# Bounded inverse theorem

In mathematics, the bounded inverse 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.

It is necessary that the spaces in question be Banach spaces. 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

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

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

$x^{(n)} = \left( 1, \frac1{2}, \dots, \frac1{n}, 0, 0, \dots \right)$

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

$x^{(\infty)} = \left( 1, \frac1{2}, \dots, \frac1{n}, \dots \right),$

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

The completion of X is the space $c_0$ of all sequences that converge to zero, which is a (closed) subspace of the p space(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

$x = \left( 1, \frac12, \frac13, \dots \right),$

is an element of $c_0$, but is not in the range of $T:c_0\to c_0$.

## References

• Renardy, Michael and Rogers, Robert C. (2004). An introduction to partial differential equations. Texts in Applied Mathematics 13 (Second edition ed.). New York: Springer-Verlag. p. 356. ISBN 0-387-00444-0. (Section 8.2)