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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 ?

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  • $\begingroup$ Is it true in finite dimensional spaces? $\endgroup$ Commented May 20, 2010 at 2:58
  • $\begingroup$ It is false on the real axis. $\endgroup$
    – Ady
    Commented May 20, 2010 at 3:09
  • $\begingroup$ 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. $\endgroup$
    – fedja
    Commented May 20, 2010 at 3:16
  • $\begingroup$ An exquisite remark. $\endgroup$
    – Ady
    Commented May 20, 2010 at 3:30
  • $\begingroup$ By line segment, do you mean any convex hull of two points? $\endgroup$ Commented May 21, 2010 at 7:34

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