Open mapping theorem (functional analysis)

Suppose ${\displaystyle A:X\to Y}$ is a surjective continuous linear operator. In order to prove that ${\displaystyle A}$ is an open map, it is sufficient to show that ${\displaystyle A}$ maps the open unit ball in ${\displaystyle X}$ to a neighborhood of the origin of ${\displaystyle Y.}$

Let ${\displaystyle U=B_{1}^{X}(0),V=B_{1}^{Y}(0).}$ Then

$${\displaystyle X=\bigcup _{k\in \mathbb {N} }kU.}$$

Since ${\displaystyle A}$ is surjective:

$${\displaystyle Y=A(X)=A\left(\bigcup _{k\in \mathbb {N} }kU\right)=\bigcup _{k\in \mathbb {N} }A(kU).}$$

But ${\displaystyle Y}$ is Banach so by Baire's category theorem

$${\displaystyle \exists k\in \mathbb {N}$$ :\qquad \left({\overline {A(kU)}}\right)^{\circ }\neq \varnothing .}

That is, we have ${\displaystyle c\in Y}$ and ${\displaystyle r>0}$ such that

$${\displaystyle B_{r}(c)\subseteq \left({\overline {A(kU)}}\right)^{\circ }\subseteq {\overline {A(kU)}}.}$$

Let ${\displaystyle v\in V,}$ then

$${\displaystyle c,c+rv\in B_{r}(c)\subseteq {\overline {A(kU)}}.}$$

By continuity of addition and linearity, the difference ${\displaystyle rv}$ satisfies

$${\displaystyle rv\in {\overline {A(kU)}}+{\overline {A(kU)}}\subseteq {\overline {A(kU)+A(kU)}}\subseteq {\overline {A(2kU)}},}$$

and by linearity again,

$${\displaystyle V\subseteq {\overline {A(LU)}}}$$

where we have set ${\displaystyle L=2k/r.}$ It follows that for all ${\displaystyle y\in Y}$ and all ${\displaystyle \epsilon >0,}$ there exists some ${\displaystyle x\in X}$ such that

$${\displaystyle \qquad \|x\|_{X}\leq L\|y\|_{Y}\quad {\text{and}}\quad \|y-Ax\|_{Y}<\epsilon .\qquad (1)}$$

Our next goal is to show that ${\displaystyle V\subseteq A(2LU).}$

Let ${\displaystyle y\in V.}$ By (1), there is some ${\displaystyle x_{1}}$ with ${\displaystyle \left\|x_{1}\right\|<L}$ and ${\displaystyle \left\|y-Ax_{1}\right\|<1/2.}$ Define a sequence ${\displaystyle \left(x_{n}\right)}$ inductively as follows. Assume:

$${\displaystyle \|x_{n}\|<{\frac {L}{2^{n-1}}}\quad {\text{and}}\quad \left\|y-A\left(x_{1}+x_{2}+\cdots +x_{n}\right)\right\|<{\frac {1}{2^{n}}}.\qquad (2)}$$

Then by (1) we can pick ${\displaystyle x_{n+1}}$ so that:

$${\displaystyle \|x_{n+1}\|<{\frac {L}{2^{n}}}\quad {\text{and}}\quad \left\|y-A\left(x_{1}+x_{2}+\cdots +x_{n}\right)-A\left(x_{n+1}\right)\right\|<{\frac {1}{2^{n+1}}},}$$

so (2) is satisfied for ${\displaystyle x_{n+1}.}$ Let

$${\displaystyle s_{n}=x_{1}+x_{2}+\cdots +x_{n}.}$$

From the first inequality in (2), ${\displaystyle \left(s_{n}\right)}$is a Cauchy sequence, and since ${\displaystyle X}$ is complete, ${\displaystyle s_{n}}$ converges to some ${\displaystyle x\in X.}$ By (2), the sequence ${\displaystyle As_{n}}$ tends to ${\displaystyle y}$ and so ${\displaystyle Ax=y}$ by continuity of ${\displaystyle A.}$ Also,

$${\displaystyle \|x\|=\lim _{n\to \infty }\|s_{n}\|\leq \sum _{n=1}^{\infty }\|x_{n}\|<2L.}$$

This shows that ${\displaystyle y}$ belongs to ${\displaystyle A(2LU),}$ so ${\displaystyle V\subseteq A(2LU)}$ as claimed. Thus the image ${\displaystyle A(U)}$ of the unit ball in ${\displaystyle X}$ contains the open ball ${\displaystyle V/2L}$ of ${\displaystyle Y.}$ Hence, ${\displaystyle A(U)}$ is a neighborhood of the origin in ${\displaystyle Y,}$ and this concludes the proof.