open mapping theorem

[[concept]]

Open Mapping Theorem

Let $B_{1}, B_{2}$ be two Banach spaces and let $T \in B (B_{1}, B_{2})$ be surjective. Then $T$ is an open map.

ie for all open subsets $U \subset B_{1}$ , we have $T (U) \subset B_{2}$ is open.

Proof

Note

We will first prove that if $B (0, 1) = {b \in B_{1} : | | b | | < 1}$ , then $T (B (0, 1))$ contains an open ball in $B_{2}$ centered at $0$ . Then, since $T$ is linear, we can scale everything properly to show that for any $b_{1} \in U \subset B_{1}$ open, there exists $ϵ > 0$ such that $B (T (b_{1}), ϵ) \subset T (U)$

Showing that the closure of the image

T (B (0, 1))

contains an open ball

B (0, r)

$T$ surjective means everything in $B_{2}$ gets mapped to by something in $B_{1}$ ie

B_{2} = ⋃_{n \in N} \overset{―}{T (B (0, n))}

So by the Baire Category Theorem, there exists some $n_{0} \in N$ such that $\overset{―}{T (B (0, n_{0}))}$ contains an open ball.

Now, since $T$ is a linear operator, $n_{0} \overset{―}{T (B (0, 1))}$ contains an open ball. ie, $\overset{―}{T (B (0, 1))}$ contains an open ball. Equivalently, there exists some $v_{0} \in B_{2}$ and $r > 0$ such that $B (v_{0}, 4 r) \subset \overset{―}{T (B (0, 1))}$ .

ie, we can pick some $u_{1} \in B (0, 1)$ such that $v_{1} = T (u_{1}) \in T (B (0, 1))$ such that $| | v_{0} - v_{1} | | < 2 r$

Note

we choose $4$ here to make some calculations easier later

Then

B (v_{1}, 2 r) \subset B (v_{0}, 4 r) \subset \overset{―}{T (B (0, 1))}

If $| | v | | < r$ , then

\frac{1}{2} (2 v + v_{1}) \in \frac{1}{2} B (v_{1}, 2 r) \subset \frac{1}{2} \overset{―}{T (B (0, 1))} = \overset{―}{T (B (0, \frac{1}{2}))}

And this means

\begin{aligned} v = - T (\frac{u_{1}}{2}) + \frac{1}{2} (2 v + v_{1}) & \in - T (\frac{u_{1}}{2}) + \overset{―}{T (B (0, \frac{1}{2}))} \\ = \overset{―}{T \underset{\subset B (0, 1) since u_{1} \in B (0, 1)}{\underset{⏟}{(- \frac{u_{1}}{2} + B (0, \frac{1}{2}))}}} \\ \subset \overset{―}{T (B (0, 1))} \end{aligned}

ie, (in $B_{2}$ ), we have $B (0, r) \subset \overset{―}{T (B (0, 1))}$ . This implies that

B (0, 2^{- n} r) = 2^{- n} B (0, r) \subset 2^{- n} \overset{―}{T (B (0, 1))} = \overset{―}{T (B (0, 2^{- n}))}

for all $n \in N$

Now we prove that $B (0, \frac{r}{2}) \subset T (B (0, 1))$ .

Showing that the image

T (B (0, 1))

contains an open ball

B (0, \frac{r}{2})

Let $| | v | | < \frac{r}{2}$ . Then $v \in \overset{―}{T (B (0, \frac{1}{2}))}$ . This implies there exists some $b_{1} \in B (0, \frac{1}{2})$ such that $| | v - T b_{1} | | < \frac{r}{4}$

ie, $v - T b_{1} \in \overset{―}{T (B (0, \frac{1}{2}))} ⟹$ there is some $b_{2} \in B (0, \frac{1}{4})$ such that

| | v - T b_{1} - T b_{2} | | = | | v - T (b_{1} - b_{2}) | | < \frac{r}{4} \cdot \frac{1}{2} = \frac{r}{8}

continuing in this manner, we obtain a sequence ${b_{k}}_{k \in N} \subset B_{1}$ such that

$| | b_{k} | | < 2^{- k}$
$| | v - \sum_{k = 1}^{n} T b_{k} | | < 2^{- n - 1} r$

The series $\sum_{k = 1}^{\infty} b_{k}$ is absolutely summable (since the norm of the sums are bounded). And since $B_{1}$ is Banach, this means there exists some $b \in B_{1}$ such that $b = \sum_{k = 1}^{\infty} b_{k}$

Further, by the triangle inequality, we have

| | b | | = lim_{n \to \infty} | | \sum_{k = 1}^{n} b_{k} | | \leq lim_{n \to \infty} \sum_{k = 1}^{\infty} | | b_{k} | |

And we can bound this

= \sum_{k = 1}^{\infty} | | b_{k} | | < \sum_{k = 1}^{\infty} 2^{- k} = 1

And since $T$ is a bounded linear operator (and thus continuous), we have

T b = lim_{n \to \infty} T (\sum_{k = 1}^{n} b_{k}) = lim_{n \to \infty} \sum_{k = 1}^{\infty} T b_{k} = v

ie, $v \in T (B (0, 1))$ . Thus $B (0, \frac{r}{2}) \subset T (B (0, 1))$ in $B_{2}$ .

Thus, if $0$ was an interior point of $U$ in the domain, then it is still an interior point in the image of $U$ .

Generalizing to any open set

Suppose $U \subset B_{1}$ is open and $b_{2} = T b_{1} \in T (U)$ . Then there exists some $ϵ > 0$ such that $b_{1} + B (0, ϵ) = B (b_{1}, ϵ) \subset U$ . And, from above, there exists some $δ > 0$ such that $B (0, δ) \subset T (B (0, 1))$ . ie

\begin{aligned} B (b_{2}, ϵ δ) & = b_{2} + ϵ B (0, δ) \\ \subset b_{2} + ϵ T (B (0, 1)) \\ = T (b_{1}) + ϵ T (B (0, 1)) \\ = T (b_{1} + ϵ B (0, 1)) \\ = T (b_{1} + B (0, ϵ)) \\ \subset T (U) \end{aligned}

Since $b_{1} + B (0, ϵ)$ is contained in $U$ . Thus we've found a ball around any $b_{2} \in T (U)$ .

\begin{matrix} ◼ \end{matrix}

Note

closed graph theorem $⟺$ open mapping theorem

References

Mentions

File	Last Modified
bijective bounded linear operators have bounded linear inverses	2025-11-09
closed graph theorem	2025-11-09
Functional Analysis Lecture 4	2025-11-09

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