# bounded inverse theorem

The next result is a corollary of the open mapping theorem^{}. It is often called the bounded inverse theorem or the inverse mapping theorem.

Theorem - Let $X,Y$ be Banach spaces^{}. Let $T:X\u27f6Y$ be an invertible bounded operator^{}. Then ${T}^{-1}$ is also .

Proof : $T$ is a surjective continuous operator between the Banach spaces $X$ and $Y$. Therefore, by the open mapping theorem, $T$ takes open sets to open sets.

So, for every open set $U\subseteq X$, $T(U)$ is open in $Y$.

Hence ${({T}^{-1})}^{-1}(U)$ is open in $Y$, which proves that ${T}^{-1}$ is continuous, i.e. bounded^{}. $\mathrm{\square}$

## 0.0.1 Remark

It is usually of great importance to know if a bounded operator $T:X\u27f6Y$ has a bounded inverse. For example, suppose the equation

$$Tx=y$$ |

has unique solutions $x$ for every given $y\in Y$. Suppose also that the above equation is very difficult to solve (numerically) for a given ${y}_{0}$, but easy to solve for a value $\stackrel{~}{y}$ ”near” ${y}_{0}$. Then, if ${T}^{-1}$ is continuous, the correspondent solutions ${x}_{0}$ and $\stackrel{~}{x}$ are also ”near” since

$$\parallel {x}_{0}-\stackrel{~}{x}\parallel =\parallel {T}^{-1}{y}_{0}-{T}^{-1}\stackrel{~}{y}\parallel \le \parallel {T}^{-1}\parallel \parallel {y}_{0}-\stackrel{~}{y}\parallel $$ |

Therefore we can solve the equation for a ”near” value $\stackrel{~}{y}$ instead, without obtaining a significant error.

Title | bounded inverse theorem |
---|---|

Canonical name | BoundedInverseTheorem |

Date of creation | 2013-03-22 17:30:51 |

Last modified on | 2013-03-22 17:30:51 |

Owner | asteroid (17536) |

Last modified by | asteroid (17536) |

Numerical id | 11 |

Author | asteroid (17536) |

Entry type | Corollary |

Classification | msc 47A05 |

Classification | msc 46A30 |

Synonym | inverse mapping theorem |