---
title: Open Mapping Theorem
parent: The Fundamental Four
grand_parent: Functional Analysis Basics
nav_order: 3
---

# {{ page.title }}

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 $Y$.

{% theorem * Open Mapping Theorem %}
A bounded linear operator between Banach spaces is open
if and only if it is surjective.
{% endtheorem %}

{% 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]({% link pages/general-topology/baire-spaces.md %})
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 $n \to \infty$

$$
Ts_n = \sum_{k=0}^{n} Tx_k = \sum_{k=0}^{n} y_k - y_{k+1}
= y_0 - y_{n+1} \to y_0 = y,
$$

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
{% endproof %}

---

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 {{ page.title }} we obtain the following:

{% corollary * Bounded Inverse Theorem %}
If a bounded linear operator between Banach spaces is bijective,
then its inverse is bounded.
{% endcorollary %}

Also known as *Inverse Mapping Theorem*.

{% corollary %}
Let $T: X \to Y$ be a bounded linear operator between Banach spaces
and suppose that $T$ is injective, so that the inverse $T^{-1} : R(T) \to X$
is defined on the range of $T$.
The linear operator $T^{-1}$ is bounded if and only if $R(T)$ is closed in $X$.
{% endcorollary %}