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 $T:X⟶Y$ 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. $\mathrm{\square }$

0.0.1 Remark

It is usually of great importance to know if a bounded operator $T:X⟶Y$ 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

$$\parallel x_0-\tilde{x}\parallel =\parallel T^{-1}{y}_0-T^{-1}\tilde{y}\parallel \le \parallel T^{-1}\parallel \parallel y_0-\tilde{y}\parallel$$

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