While reading Complex and Real Analysis, I was surprised by Lemma 7.23 [1]: its conclusion reminded me of the Open Mapping Theorem (Theorem 5.9 [1]). Both results assert that, under certain conditions, the image of a map contains a full ball in its codomain. Interestingly, a similar phenomenon appears in Theorem 13.2 of Apostol’s Mathematical Analysis [2].

Although these theorems come from senarios, they share a common geometric theme: continuity (and additional hypotheses) forces the image to be “thick” enough to fit an open ball. This recurring pattern hints at a deeper structural property of continuous maps, and it is tempting to compare these results side by side.

We begin by a conclusion from Brouwer’s fix point theorem.

Continuous function with open mapping ([1], Lemma 7.23)

Lemma 1. Let $\partial B=\{x\in \mathbb{R}^k:|x|=1\}$ be sphere in $\mathbb{R}^k$, and $B$ be an open unit ball in $\mathbb{R}^k$ and so $\overline{B}=B\cup\partial B$. If $F:\overline{B}\rightarrow \mathbb{R}^k$ is continuous, $0<\epsilon<1$, and
$$
|F(x)-x|<\epsilon, \quad \text{if}\quad x\in \partial B \tag{1}
$$
Then $F(B)\supseteq B(0, 1-\epsilon)$.
Remark. This theorem is essentially saying that for every continuous map that the boundary of $B(0, 1)$ only differs by $\epsilon$, then $B(0, 1-\epsilon)$ can fit in the image $F(B)$.
Proof. By way of contradiction, suppose $a\in B(0, 1-\epsilon)$ but $a\notin F(B)$. For each $x\in \partial B$, $|x|-|F(x)|\leq|F(x)-x|<\epsilon$ so that $|F(x)|>1-\epsilon>|a|$. This show that $a\notin F(\overline{B})$. Define the map $G:\overline{B}\rightarrow \overline{B}$ by
$$
G(x)=\frac{a-F(x)}{|a-F(x)|}, \quad x\in \overline{B}
$$
Note that
(1) $G$ is well define since $a-F(x)\ne 0$ for each $x\in \overline{B}$.
(2) $|G(x)|=1$ so that $G(x)\in S$ for all $x\in \overline{B}$.
(3) $G$ is a continuous function.
(4) If $x\in S$, then

$$
(a - F(x)) \cdot x = a \cdot x + (x - F(x)) \cdot x - 1
$$

$$
< |a|\,|x| + |x - F(x)|\,|x| - 1
$$

$$
< (1-\epsilon) \cdot 1 + \epsilon \cdot 1 - 1 = 0
$$
Hence $G(x)\neq x$ for $x\in S$. By (2), $G(x)\neq x$ for all $x\in \overline{B}$ which contradicts Brouwer’s fixed point theorem.
The next theorem is chapter on Inverse and Implict function theorem.

Differentiable open mapping ([2], Theorem 13.2)

Theorem 2. Let $B=B(a, r)$ and $\partial B$ denotes again the boundary of $B$. Let $f=(f_1, \dots, f_n)$ be continuous function on $\overline{B}$. If

  1. All partial derivative $\frac{\partial f_j}{\partial x_i}(x)$ exists if $x\in B$.
  2. $f(a)\ne f(x)$ if $x\in \partial B$
  3. $J_f(x)\ne 0$ for each $x\in B$

Then there is $\delta>0$ such that $f(B)\supseteq B(f(a), \delta)$.
Remark. Here, the assumption is stronger than the open mapping theorem but applies to differentiable functions rather than bounded linear functional.
Note that we can assume $f(a)=0$ by letting $f(x)$ to be $f(x)-f(a)$. Also, we can translate the function so that $a=0$. Note that differentiability of $f$ is not assumed.
Proof.

  1. Define $g(x)=\|f(x)-f(a)\|_2$ , then $g(x)>0$ for $x\in \partial B$ and $g(x)$ is continuous. Since $\partial B$ is compact, then $g(x)$ take minimum

    $$
    m=\inf_{x\in \partial B}g(x)=g(x^*)>0
    $$
    Let $T=B(f(a), \frac{m}{2})$, We shall prove that $T\subseteq f(B)$ which completes the proof.

