I am trying to prove the following:
Let X be a normed linear space satisfying the property: ∀{xn},{yn}⊆X, we have
‖xn‖=‖yn‖=1,‖xn+yn‖→2⇒‖xn−yn‖→0.
If {zn}⊆X converges to z∈X weakly (meaning limn→∞f(zn)=z for all f∈X∗) and ‖zn‖→‖z‖, then ‖zn−z‖→0.
Here is what I am trying to do:
I can consider {z} as a sequence in X. I want to show that ‖zn+z‖→2. Well, since ‖zn‖→‖z‖=1, then since ‖zn+z‖≤‖zn‖+‖z‖, then limn→∞‖zn+z‖≤2‖z‖=2.
I can't figure out how to possibly show that limn→∞‖zn+z‖≥2. How would I even incorporate the weak convergence assumption? Any help would be greatly appreciated! Thank you.
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