The open mapping, bounded inverse, and closed graph theorems concern a single linear operator between Banach spaces, and each is proved from the Baire category theorem. The uniform boundedness principle concerns a family of operators and is also proved from the Baire category theorem. The Hahn–Banach theorem concerns extension of linear functionals and has no completeness hypothesis.
Open mapping, bounded inverse, and closed graph
Let \(X\) and \(Y\) be Banach spaces.
Open mapping theorem (Tag T-KQTPR)
Let \(X\) and \(Y\) be Banach spaces and let \(T\colon X\to Y\) be a continuous linear surjection. Then \(T\) is an open map: \(T(V)\) is open in \(Y\) for every open \(V\subseteq X\).
Bounded inverse theorem (Tag T-FO27T)
Let \(X\) and \(Y\) be Banach spaces and let \(T\colon X\to Y\) be a continuous linear bijection. Then the linear map \(\inverseof{T}\colon Y\to X\) is continuous, so \(T\) is a homeomorphism.
Closed graph theorem (Tag T-TTLXS)
Let \(X\) and \(Y\) be Banach spaces over the same field \(\RR\) or \(\CC\), and let \(T\colon X\to Y\) be a linear map whose graph \[ \Gamma(T)\coloneqq\theset{(x,Tx)\suchthat x\in X}\subseteq X\times Y \] is closed in \(X\times Y\) with the product topology. Then \(T\) is bounded: there exists \(C\geq0\) with \(\norm{Tx}_Y\leq C\norm{x}_X\) for all \(x\in X\).
To prove that a linear map \(T\colon X\to Y\) is bounded, it suffices by the closed graph theorem to show that \(x_n\to x\) and \(Tx_n\to y\) imply \(y = Tx\), which avoids an explicit estimate \(\norm{Tx}\le C\norm{x}\).
Uniform boundedness
Uniform boundedness principle (Tag FT-JRRRW)
Let \(X\) be a Banach space, let \(Y\) be a normed vector space, and let \(\mathcal{F}\) be a family of bounded linear operators \(T\colon X\to Y\). If \[ \sup_{T \in \mathcal{F}} \norm{Tx}_Y < \infty \quad\text{for every } x\in X, \] then \(\sup_{T\in \mathcal{F}} \norm{T} < \infty\), where \(\norm{T}\coloneqq\sup_{\norm{x}_X\le1}\norm{Tx}_Y\) is the operator norm.
Contrapositively, if \(\sup_{T\in\mathcal F}\norm{T}=\infty\), then there exists \(x\in X\) with \(\sup_{T\in\mathcal F}\norm{Tx}_Y=\infty\).
Hahn–Banach
Hahn--Banach theorem (Tag T-OW2QG)
Let \(\mathcal X\) be a real vector space and let \(p\colon \mathcal X\to\RR\) be sublinear, that is, \(p(x+y) \leq p(x) + p(y)\) and \(p(\lambda x) = \lambda p(x)\) for all \(x,y\in \mathcal X\) and all \(\lambda \geq 0\). Let \(\mathcal M \subseteq \mathcal X\) be a linear subspace and \(f\colon\mathcal M\to\RR\) a linear functional with \(f \leq p\) on \(\mathcal M\). Then there exists a linear functional \(F\colon\mathcal X\to\RR\) with \(\ro{F}{\mathcal M} = f\) and \(F \leq p\) on \(\mathcal X\).
Let \(\mathcal X\) be a complex vector space and let \(p\colon\mathcal X\to\RR\) be a seminorm, that is, \(p(x+y) \leq p(x) + p(y)\) and \(p(\lambda x) = \abs{\lambda} p(x)\) for all \(x,y\in \mathcal X\) and all \(\lambda \in\CC\). Let \(\mathcal M \subseteq \mathcal X\) be a linear subspace and \(f\colon\mathcal M\to\CC\) a linear functional with \(\abs f \leq p\) on \(\mathcal M\). Then there exists a linear functional \(F\colon\mathcal X\to\CC\) with \(\ro{F}{\mathcal M} = f\) and \(\abs F \leq p\) on \(\mathcal X\) [@Fol13].
The proof extends \(f\) one dimension at a time and applies Zorn’s lemma.
The extension \(F\) need not be unique. On \(\mathcal X=\RR^2\) let \(p(x)\coloneqq\abs{x_1}+\abs{x_2}\), \(\mathcal M\coloneqq\RR\times\theset{0}\), and \(f(x_1,0)\coloneqq x_1\). For every \(c\in[-1,1]\), the functional \(F_c(x)\coloneqq x_1+cx_2\) extends \(f\) and satisfies \(F_c\leq p\).
Let \(\mathcal X\) be a normed vector space over \(\RR\) or \(\CC\) with dual space \(\dualof{\mathcal X}\) of continuous linear functionals and dual norm.
If \(\mathcal M \subseteq \mathcal X\) is a closed linear subspace and \(x\in\mathcal X\setminus\mathcal M\), put \(\delta \coloneqq \inf_{y\in \mathcal M}\norm{x-y}\). Then there is \(f \in \dualof{\mathcal X}\) with \(\norm{f} = 1\), \(\ro f {\mathcal M} = 0\), and \(f(x) = \delta\).
For each \(x\in\mathcal X\) with \(x\neq 0\) there is \(f\in \dualof{\mathcal X}\) with \(\norm f = 1\) and \(f(x) = \norm x\). In particular, the functionals in \(\dualof{\mathcal X}\) separate the points of \(\mathcal X\).
The map \(\mathcal X \injects \mathcal X^{\vee\vee}\), \(x \mapsto \hat x\) with \(\hat x(f) \coloneqq f(x)\) for \(f\in\dualof{\mathcal X}\), is a linear isometry.
[@Fol13]