From Baire Category to Openness
The Uniform Boundedness Principle used in the previous tutorial is one consequence of the Baire Category Theorem. Another fundamental consequence concerns linear maps: under the right completeness and surjectivity assumptions, a bounded linear map cannot collapse open sets into sets with empty interior. The Open Mapping Theorem makes this precise.
The key step is not initially to find exact preimages of a whole ball. Baire category first gives a ball contained in the closure of the image of a unit ball. A successive approximation argument then replaces approximate preimages by an exact one, with the preimages forming a convergent series in the domain Banach space.
For a linear map, it is enough to understand what happens near zero: translating an open ball and scaling its radius translate and scale its image. The proof below establishes the stronger local fact that the image of the open unit ball contains a ball about zero.
The Closure of the Image Contains a Ball
Proof. For each positive integer \(n\), let \(F_n=\overline{T(\overline{B_X(0,n)})}\). Each \(F_n\) is closed in \(Y\). Surjectivity implies that \(Y=\bigcup_{n=1}^{\infty}F_n\): given \(y\in Y\), choose \(x\in X\) with \(Tx=y\), and then choose a positive integer \(n\geq\|x\|_X\). Thus \(y\in T(\overline{B_X(0,n)})\subseteq F_n\).
Since \(Y\) is complete, the Baire Category Theorem implies that some \(F_N\) has nonempty interior. Choose \(y_0\in Y\) and \(r>0\) such that \(B_Y(y_0,r)\subseteq F_N\). If \(\|z\|_Y<r\), both \(y_0+z\) and \(y_0\) belong to \(F_N\). We claim that \(z\in\overline{T(\overline{B_X(0,2N)})}\). Indeed, by the definition of closure, there are sequences \(u_j,v_j\in\overline{B_X(0,N)}\) such that \(Tu_j\to y_0+z\) and \(Tv_j\to y_0\). Then \(T(u_j-v_j)\to z\), and \(\|u_j-v_j\|_X\leq\|u_j\|_X+\|v_j\|_X\leq 2N\), proving the claim.
Now let \(\delta=r/(2N)\). If \(\|w\|_Y<\delta\), then \(\|2Nw\|_Y<r\), so \(2Nw\in\overline{T(\overline{B_X(0,2N)})}\). Scaling the approximating preimages by \(1/(2N)\) shows that \(w\in\overline{T(\overline{B_X(0,1)})}\). This proves the lemma. \(\square\)
The closure in this result matters: it means that every target in the ball can be approximated by images of vectors in the closed unit ball, but it does not yet provide an exact preimage. The next argument removes that gap.
Turning Approximation into an Exact Preimage
Proof. Apply the lemma and fix \(\delta>0\) such that \(B_Y(0,\delta)\subseteq\overline{T(\overline{B_X(0,1)})}\). By scaling, for every \(a>0\), \(B_Y(0,a\delta)\subseteq\overline{T(\overline{B_X(0,a)})}\). We will show that \(B_Y(0,\delta/4)\subseteq T(B_X(0,1))\).
Take \(y\in Y\) with \(\|y\|_Y<\delta/4\), and set \(e_0=y\). We construct \(x_k\in X\) and residuals \(e_k\in Y\) recursively. Suppose that \(\|e_{k-1}\|_Y<\delta 2^{-(k+1)}\). The scaled closure inclusion, with \(a=2^{-(k+1)}\), puts \(e_{k-1}\) in \(\overline{T(\overline{B_X(0,2^{-(k+1)})})}\). Therefore we can choose \(x_k\) with \(\|x_k\|_X\leq 2^{-(k+1)}\) and \(\|e_{k-1}-Tx_k\|_Y<\delta 2^{-(k+2)}\). Define \(e_k=e_{k-1}-Tx_k\). The required bound for the next step follows, and the initial bound holds because \(\|e_0\|_Y<\delta/4\).
The series \(\sum_{k=1}^{\infty}x_k\) converges in \(X\): the sum of the norms is at most \(\sum_{k=1}^{\infty}2^{-(k+1)}=1/2\), and \(X\) is complete. Write \(x=\sum_{k=1}^{\infty}x_k\); then \(\|x\|_X\leq1/2<1\). By linearity, \(T(\sum_{k=1}^m x_k)=y-e_m\). The residual estimates imply \(e_m\to0\), and boundedness of \(T\) implies continuity, so taking limits gives \(Tx=y\). This proves the asserted ball inclusion with \(\varepsilon=\delta/4\).
To prove openness, let \(U\subseteq X\) be open and take any \(x_0\in U\). There is an \(s>0\) such that \(x_0+B_X(0,s)\subseteq U\). By the ball inclusion and scaling, \(B_Y(0,s\varepsilon)\subseteq T(B_X(0,s))\). Linearity then gives \(Tx_0+B_Y(0,s\varepsilon)\subseteq T(U)\). Thus every point of \(T(U)\) has an open ball around it contained in \(T(U)\), so \(T(U)\) is open. \(\square\)
Worked Examples and Consequences
Worked Example: A Coordinate Projection Is Open
Let \(X\) and \(Z\) be Banach spaces, equip \(X\times Z\) with the maximum norm \(\|(x,z)\|=\max(\|x\|_X,\|z\|_Z)\), and define \(P:X\times Z\to X\) by \(P(x,z)=x\). The product is Banach, and \(P\) is linear and bounded because \(\|P(x,z)\|_X=\|x\|_X\leq\max(\|x\|_X,\|z\|_Z)\). It is surjective since \(P(x,0)=x\).
