I think Ady's question has a negative answer.
Recall that the Mazur map, T, from the unit ball of ell2 onto the unit ball of ell1, is defined by
T(sumaiei)=suma2ibiei, where bi is the sign of ai. It is a uniform homeomorphism in the norm topologies and obviously is coordinatewise continuous, which means that it is weak (= weak∗ in ell2) to weak∗ continuous (since on the unit ball of ellp with p finite the
weak* topology is the topology of coordinatewise convergence).
T is not weak to weak continuous since the unit vector basis converges weakly to zero in ell2 but not in ell1.
The problem is to extend T to all of ell2. I think that the easiest way to do this is to show that there is a retraction R from ell2 onto its unit ball which is both norm to norm continuous and weak to weak continuous. Define R on the complement of the unit ball by
R(sumaiei)=sumni=1aiei+ten+1,
where sumni=1a2ile1<sumn+1i=1a2i and t is chosen to have the same sign as an+1 and to make the image vector have norm one. R is obviously continuous (even Lipschitz) in the norm topology and is continuous in the topology of coordinatewise convergence, hence is weak to weak continuous (since in all of ell2, the weak topology is stronger than the topology of coordinatewise convergence).
To get a counterexample that maps one dual space to itself, work with the space ell2oplusell1.
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