# Open Mapping Theorem

Recall that a mapping $T : X \to Y$, where $X$ and $Y$ are topological spaces, is called *open* if the image under $T$ of each open set of $X$ is open in $Y$.

Open Mapping Theorem

A bounded linear operator between Banach spaces is open if and only if it is surjective.

**Proof:** Let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a bounded linear operator. Let $B_X$ and $B_Y$ denote the open unit balls in $X$ and $Y$, respectively.

First, suppose that $T$ is surjective. The balls $m B_X$, $m \in \NN$, cover $X$. Since $T$ is surjective, their images $mTB_X$ cover $Y$. This remains true, if we take closures: $\bigcup \overline{mTB_X} = Y$. Hence, we have written the space $Y$, which is assumed to have a complete norm, as the union of countably many closed sets. It follows form the Baire Category Theorem that $\overline{mTB_X}$ has nonempty interior for some $m$. Thus there are $q \in Y$ and $\alpha > 0$ such that $q + \alpha B_Y \subset \overline{mTB_X}$. Choose a $p \in X$ with $Tp=q$. It is a well known fact, that in a normed space the translation by a vector and the multiplication with a nonzero scalar are homeomorphisms and thus compatible with taking the closure. We conclude $\alpha B_Y \subset \overline{T(mB_X-q)}$. Since $mB_X-q$ is a bounded set, it is contained in a ball $\beta B_X$ for some $\beta > 0$. Thus, $\alpha B_Y \subset \overline{T \beta B_X} = \beta \overline{TB_X}$. With $\gamma := \alpha / \beta > 0$ we obtain $\gamma B_Y \subset \overline{TB_X}$.

Clearly, every $y \in \gamma B_Y$ is the limit of a sequence $(Tx_n)$, where $x_n \in B_X$. However, the sequence $(x_n)$ *may not converge*! We show that it is possible to find a *convergent* sequence $(s_n)$ in $4B_X$ such that $Ts_n \to y$. To construct $(s_n)$, we recursively define a sequence $(y_k)$ with $y_k \in 2^{-k} \gamma B_Y$ for $k \in \NN_0$. The sequence starts with $y_0 := y \in 2^0 \gamma B_Y$. Given $y_k \in 2^{-k} \gamma B_Y$, one has $y_k \in \overline{T 2^{-k} B_X}$. By the definition of closure, there exists a $x_k \in 2^{-k} B_X$ such that $Tx_k$ lies in the open $2^{-(k+1)} \gamma$-ball about $y_k$. This means that $y_{k+1} := y_k - Tx_k \in 2^{-(k+1)}\gamma B_Y$. Now define $s_n$ as the $n$-th partial sum of the series $\sum_{k=0}^{\infty} x_k$. The series converges, because it converges absolutely (Here we use the completeness of $X$). The latter is true because $\sum \norm{x_k} \le \sum 2^{-k} = 3$. This also shows that each $s_n$ and the limit $x := \lim s_n$ lie in $4B_X$. The auxiliary sequence $(y_n)$ converges to $0$ by construction. Therefore, in the limit $n \to \infty$

as desired. It follows from the continuity of $T$ that $Ts_n \to Tx$, thus $Tx = y$.

In the preceding paragraph it was proven that $\gamma B_Y \subset 4TB_X$. Hence, $\delta B_Y \subset TB_X$ where $\delta := \gamma/4$. To show that $T$ is open, consider any open set $U \subset X$. If $y$ lies in $TU$, there exists a $x \in U$ such that $Tx=y$. Since $U$ is open, there is an $\epsilon > 0$ such that $x+\epsilon B_X \subset U$. Applying $T$, we find $y + \epsilon TB_X \subset TU$. Combine with $\delta B_Y \subset TB_X$ to see $y + \epsilon \delta B_X \subset TU$. Hence, $TU$ is open. This shows that $T$ is indeed an open mapping.

Conversely, suppose that $T$ is open. TODO $\square\enspace$

For a bijective mapping between topological spaces, to say that it is open, is equivalent to saying that its inverse is continuous. The inverse of a bijective linear map between normed spaces is automatically linear and thus continuous if and only if it is bounded. As a corollary to the Open Mapping Theorem we obtain the following:

Bounded Inverse Theorem

If a bounded linear operator between Banach spaces is bijective, then its inverse is bounded.

Also known as *Inverse Mapping Theorem*.