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Bill Johnson
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Self-dual finite-dimensional complex normed spaces

Suppose $X$ is a complex normed space of dimension 2 or 3 and $X$ is isometrically isomorphic to its dual. Is $X$ a Hilbert space?

Remarks: There are easy counterexamples in the real case, and in higher dimensions one can construct counterexamples from sums of 2-dimensional spaces which are not isometric to their duals. Similarly a 3-dimensional counterexample can be constructed from a 2-dimensional counterexample.

Mark Meckes
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