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Consider a separable Hilbert space $(H, \left<\cdot, \cdot\right>)$, and denote with $\mathcal{T}^s$ and $\mathcal{T}^w$ the topology induced by the norm and the weak topology, respectively. Is it true that if a compact (that is, continuous (not necessarily bounded) which sends bounded sets into relatively compact set) operator $T: H \to H$ is continuous from $(H, \mathcal{T}^s)$ to $(H, \mathcal{T}^s)$, then is continuous as a map from $(H, \mathcal{T}^w)$ to $(H, \mathcal{T}^w)$?

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Closely related question:… – Nate Eldredge Oct 7 '12 at 16:16
A compact linear operator is necessarily bounded (which is equivalent to being norm-norm continuous). Did you mean something else? Also, a (linear) operator between normed spaces is norm-norm continuous if and only if it is weak-weak continuous (see here, e.g.). – David Mitra Oct 7 '12 at 16:18
Thank you! But I'm interested in the non-linear case! – Francis Oct 7 '12 at 18:36

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