# 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 $T:X\longrightarrow Y$ 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. $\square$

## 0.0.1 Remark

It is usually of great importance to know if a bounded operator $T:X\longrightarrow Y$ 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 $\tilde{y}$ ”near” $y_{0}$. Then, if $T^{-1}$ is continuous, the correspondent solutions $x_{0}$ and $\tilde{x}$ are also ”near” since

 $\|x_{0}-\tilde{x}\|=\|T^{-1}y_{0}-T^{-1}\tilde{y}\|\leq\|T^{-1}\|\|y_{0}-% \tilde{y}\|$

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

Title bounded inverse theorem BoundedInverseTheorem 2013-03-22 17:30:51 2013-03-22 17:30:51 asteroid (17536) asteroid (17536) 11 asteroid (17536) Corollary msc 47A05 msc 46A30 inverse mapping theorem