Uniform Boundedness, Open Mapping & Closed Graph

Three Baire-category theorems turn pointwise or qualitative information into uniform operator control. This chapter states their exact hypotheses, shows how they connect, and diagnoses the incomplete-space counterexamples that make those hypotheses necessary.

budding working of working confidence: certain planted 2026-08-21 tended 2026-08-25 banach-steinhausopen-mappingclosed-graph

The three fundamental Banach-space theorems all use completeness to upgrade weak information. Uniform boundedness upgrades one bound per vector to one bound for an operator family. Open mapping upgrades surjectivity to stable solvability. Closed graph upgrades a closed input/output relation to continuity.

The theorem map #

Theorem	Hypotheses	Qualitative input	Quantitative conclusion
Uniform boundedness	$X$ Banach, $Y$ normed, $T_i\in\mathcal B(X,Y)$	$\sup_i\|T_ix\|<\infty$ for every $x$	$\sup_i\|T_i\|<\infty$
Open mapping	$X,Y$ Banach, $T\in\mathcal B(X,Y)$ surjective	every $y$ has a preimage	$T$ maps open sets to open sets
Bounded inverse	$X,Y$ Banach, $T$ bounded bijective	algebraic inverse exists	$T^{-1}$ is bounded
Closed graph	$X,Y$ Banach, $T:X\to Y$ linear and everywhere-defined	$G(T)$ is closed	$T$ is bounded

The domain completeness in uniform boundedness and both-space completeness in the other statements are load-bearing.

Completeness is the common engine: pointwise family control becomes an operator-norm bound, surjectivity becomes stable solvability, and a closed graph becomes boundedness.

Uniform boundedness: where the common ball comes from #

For $n\in\mathbb N$ , define

E_n=\{x\in X:\sup_i\|T_ix\|\le n\}.

Pointwise boundedness says $X=\bigcup_nE_n$ . Each $E_n$ is closed. Baire's theorem says that some $E_n$ contains a ball $B_r(x_0)$ . Linearity and symmetry then transfer the bound to a ball around zero, and scaling turns it into a global operator-norm bound.

The contrapositive is often more useful: if $\sup_i\|T_i\|=\infty$ , there is some $x$ for which $\sup_i\|T_ix\|=\infty$ . This proves the existence of a single bad input without constructing it explicitly.

Open mapping and stable solvability #

Surjectivity gives

Y=\bigcup_{n=1}^{\infty}T(nB_X).

Baire ensures that the closure of one image contains a ball. A geometric series correction then removes the closure: a smaller ball lies in $T(B_X)$ . Scaling and translation show that every open set has open image.

Quantitatively, there is $C>0$ such that every $y\in Y$ has some preimage $x$ with

Tx=y,qquad \|x\|\le C\|y\|.

If $T$ is also injective, that preimage is $T^{-1}y$ , proving the bounded inverse theorem.

Closed graph follows through a product space #

If $G(T)$ is closed in the Banach product $X\times Y$ , then $G(T)$ is Banach. The coordinate projection

\pi_X:G(T)\to X,qquad (x,Tx)\mapsto x

is bounded and bijective. Bounded inverse says $x\mapsto(x,Tx)$ is bounded, which bounds $T$ . This proof explains why an everywhere-defined domain is essential: otherwise $\pi_X$ maps only onto $D(T)$ , not all of $X$ .

Counterexample when completeness is missing #

Take $X=C^1[0,1]$ with only the sup norm. It is an incomplete normed space. Differentiation

D:X\to C[0,1],\qquad Df=f'

is everywhere-defined on $X$ , linear, closed, and unbounded. Thus closed graph fails exactly because its domain is not Banach. Equipping $X$ with $\|f\|_\infty+\|f'\|_\infty$ completes the domain and makes $D$ bounded.

Which theorem should you reach for? #

Many operators, one input at a time: uniform boundedness.
One surjective operator and existence with norm control: open mapping.
A bounded bijection whose inverse needs continuity: bounded inverse.
A limit argument proves the graph is closed but no direct estimate is available: closed graph.

Before invoking one, write the exact spaces and mark which are Banach. A theorem name without this hypothesis audit is not a proof.