The Open Mapping Theorem

# The Open Mapping Theorem

Recall from the Open and Closed Mappings page that if $X$ and $Y$ are topological spaces then a function $f : X \to Y$ is said to be an open mapping if for every open set $U$ in $X$ the image, $f(U)$ is an open set in $T(X)$.

We are now ready to prove the very important Open Mapping theorem.

 Theorem 1 (The Open Mapping Theorem): Let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a bounded linear operator. Then the range $T(X)$ is closed if and only if $T$ is an open mapping.
• Proof: From the theorem on the Second IFF Criterion for the Range of a BLO to be Closed when X and Y are Banach Spaces page, we have 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 there exists a positive constant $M' \in \mathbb{R}$, $M' > 0$ such that for all $x \in X$ with $\| T(x) \| < 1$ we have that there exists an $x' \in X$ such that $T(x) = T(x')$ and $\| x' \| < M'$.
• So if $\displaystyle{r = \frac{1}{M'} > 0}$ then equivalently, $T(X)$ is closed if and only if for every $x \in X$ with $\| T(x) \| < r$ there exists an $x' \in X$ such that $T(x) = T(x')$ and $\| x' \| < 1$.
• $\Rightarrow$ Suppose that $T(X)$ is closed. Then the condition above holds. Let $U$ be any open set in $X$. We want to show that $T(U)$ is an open set in $T(X)$. Now since $Y$ is open in $X$, for every $u \in U$ there exists an open ball centered at $u$ fully contained in $U$. In other words, there exists an $r_u > 0$ such that:
(1)
\begin{align} \quad B(u, r_u) \subseteq U \end{align}
• Let $x \in X$ be such that:
(2)
\begin{align} \quad \| T(x - u) \| = \| T(x) - T(u) \| < rr_u \end{align}
• Then by the condition above, there exists an $x' \in X$ such that $T(x') = T(x - u) = T(x) - T(u)$ and $\| x' \| < r_u$. Therefore:
(3)
\begin{align} \quad T(x) = T(x') + T(u) = T(x' + u) \quad , \quad \| (x' + u) - u \| = \| x' \| < r_u \end{align}
• So $(x' + u) \in B(u, r_u)$. So $T(x) = T(x' + u) \in T(B(u, r_u))$. But since $B(u, r_u) \subseteq U$ we have that:
(4)
\begin{align} \quad T(B(u, r_u)) \subseteq T(U) \end{align}
• Therefore:
(5)
\begin{align} \quad T(u) \in B(T(u), rr_u) \cap T(X) \subseteq T(U) \end{align}
• Hence $T(U)$ is open in $T(X)$. So for every open set $U$ in $X$ we have that $T(U)$ is open in $T(X)$. Hence $T$ is an open mapping.
• $\Leftarrow$ Suppose that $T$ is an open mapping. Let $B_X$ denote the open unit ball in $X$. Then $T(B_X)$ is open in $T(X)$. Since $0 \in T(B_X)$, there exists an $r > 0$ such that:
(6)
\begin{align} \quad B(0, r) \cap T(X) \subseteq T(B_X) \end{align}
• So if $\| T(x) \| < r$ then we have that:
(7)
\begin{align} \quad T(x) \in B(0, r) \cap T(X) \subseteq T(B_X) \end{align}
• So there exists an $x' \in B_X$ such that $T(x) = T(x')$ and $\| x' \| < 1$. So $T(X)$ is closed. $\blacksquare$