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 $m{B}_X$, $m\in \mathbb{N}$, cover $X$. Since $T$ is surjective, their images $mT{B}_X$ cover $Y$. This remains true, if we take closures: $\bigcup \overline{mT{B}_X}=Y$. 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 $\overline{mT{B}_X}$ has nonempty interior for some $m$. Thus, there are $q\in Y$ and $\alpha >0$ such that $q+\alpha B_Y\subset \overline{mT{B}_X}$. 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 $\alpha B_Y\subset \overline{T(m{B}_X-q)}$. Since $m{B}_X-q$ is a bounded set, it is contained in a ball $\beta B_X$ for some $\beta >0$. Thus, $\alpha B_Y\subset \overline{T\beta B_X}=\beta \overline{T{B}_X}$. With $\gamma :=\alpha /\beta >0$ we obtain $\gamma B_Y\subset \overline{T{B}_X}$.

Clearly, every $y\in \gamma B_Y$ is the limit of a sequence $(T{x}_n)$, 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 $T{s}_n\to y$. To construct $(s_n)$, we recursively define a sequence $(y_k)$ with $y_k\in 2^{-k}\gamma B_Y$ for $k\in {\mathbb{N}}_0$. 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 $T{x}_k$ lies in the open $2^{-(k+1)}\gamma$-ball about $y_k$. This means that $y_{k+1}:=y_k-T{x}_k\in 2^{-(k+1)}\gamma B_Y$. 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 $\sum \Vert x_k\Vert \le \sum 2^{-k}=3$. 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 $n\to \infty$

$$T{s}_n=\sum _{k=0}^{n}T{x}_k=\sum _{k=0}^n{y}_k-y_{k+1}=y_0-y_{n+1}\to y_0=y,$$

as desired. It follows from the continuity of $T$ that $T{s}_n\to Tx$, thus $Tx=y$.

In the preceding paragraph it was proven that $\gamma B_Y\subset 4T{B}_X$. Hence, $\delta B_Y\subset T{B}_X$ 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 $ϵ>0$ such that $x+ϵB_X\subset U$. Applying $T$, we find $y+ϵT{B}_X\subset TU$. Combine with $\delta B_Y\subset T{B}_X$ to see $y+ϵ\delta B_X\subset TU$. Hence, $TU$ is open. This shows that $T$ is indeed an open mapping.

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

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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 $T:X\to Y$ 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 $X$.