# Open Mapping Theorem

Recall that a mapping $T:X→Y$, 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 $T:X→Y$ 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 $mB_{X}$, $m∈N$, cover $X$. Since $T$ is surjective, their images $mTB_{X}$ cover $Y$. This remains true, if we take closures: $⋃mTB_{X} =Y$. 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 from the Baire Category Theorem that $mTB_{X} $ has nonempty interior for some $m$. Thus, there are $q∈Y$ and $α>0$ such that $q+αB_{Y}⊂mTB_{X} $. Choose a $p∈X$ 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 $αB_{Y}⊂T(mB_{X}−q) $. Since $mB_{X}−q$ is a bounded set, it is contained in a ball $βB_{X}$ for some $β>0$. Thus, $αB_{Y}⊂TβB_{X} =βTB_{X} $. With $γ:=α/β>0$ we obtain $γB_{Y}⊂TB_{X} $.

Clearly, every $y∈γB_{Y}$ is the limit of a sequence $(Tx_{n})$, where $x_{n}∈B_{X}$. 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 $Ts_{n}→y$. To construct $(s_{n})$, we recursively define a sequence $(y_{k})$ with $y_{k}∈2_{−k}γB_{Y}$ for $k∈N_{0}$. The sequence starts with $y_{0}:=y∈2_{0}γB_{Y}$. Given $y_{k}∈2_{−k}γB_{Y}$, one has $y_{k}∈T2_{−k}B_{X} $. By the definition of closure, there exists a $x_{k}∈2_{−k}B_{X}$ such that $Tx_{k}$ lies in the open $2_{−(k+1)}γ$-ball about $y_{k}$. This means that $y_{k+1}:=y_{k}−Tx_{k}∈2_{−(k+1)}γB_{Y}$. Now define $s_{n}$ as the $n$-th partial sum of the series $∑_{k=0}x_{k}$. The series converges, because it converges absolutely (Here we use the completeness of $X$). The latter is true because $∑∥x_{k}∥≤∑2_{−k}=3$. This also shows that each $s_{n}$ and the limit $x:=lims_{n}$ lie in $4B_{X}$. The auxiliary sequence $(y_{n})$ converges to $0$ by construction. Therefore, in the limit $n→∞$

as desired. It follows from the continuity of $T$ that $Ts_{n}→Tx$, thus $Tx=y$.

In the preceding paragraph it was proven that $γB_{Y}⊂4TB_{X}$. Hence, $δB_{Y}⊂TB_{X}$ where $δ:=γ/4$. To show that $T$ is open, consider any open set $U⊂X$. If $y$ lies in $TU$, there exists a $x∈U$ such that $Tx=y$. Since $U$ is open, there is an $ϵ>0$ such that $x+ϵB_{X}⊂U$. Applying $T$, we find $y+ϵTB_{X}⊂TU$. Combine with $δB_{Y}⊂TB_{X}$ to see $y+ϵδB_{X}⊂TU$. Hence, $TU$ is open. This shows that $T$ is indeed an open mapping.

Conversely, suppose that $T$ is open. TODO $□$

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*.

Let $T:X→Y$ be a bounded linear operator between Banach spaces and suppose that $T$ is injective, so that the inverse $T_{−1}:R(T)→X$ is defined on the range of $T$. The linear operator $T_{−1}$ is bounded if and only if $R(T)$ is closed in $X$.