This theorem may not hold for normed spaces that are not complete.
For example, consider the space X of sequences x : N → R with only finitely many non-zero terms equipped with the supremum norm. The map T : X → X defined by
is bounded, linear and invertible, but T−1 is unbounded.
This does not contradict the bounded inverse theorem since X is not complete, and thus is not a Banach space.
To see that it's not complete, consider the sequence of sequences x(n) ∈ X given by
converges as n → ∞ to the sequence x(∞) given by
which has all its terms non-zero, and so does not lie in X.
The completion of X is the space of all sequences that converge to zero, which is a (closed) subspace of the ℓp space ℓ∞(N), which is the space of all bounded sequences.
However, in this case, the map T is not onto, and thus not a bijection. To see this, one need simply note that the sequence
is an element of , but is not in the range of .