The Open Mapping Theorem therefore says that \(P\) maps open sets to open sets. In this example the ball calculation can also be checked directly: \(P(B_{X\times Z}((x_0,z_0),r))=B_X(x_0,r)\). One inclusion follows from \(\|x-x_0\|_X<r\) for any \((x,z)\) in the product ball. For the other, if \(\|x-x_0\|_X<r\), then \((x,z_0)\) belongs to the product ball and projects to \(x\). Every open set is a union of open balls, so its image under \(P\) is open.
Proof. The Open Mapping Theorem says that \(T\) maps open sets to open sets. Since \(T\) is bijective, this says exactly that \(T^{-1}\) is continuous. To see explicitly why continuity at zero gives boundedness, choose \(r>0\) such that \(\|y\|_Y<r\) implies \(\|T^{-1}y\|_X<1\). For \(y\neq0\), apply this implication to \(ry/(2\|y\|_Y)\), whose norm is \(r/2\). Linearity gives \(\|T^{-1}y\|_X<(2/r)\|y\|_Y\). The same bound holds for \(y=0\), so \(T^{-1}\) is bounded. \(\square\)
Worked Example: Equivalent Norms on a Finite-Dimensional Space
On \(\mathbb{R}^2\), consider the norms \(\|(a,b)\|_1=|a|+|b|\) and \(\|(a,b)\|_\infty=\max(|a|,|b|)\). The identity map from \((\mathbb{R}^2,\|\cdot\|_1)\) to \((\mathbb{R}^2,\|\cdot\|_\infty)\) is linear, bijective, and bounded because \(\|(a,b)\|_\infty\leq\|(a,b)\|_1\). Both spaces are Banach, so the bounded inverse theorem shows that the inverse identity map is bounded as well.
Concretely, \(|a|\leq\|(a,b)\|_\infty\) and \(|b|\leq\|(a,b)\|_\infty\), so \(\|(a,b)\|_1\leq2\|(a,b)\|_\infty\). Equality holds for \((a,b)=(1,1)\). Thus the two norms control each other, and the identity map carries open sets for one norm to open sets for the other.
Worked Example: Solving a First-Order Differential Equation
Let \(Y=C([0,1])\) with the supremum norm, and let \(X=\{f\in C^1([0,1]):f(0)=0\}\) with \(\|f\|_X=\|f\|_\infty+\|f'\|_\infty\). This is a Banach space: if \((f_n)\) is Cauchy in this norm, then \(f_n\to f\) uniformly and \(f_n'\to g\) uniformly for continuous functions \(f,g\). Since \(f_n(t)=\int_0^t f_n'(s)\,ds\), uniform convergence gives \(f(t)=\int_0^t g(s)\,ds\). Hence \(f\in C^1([0,1])\), \(f'=g\), and \(f(0)=0\).
Define \(T:X\to Y\) by \(Tf=f'+f\). It is linear and bounded since \(\|Tf\|_\infty\leq\|f'\|_\infty+\|f\|_\infty=\|f\|_X\). For any \(g\in Y\), set \(f(t)=\int_0^t e^{-(t-s)}g(s)\,ds\). This function satisfies \(f(0)=0\), and differentiation gives \(f'(t)=g(t)-f(t)\), so \(Tf=g\). The solution is unique: if \(f'+f=0\) and \(f(0)=0\), then \((e^t f(t))'=0\), so \(e^t f(t)\) is constant and must be zero.
Thus \(T\) is a bounded linear bijection between Banach spaces. Its inverse is bounded, as the theorem guarantees. Directly, the integral formula gives \(\|f\|_\infty\leq\|g\|_\infty\int_0^t e^{-(t-s)}\,ds\leq\|g\|_\infty\), and \(f'=g-f\) gives \(\|f'\|_\infty\leq2\|g\|_\infty\). Therefore \(\|T^{-1}g\|_X\leq3\|g\|_\infty\).
Why the Hypotheses Matter
Surjectivity is essential to the theorem: it is what lets the closed sets \(F_n\) cover the whole target space. For instance, the inclusion \(i:\mathbb{R}\to\mathbb{R}^2\), \(i(t)=(t,0)\), is bounded and linear, but it is not surjective. The image of the open interval \((-1,1)\) is a line segment, which is not open in \(\mathbb{R}^2\). Completeness is also essential to this proof, since the Baire Category Theorem is applied in \(Y\), and the convergent series of approximate preimages is summed in \(X\).
A common point of confusion is to read “open mapping” as saying that every bounded linear map between Banach spaces is open. The theorem requires surjectivity. Its strongest practical consequence is often the bounded inverse theorem: a bounded linear bijection between Banach spaces has a bounded inverse, even when an estimate for that inverse is not apparent from its definition.
Check Your Understanding
Use the Baire-category and successive-approximation arguments to answer the following questions.
- Why do the closed sets \(F_n=\overline{T(\overline{B_X(0,n)})}\) cover \(Y\) when \(T\) is surjective?
- How does the proof show that a ball about zero lies in the closure of the image of the unit ball?
- Why does the sum of the approximate preimages converge in \(X\), and why does its image equal the target vector?
- Which hypotheses in the Open Mapping Theorem are used to apply the Baire Category Theorem?
- How does openness of a bounded linear bijection imply boundedness of its inverse?