  2. Fix $y\in T$, define $h:\overline{B}\rightarrow \mathbb{R}$ by
    $$
    h(x)=\|f(x)-y\|
    $$
    which is continuous. Note that $h(a)=\|f(a)-y\|<\frac{m}{2}$ since $y\in T$. If $x\in S$,
    $$
    \|f(x)-y\|\ge \|f(x)-f(a)\|-\|y-f(a)\|\ge m-\frac{m}{2}=\frac{m}{2}
    $$
    so that the of $h$ is attained $B$, the point is call $c$. Since
    $$
    \frac{\partial}{\partial x_i}h^2( c )=2\sum_{j=1}^n(f_j( c )-y_j)\frac{\partial f_j}{\partial x_i}( c )=0, \quad i=1, \dots, n
    $$
    Then this from a system of equations,
    $$
    Df( c )(f( c )-y)=0
    $$
    Since $Df( c )\ne 0$ given that $c\in B$, we conclude that $f( c )=y$. This proves the claim.

Definition. (Open mapping) Let $X, Y$ be metric spaces, a function $f:X\rightarrow Y$ is called an open mapping if for every open set $W$ in $X$, $f(W)$ is open in $Y$.

If we assume $f$ to be differentiable, we can actually restrict to fewer conditions.

Theorem 3. If $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is differentiable and , then there is $\delta>0$ such that
$$f(B(a, r))\supseteq B(f(a), \delta)$$
Proof. Assume $a=0$ and $f(a)=0$ as remark of previous theorem. Put $f’(0)=A$. Let $g=A^{-1}f$, then $g’(0)=I$. Since $g$ is differentiable, there is $\delta_1>0$ such that $$
|g(x)-x|<\frac{1}{2}|x|
$$
If $0<r<\min\{\delta_1, 1\}$, then by Lemma 1. above with $\epsilon=\frac{1}{2}$, we have
$$
B(0, \frac{1}{2}r)\subseteq g(B(0, r))\subseteq g(B(0, 1))
$$
Note that since $A$ is invertible, then $A$ is a homeomorphism, so that $A(B(0, \frac{r}{2}))$ is open. Hence, there exists a open ball $B(0, \delta)\subseteq A(B(0, \frac{r}{2}))\subseteq Ag(B(0, 1))=f(B(0, 1))$. This completes the proof.

We now show that one-to-one plays an important role in open mapping theorems
Theorem 4. Let $A$ be an open subset of $\mathbb{R}^k$ and assume that $f:A\rightarrow \mathbb{R}^n$ is continuous and has finite partial derivatives $\frac{\partial f_i}{\partial x_j}$ on $A$. If $f$ is one-to-one on $A$ and if $J_f(x)\ne 0$ for each $x\in A$, then $f(A)$ is open.
Proof. If $b\in f(A)$, then there is some $a\in A$ such that $b=f(a)$. Since $f$ is one-to-one, there exists some ball $B$ satisfy the hypothesis of Theorem 2., hence $f(B)$ contains a ball center at $b$, hence $f(B)$ is open.

Finally, we dealt with The Open mapping theorem in general space.

Open mapping theorem

Theorem 5. Let $U$, $V$ be unit open ball in Banach spaces $X$ and $Y$. To every bouned linear transformation $\Lambda$ of $X$ onto $Y$, there corresponds a $\delta>0$ such that
$$
\Lambda(U)\supseteq \delta V
$$
Remark. This theorem essentially asserts that bounded surjective linear functional $\Lambda$ is an open mapping.
Proof.

