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Let H be an infinite dimensional real Hilbert space.

A [not necessarily linear] mapping of H into itself is said to be hemicontinuous if it is continuous from each line segment of H to the weak topology of H (F.E. Browder / G.J. Minty). [Obviously, any linear operator is hemicontinuous.] Intuitively speaking, this is an extremely weak continuity requirement, still very useful, e.g., in the study of nonlinear elliptic boundary value problems.

Now, here is my problem. Let U be an arbitrary (i.e., possibly discontinuous) selfmap of H. Is it true that there exists some hemicontinuous selfmap of H, say V, such that U 2 = V 2 ?

share|improve this question
    
Is it true in finite dimensional spaces? –  Andrey Rekalo May 20 '10 at 2:58
    
It is false on the real axis. –  Ady May 20 '10 at 3:09
    
No. Hemicontinuous maps are too few in finite-dimensional spaces (just continuum cardinality) because they are completely defined by their values at points with rational coordinates. –  fedja May 20 '10 at 3:16
    
An exquisite remark. –  Ady May 20 '10 at 3:30
    
By line segment, do you mean any convex hull of two points? –  Kevin Ventullo May 21 '10 at 7:34
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