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