Ref:

• Functional Analysis lectures by Casey Rodriguez. Lecture on the Open Mapping Theorem. Link to the lecture: Link.

• “Introductory Functional Analysis” by Kreyszig.

Let ${ B _1, B _2 }$ be Banach spaces. Let ${ T \in \mathcal{B}(B _1, B _2) }$ be a continuous linear operator.
Suppose ${ T }$ is bijective. Is ${ T }$ a toplinear isomorphism?

It suffices to show ${ T ^{-1} }$ is continuous, that is ${ T }$ is an open map.

It turns out surjectivity itself guarantees ${ T }$ is an open map.

Theorem [Open Mapping Theorem]

Let ${ B _1, B _2 }$ be Banach spaces. Let ${ T \in \mathcal{B}(B _1, B _2) }$ be a bounded linear operator.
Suppose ${ T }$ is surjective. Then ${ T }$ is an open map.

Proof: We will first study the image of the unit ball, ${ T(B(0,1)) . }$ We will show every point in the image is an interior point. We will show ${ 0 }$ is an interior point.

Note that by surjectivity

\[{ B _2 = \bigcup _{n} \overline{T(B(0,n))} . }\]

Hence by Baire Category Theorem, there is a ${ \overline{T(B(0,n _0))} }$ containing an open ball.

Note that

\[{ \overline{T(B(0,n _0))} = n _0 \overline{T(B(0,1))} . }\]

Hence ${ \overline{T(B(0,1))} }$ contains an open ball ${ B(v _0, 4r) . }$

We will now show

\[{ \boxed{\text{To show:} \quad B(0, r) \subseteq \overline{T(B(0,1))}} . }\]

Since ${ v _0 \in \overline{T(B(0,1))} ,}$ there exists a point ${ v _1 \in B(v _0, 2r) \cap T(B(0,1)). }$ Note that there exists a ${ u _1 \in B(0,1) }$ such that ${ T(u _1) = v _1 . }$

Note that

\[{ B(v _1, 2r) \subseteq B(v _0, 4r) \subseteq \overline{T(B(0,1))}. }\]

If ${ v \in B(0, r) , }$ note that the operated point

\[{ (2v + v _1) \in B(v _1, 2r) . }\]

Hence if ${ v \in B(0, r) , }$ note that

\[{ {\begin{aligned} &\, \frac{1}{2} (2v + v _1) \\ \in &\, \frac{1}{2} B(v _1, 2r) \\ \subseteq &\, \frac{1}{2} \overline{T(B(0,1))} \\ = &\, \overline{T(B(0, 2 ^{-1}))} . \end{aligned}} }\]

We used the fact that closure respects translation and scalar multiplication of sets.

Hence if ${ v \in B(0, r), }$ note that

\[{ {\begin{aligned} &\, v \\ = &\, - \frac{v _1}{2} + \frac{1}{2} (2v + v _1) \\ \in &\, T\left(- \frac{u _1}{2}\right) + \overline{T(B(0, 2 ^{-1}))} \\ = &\, \overline{T\left(- \frac{u _1}{2} + B(0, 2 ^{-1}) \right)} \\ \subseteq &\, \overline{T(B(0, 1))} . \end{aligned}} }\]

Hence

\[{ B(0, r) \subseteq \overline{T(B(0, 1))} }\]

as needed.

We will now show

\[{ \boxed{\text{To show:} \quad B(0, \frac{r}{2}) \subseteq T(B(0,1)) }. }\]

Let ${ v \in B(0, \frac{r}{2}) . }$

Note that

\[{ v \in \frac{1}{2} B(0, r) \subseteq \overline{T(B(0, 2 ^{-1}))} . }\]

Hence there exists a ${ b _1 \in B(0, 2 ^{-1}) }$ such that ${ T(b _1) \in B(v, r 2 ^{-2}) . }$

Hence

\[{ v - T (b _1) \in B(0, r 2 ^{-2}) \subseteq \overline{T(B(0, 2 ^{-2}))} . }\]

Hence there exists a ${ b _2 \in B(0, 2 ^{-2}) }$ such that ${ T(b _2) \in B( v - T(b _1), r2 ^{-3}) .}$

And so on.

Hence we get a sequence ${ (b _n) }$ such that

• ${ b _n \in B(0, 2 ^{-n}) . }$

• ${ v - T(b _1) - \ldots - T(b _n) \in B(0, r 2 ^{-(n+1)}) . }$

Note that ${ \sum b _n }$ is absolutely summable, hence ${ \sum b _n }$ is summable. Hence we can consider

\[{ b = \sum _{n = 1} ^{\infty} b _n . }\]

Note that

\[{ {\begin{aligned} &\, \lVert b \rVert \\ = &\, \lim _{n \to \infty} \left\lVert \sum _{k = 1} ^{n} b _k \right\rVert \\ \leq &\, \lim _{n \to \infty} \sum _{k = 1} ^{n} \lVert b _k \rVert \\ \leq &\, \lim _{n \to \infty} \sum _{k = 1} ^{n} 2 ^{-k} \\ = &\, 1 . \end{aligned}} }\]

Hence

\[{ b \in B(0, 1) . }\]

Note that

\[{ {\begin{aligned} &\, T(b) \\ = &\, \lim _{n \to \infty} \sum _{k = 1} ^{n} T(b _k) \\ = &\, v . \end{aligned}} }\]

Hence

\[{ v = T(b) \in T(B(0,1)) . }\]

Hence

\[{ v \in B(0, \frac{r}{2}) \implies v \in T(B(0,1)) . }\]

Hence

\[{ B(0, \frac{r}{2}) \subseteq T(B(0,1)) }\]

as needed.

We will now show

\[{ \boxed{\text{To show:} \quad T \text{ is an open map.}} }\]

Let ${ U \subseteq B _1 }$ be open. Let ${ b _2 = T(b _1) \in T(U) . }$

We are to show there is an open ball around ${ b _2 }$ contained in ${ T(U) . }$

Note that there is a ball

\[{ b _1 + B(0, \varepsilon) \subseteq U . }\]

Note that there is a ball

\[{ B(0, \delta) \subseteq T(B(0,1)) . }\]

Consider the ball

\[{ b _2 + B(0, \varepsilon \delta) . }\]

Note that

\[{ {\begin{aligned} &\, b _2 + B(0, \varepsilon \delta) \\ = &\, b _2 + \varepsilon B(0, \delta) \\ \subseteq &\, b _2 + T(B(0, \varepsilon)) \\ = &\, T(b _1 + B(0, \varepsilon)) \\ \subseteq &\, T(U) . \end{aligned}} }\]

Hence

\[{ b _2 + B(0, \varepsilon \delta) \subseteq T(U) }\]

as needed.