  1. Given $y\in Y$, there exists $x\in X$ such that $\Lambda x= y$. Since $\|x\|< \infty$, there is $k\in \mathbb{N}$ such that $y\in \Lambda(kU)$. Hence $\cup_{k=1}^\infty \Lambda(kU)=Y$. By Baire’s theorem, there is non-empty open set $W\subseteq \overline{\Lambda(kU)}$ for some $k$ since $Y$ is complete. Choose $y_0\in W$, and choose $\eta>0$ so that $y+y_0\in W$ if $\|y\|<\eta$, given that $W$ is open. There exists sequence $\{x_i’\}$, $\{x_i’’\}$ in $kU$ such that
    $$
    \Lambda x_i’\rightarrow y_0, \quad \Lambda x_i’’\rightarrow y_0+y, \quad\text{as}\quad i\rightarrow \infty
    $$
    Put $x_i=x_i’’-x_i’$, then $\Lambda x_i\rightarrow y$ and $\|x_i\|\le \|x_i’’\|+\|x_i’\|<2k$. Put $\delta= \frac{\eta}{2k}$. We have shown the follwing: To each $y\in Y$ and $\epsilon >0$, there corresponds an $x\in X$ such that
    $$
    \|x\|\le \delta^{-1}\|y\|\quad \text{and}\quad \|\Lambda x -y\|<\epsilon
    $$
    This is almost the desired conclusion expect that we need $\epsilon$ to be $0$.
  2. If $|y|<\delta$, then there is $x\in X$ such that $\|x_1\|<1$ and $\|\Lambda x_1- y\|<\frac{1}{2}\epsilon\delta$. Now, apply the argument on $\Lambda x_1-y$, there is $x_2\in X$ such that $\|x_2\|<\frac{\epsilon}{2}$ and $\|y-\Lambda x_1 -\Lambda x_2\|<\frac{1}{4}\delta \epsilon$. Continue the process, we see that if $s_n=x_1+\dots+x_n$, then $$
    \|y-\Lambda s_n\|<\frac{1}{2^n}\delta \epsilon \quad \text{and}\quad \|x_n\|<\frac{1}{2^n}
    $$
    Since $\Lambda$ is continuos and $X$ is complete, then there is $s\in X$ such that $s_n\rightarrow s$ and hence $y=\Lambda s$. Note that $|s_n|\le 1+\epsilon$ so continuity of norm shows that $\|s\|\le 1+\epsilon$.
  3. We have prove that
    $$
    \Lambda ((1+\epsilon)U)\supseteq \delta V \implies \Lambda (U)\supseteq (1+\epsilon)^{-1}\delta V
    $$
    Taking union over all $\epsilon$, since $\cup_{\epsilon>0}(1+\epsilon)^{-1}\delta V=V$, we get the desire solution.

Here is another one-to-one enhancement of the open mapping theorem.
Theorem 6. If $X$ and $Y$ are Banach spaces and if $\Lambda$ is a bijective linear transformation of $X$ onto $Y$, then there is $\delta>0$ such that
$$
\|\Lambda x\|\ge \delta \|x\|
$$
That is, $\Lambda^{-1}$ is a bounded linear transformation of $Y$ onto $X$.
Remark. Let $X$ be a Banach space if there are two norms such that $\|x\|_1\le C\|x\|_2$, then they are equivalent, by taking $\Lambda =I$. This theorem is used for example, if $\|x|\|$ blows up, then $\|\Lambda x\|$ blows up and vise versa.
Proof. By previous theorem and $\Lambda$ is one-to-one there is a $\delta>0$ such that if $\|\Lambda x\|<\delta$, then $\|x\|<1$. Hence $\|x\|\ge 1$ implies that $\|\Lambda x\|\ge \delta$. This gives the desired result and $\|\Lambda^{-1}\|\le \delta^{-1}$.

Conclusion

In this post, we have dealt with the detailed proof of the three open mapping theorems. The technique of proving those theorems really differs from one to another. The first lemma uses the Brouwer fixed point theorem, which confines the deformation of the continuous map. The second theorem concerns differentiable mappings; the goal to prove a ball $B$ inside of the image is then transformed to finding the minimum of $\|f(x)-y\|$ to discover actually $f(x)=y$. The third theorem is a well-known theorem in functional analysis; the main component is to use Baire’s theorem and the completeness of the spaces. Theorem 3 (developed by the post author) is not in the references and serves as a connection between Theorem 1 and 2.

References

[1] Rudin, W.: Real and Complex Analysis, 3rd edn. McGraw-Hill Inc, USA (1987)
[2] T. M. Apostol, “Mathematical Analysis,” 3rd Edition, Addison-Wesley, Boston, 1957.