Open Mapping and Closed Graph Theorems — Mathematics Knowledge System

Overview

The open mapping theorem (Banach–Schauder theorem) and the closed graph theorem are two of the three fundamental principles of functional analysis, alongside the Hahn–Banach theorem. The open mapping theorem states that a surjective bounded linear operator between Banach spaces is an open map—it sends open sets to open sets. The closed graph theorem provides a complementary criterion: a linear operator between Banach spaces is bounded if and only if its graph is closed. Both theorems rest on the Baire category theorem applied to the Banach space structure and have profound consequences for operator theory and PDE analysis.

Intuition

The open mapping theorem says: if you can hit every target (surjective), then you can do so with uniform efficiency—nearby targets are hit by nearby inputs. The closed graph theorem says: if whenever $x_n \to x$ and $Tx_n \to y$ we must have $y = Tx$ (the graph is closed), then $T$ is automatically bounded. This is remarkable: a natural continuity condition on the graph replaces the need to directly verify boundedness.

Formal Definition

Definition

A map $T: X \to Y$ is open if $T(U)$ is open in $Y$ for every open $U \subset X$. The graph of $T$ is $\Gamma(T) = \{(x, Tx) : x \in X\} \subset X \times Y$.

T(U) \text{ open in } Y \quad \text{for every open } U \subset X

Open map

\Gamma(T) = \{(x, Tx) : x \in X\} \subset X \times Y

Graph of T

Proofs

Open Mapping Theorem (sketch via Baire)

$Y = \bigcup_{n=1}^\infty T(\overline{B_X(0,n)}) = \bigcup_{n=1}^\infty n \cdot T(\overline{B_X(0,1)})$ (Surjectivity of $T$.)
$\text{By Baire, some } n_0 \cdot \overline{T(B_X(0,1))} \text{ has nonempty interior in } Y.$ ($Y$ is a complete metric space; Baire category theorem.)
$\text{Hence } \overline{T(B_X(0,1))} \text{ contains an open ball } B_Y(y_0, r).$ (Scaling and homogeneity.)
$B_Y(0, r/2) \subset T(B_X(0,1)) \text{ (passing from closure to image using completeness of } X\text{)}.$ (Standard Cauchy series argument exploiting completeness of $X$.)

Properties

Quotient map is open

The canonical quotient map \pi: X \to X/M onto a closed subspace is open.

Condition: Immediate from the open mapping theorem applied to the quotient.

Isomorphism criterion

T \in \mathcal{B}(X,Y) is a topological isomorphism iff it is bijective.

Condition: Requires both X and Y to be Banach spaces.

Theorems

Theorem 4.1: Open Mapping Theorem (Banach–Schauder)

Let X and Y be Banach spaces and T \in \mathcal{B}(X,Y) be surjective. Then T is an open map; in particular, there exists \delta > 0 such that B_Y(0,\delta) \subset T(B_X(0,1)).

Theorem 4.2: Bounded Inverse Theorem

If T \in \mathcal{B}(X,Y) is a bijection between Banach spaces, then T^{-1} \in \mathcal{B}(Y,X). Equivalently, T is a topological isomorphism.

Theorem 4.3: Closed Graph Theorem

Let X and Y be Banach spaces and T: X \to Y a linear operator. If the graph \Gamma(T) is closed in X \times Y (with product norm), then T is bounded.

Worked Examples

Bound: $|Tf(t)| \le \int_0^1 |f(s)|\,ds \le \|f\|_\infty$, so $\|T\| \le 1$. Boundedness confirmed.
$\|Tf\|_\infty \le \|f\|_\infty$
Every function in the range of $T$ is differentiable (in fact absolutely continuous with $F'(0)=0$). The constant function $g \equiv 1$ is not differentiable at 0 when expressed as an indefinite integral of a continuous function, but more simply: $Tf(0) = 0$ for all $f$, so $g \equiv 1$ is not in the range.
$Tf(0) = \int_0^0 f = 0 \quad \forall f$
Since $T$ is not surjective, the open mapping theorem does not apply; indeed $T$ is injective but its inverse is the differentiation operator, which is unbounded.

Answer: $T$ is bounded but not surjective; the open mapping theorem's hypothesis fails and the inverse is unbounded.

Practice Problems

Difficulty 8/10

Two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a vector space $X$ make $X$ a Banach space. If $\|x\|_1 \le C\|x\|_2$ for all $x$, prove that the norms are equivalent.

Difficulty 9/10

Explain why the closed graph theorem fails if $Y$ is not complete. Give a concrete counterexample.

Difficulty 9/10

Let $T: X \to Y$ be a bounded bijection between Banach spaces. Show directly (without quoting Theorem 4.2) that $\|T^{-1}\|$ is finite.

Common Mistakes

Common Mistake

The open mapping theorem holds for all bounded operators

Surjectivity is essential. The inclusion $\ell^1 \hookrightarrow \ell^2$ is bounded and injective but not surjective, and its range is not open.

Common Mistake

A map with closed graph is always continuous

This requires both domain and codomain to be Banach spaces (or at least that the domain is Baire and the codomain is metrizable). The example of differentiation on $C^1$ with the $L^\infty$ norm shows the theorem fails without completeness.

Quiz

Summary

The open mapping theorem: a surjective bounded linear operator between Banach spaces sends open sets to open sets.
Equivalently, there exists $\delta > 0$ such that $B_Y(0,\delta) \subset T(B_X(0,1))$.
The bounded inverse theorem is the corollary: a bijective bounded linear operator between Banach spaces has a bounded inverse.
The closed graph theorem: a linear operator $T: X \to Y$ between Banach spaces is bounded if and only if its graph $\{(x,Tx)\}$ is closed in $X \times Y$.
All three fundamental principles (Hahn–Banach, open mapping, uniform boundedness) ultimately rely on the Baire category theorem.

References

BookRudin, W. Functional Analysis. 2nd ed., McGraw-Hill, 1991. Chapter 2.
BookKreyszig, E. Introductory Functional Analysis with Applications. Wiley, 1978.
WebsiteWikipedia — Open mapping theorem (functional analysis)