In mathematics, the bounded inverse theorem ( also called inverse mapping theorem or Banach isomorphism theorem) is a result in the theory of bounded linear operators on Banach spaces. It states that a bijective bounded linear operator T from one Banach space to another has bounded inverse T⁻¹. It is equivalent to both the open mapping theorem and the closed graph theorem.

Generalization

Counterexample

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

${\displaystyle Tx=\left(x_{1},{\frac {x_{2)){2)),{\frac {x_{3)){3)),\dots \right)}$

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

${\displaystyle x^{(n)}=\left(1,{\frac {1}{2)),\dots ,{\frac {1}{n)),0,0,\dots \right)}$

converges as n → ∞ to the sequence x^(∞) given by

${\displaystyle x^{(\infty )}=\left(1,{\frac {1}{2)),\dots ,{\frac {1}{n)),\dots \right),}$

which has all its terms non-zero, and so does not lie in X.

The completion of X is the space ${\displaystyle c_{0))$ of all sequences that converge to zero, which is a (closed) subspace of the ℓ_p 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

${\displaystyle x=\left(1,{\frac {1}{2)),{\frac {1}{3)),\dots \right),}$

is an element of ${\displaystyle c_{0))$, but is not in the range of ${\displaystyle T:c_{0}\to c_{0))$.