Editing Reflexive space (section)

=== Examples ===

# Every finite-dimensional normed space is reflexive, simply because in this case, the space, its dual and bidual all have the same linear dimension, hence the linear injection <math>J</math> from the definition is bijective, by the [[rank–nullity theorem]].
# The Banach space [[Sequence space#c and c0|<math>c_0</math>]] of scalar sequences tending to 0 at infinity, equipped with the supremum norm, is not reflexive.  It follows from the general properties below that [[Sequence space#.E2.84.93p spaces|<math>\ell^1</math> and <math>\ell^{\infty}</math>]] are not reflexive, because <math>\ell^1</math> is isomorphic to the dual of <math>c_0</math> and <math>\ell^{\infty}</math> is isomorphic to the dual of <math>\ell^1.</math>
# All [[Hilbert space]]s are reflexive, as are the [[Lp space]]s <math>L^p</math> for <math>1 < p < \infty.</math> More generally: all [[uniformly convex space|uniformly convex]] Banach spaces are reflexive according to the [[Milman–Pettis theorem]].  The <math>L^1(\mu)</math> and <math>L^{\infty}(\mu)</math> spaces are not reflexive (unless they are finite dimensional, which happens for example when <math>\mu</math> is a measure on a finite set).  Likewise, the Banach space <math>C([0, 1])</math> of continuous functions on <math>[0, 1]</math> is not reflexive.
# The spaces <math>S_p(H)</math> of operators in the [[Schatten class operator|Schatten class]]  on a Hilbert space <math>H</math> are uniformly convex, hence reflexive, when <math>1 < p < \infty.</math>  When the dimension of <math>H</math> is infinite, then <math>S_1(H)</math> (the [[trace class]]) is not reflexive, because it contains a subspace isomorphic to <math>\ell^1,</math> and <math>S_{\infty}(H) = L(H)</math> (the bounded linear operators on <math>H</math>) is not reflexive, because it contains a subspace isomorphic to <math>\ell^{\infty}.</math>  In both cases, the subspace can be chosen to be the operators diagonal with respect to a given orthonormal basis of <math>H.</math>