Isomorphisms Between Banach Spaces

Recall from The Open Mapping Theorem page the Open Mapping theorem which states that if $X$ and $Y$ are Banach spaces and $T : X \to Y$ is a bounded linear operator then the range $T(X)$ is closed if and only if $T$ is an open mapping.

We now present a very important corollary which tells us exactly when a bounded linear operator $T : X \to Y$ is an isomorphism when $X$ and $Y$ are both Banach spaces.

Corollary 1: Let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a bounded linear operator. Then $T$ is an isomorphism if and only if $T$ is bijective.

Recall from the Isomorphism Linear Operators on Normed Linear Spaces page that an isomorphism from $X$ to $Y$ is a bijective bounded linear operator $T : X \to Y$ such that $T^{-1}$ is also bounded or equivalently, $T$ is a bijective continuous function such that $T^{-1}$ is continuous.

• Proof: $\Rightarrow$ Suppose that $T$ is an isomorphism. Then by definition, $T$ is bijective.

• $\Leftarrow$ Suppose that $T$ is bijective. Then $T(X) = Y$ is a closed subspace of $Y$. So by the Open Mapping theorem we have that $T$ is an open mapping.

• But this implies that $T^{-1}$ is continuous.

• So $T : X \to Y$ is a bijection that is continuous with continuous inverse. Hence $T$ is an isomorphism from $X$ to $Y$. $\blacksquare$