1\}.
$$
Let $\varepsilon> 0$. Then there exists $f\in X^*$ with $\|f\|=1$ and $f(z)>1-\varepsilon/4$. For all $n$ big enough,  $|f(z_n)-f(z)|<\varepsilon/2$. So $$|f(z_n+z)|=|f(z_n)+f(z)|=|f(z_n)-f(z)+2f(z)|> 2f(z)-\varepsilon/2> 2-\varepsilon. 
$$
Thus $\|z_n+z\|> 2-\varepsilon$, implying that $\lim\|z_n+z\|=2$. Now a little more work is needed, because possibly $\|z_n\|\ne1$ for some or all $n$. But as $\|z_n\|\to1$, this is easily arranged by using $z_n/\|z_n\|$.
A: To fix the fact that $\|z_n\|$ might not be $1$, as in Martin Argerami's answer: suppose that $z\neq 0$, so that, for large $n$, $\|z_n\|\geq\varepsilon$ for some fixed $\varepsilon$. Set $y_n=\frac{z_n}{\|z_n\|}$, and $y=\frac{z}{\|z\|}$. Then, if $f\in X^*$, $$|f(y_n-y)|=\left|f\left(\frac{z_n\|z\|-z\|z_n\|}{\|z_n\|\|z\|}\right)\right|\leq\frac{1}{\varepsilon\|z\|}|f(z_n\|z\|-z\|z_n\|)|\leq$$$$\frac{1}{\varepsilon}|f(z_n-z)|+\frac{|\|z_n\|-\|z\||}{\varepsilon\|z\|}|f(z)|,$$ which converges to $0$. Then, apply Martin Argerami's argument to $y_n$ and $y$.
A: Use the norm identity 
$$ \|x\|=\sup_{\substack{f\in X^*,\\ \|f\|_*=1}}|f(x)|$$
Then for a given $n$,
$$ \| z_n - z\| = \sup_{\substack{f\in X^*,\\ \|f\|_*=1}}|f(z_n - z)| \ge |f(z_n-z)|$$
for every $f \in X^\ast$. In particular, by the Hahn-Banach theorem there exists a linear functional $f_n \in X^\ast$ with the property $f_n(z_n -z) = 2$. To see this let $u$ be any element in $X$ not in the closure of the linear span  of $z,z_n$. Then the space $U$ generated by $u$ is a subspace of $X$ such that $z_n,z$ are not in its closure. Hence there exists a linear functional $f'_n$ with $f'_n(z_n-z) = 1$ and hence $f_n(x) = 2f'_n(x)$ is a linear functional with $f_n(z_n-z) = 2$. Hence for every $n$, $\|z_n -z\|\ge 2$ and hence $\lim \|z_n-z\| \ge 2$.
