The Baire Category Theorem
Three of the cornerstones of functional analysis share a single engine: the open mapping
theorem, the closed graph theorem, and the principle of uniform
boundedness. Each converts a hypothesis that holds pointwise or
set-theoretically into a conclusion that holds uniformly, and in every case the
conversion runs on one structural fact about complete spaces. Such a space cannot be assembled
from countably many negligible pieces. That fact is the
Baire category theorem. The fourth cornerstone, the
Hahn-Banach
theorem, runs on a different engine entirely, a transfinite extension argument, and that is why it was
treated separately.
The present page builds the Baire engine and then drives the three theorems that depend on it.
The theorem comes in two incarnations, one for locally compact Hausdorff spaces and one for complete metric
spaces, resting on the same nested-shrinking construction. The
locally compact Hausdorff version
was established earlier, where the nested objects were precompact open sets and the non-empty intersection came
from compactness. Functional analysis needs the other incarnation: the one for
complete metric spaces,
where the nested objects are closed balls of shrinking radius and the non-empty intersection comes from
completeness. We prove it in full here, discharging the forward reference made when the locally compact version
was given.
Nowhere Dense Sets and the Statement
A subset \(E\) of a metric space is nowhere dense if the interior of its closure is empty:
\(\operatorname{int}(\overline{E}) = \varnothing\). Two equivalent formulations are that
\(\overline{E}\) contains no ball, and that every non-empty
open set contains a non-empty open subset disjoint from \(E\). Such sets are the “thin”
pieces of the theorem. The content is that a complete space is too substantial to be a countable union of them.
Theorem (Baire Category Theorem, Complete Metric Space Version)
Let \((\mathcal{X}, d)\) be a non-empty complete metric space. If \(\{E_n\}_{n \geq 1}\) are nowhere dense
subsets, then their union \(\bigcup_{n=1}^{\infty} E_n\) has empty interior. In particular it
cannot be all of \(\mathcal{X}\).
It is equivalent, and more convenient for the proof, to argue the dual form. A set is nowhere
dense exactly when the complement of its closure is dense and open, so passing to complements
turns the empty-interior statement into
“a countable intersection of dense open sets is dense.” This dual form is also the one the three pillars
invoke directly. We prove it.
Proof by Nested Closed Balls
Proof
Let \(\{G_n\}_{n \geq 1}\) be a sequence of dense open subsets of \(\mathcal{X}\). We show
that \(\bigcap_{n=1}^{\infty} G_n\) is dense, i.e. that it meets every non-empty open set
\(W_0\). Fix such a
\(W_0\). We construct a nested sequence of closed balls whose radii tend to zero, each contained in the next
larger open set intersected with \(G_n\), and locate a point of the intersection at their common point.
Construction.
Since \(G_1\) is dense and open, \(W_0 \cap G_1\) is non-empty and open. Pick a point \(x_1\)
in it and a radius \(0 \lt r_1 \leq 1\) small enough that the closed ball satisfies
\[
\overline{B}(x_1, r_1) \subseteq W_0 \cap G_1 .
\]
Such a radius exists because an open set containing \(x_1\) contains an open ball \(B(x_1, \rho)\), and then
\(\overline{B}(x_1, \rho/2) \subseteq B(x_1, \rho)\). Inductively, suppose \(x_n\) and \(r_n\) have been chosen
with the open ball \(B(x_n, r_n)\) non-empty. Because \(G_{n+1}\) is dense and open,
\(B(x_n, r_n) \cap G_{n+1}\) is non-empty and open.
Pick \(x_{n+1}\) in it and a radius
\[
0 \lt r_{n+1} \leq \tfrac{1}{2} r_n, \quad r_{n+1} \leq \tfrac{1}{n+1},
\]
small enough that
\[
\overline{B}(x_{n+1}, r_{n+1}) \subseteq B(x_n, r_n) \cap G_{n+1} .
\]
The first constraint forces \(r_n \leq 2^{-(n-1)} r_1 \to 0\). The second guarantees \(r_n \to
0\) independently. By construction the closed balls are nested,
\[
\overline{B}(x_1, r_1) \supseteq \overline{B}(x_2, r_2) \supseteq \cdots ,
\]
since \(\overline{B}(x_{n+1}, r_{n+1}) \subseteq B(x_n, r_n) \subseteq \overline{B}(x_n, r_n)\).
The centers form a Cauchy sequence.
If \(m \geq n\), then \(x_m \in \overline{B}(x_n, r_n)\)
by nesting, so \(d(x_m, x_n) \leq r_n\). Given \(\varepsilon \gt 0\), choose \(N\) with
\(r_N \lt \varepsilon\), which is possible
since \(r_n \to 0\). Then for all \(m, n \geq N\),
\[
d(x_m, x_n) \leq d(x_m, x_N) + d(x_N, x_n) \leq r_N + r_N \lt 2\varepsilon .
\]
Hence \(\{x_n\}\) is Cauchy. Here completeness enters. The sequence converges
to some limit \(x \in \mathcal{X}\).
The limit lies in every ball.
Fix \(n\). For every \(m \geq n\) we have
\(x_m \in \overline{B}(x_n, r_n)\), and a closed ball is a closed set, so the limit of the tail also lies in
it: \(x \in \overline{B}(x_n, r_n)\). This holds for every \(n\).
Conclusion.
For each \(n\), the inclusion built into the construction gives
\(\overline{B}(x_n, r_n) \subseteq G_n\) (for \(n = 1\) directly, and for \(n \geq 2\) because
\(\overline{B}(x_n, r_n) \subseteq B(x_{n-1}, r_{n-1}) \cap G_n \subseteq G_n\)). Therefore \(x \in G_n\) for
every \(n\), so \(x \in \bigcap_{n=1}^{\infty} G_n\). Moreover \(x \in \overline{B}(x_1, r_1) \subseteq W_0\).
Thus \(W_0\) meets the intersection. Since \(W_0\) was an arbitrary non-empty open set,
\(\bigcap_{n=1}^{\infty} G_n\) is dense.
For the nowhere-dense form, suppose \(\mathcal{X} = \bigcup_n E_n\) with each \(E_n\) nowhere
dense. Then each \(G_n := \mathcal{X} \setminus \overline{E_n}\) is dense and open, and
\(\bigcap_n G_n = \mathcal{X} \setminus \bigcup_n \overline{E_n} \subseteq \mathcal{X} \setminus \bigcup_n E_n
= \varnothing\). A dense set cannot be empty in a non-empty space, contradicting the density just proved.
Hence no such cover exists.
The mechanism is worth isolating, because the three pillars use it through one recurring move. To
prove a uniform conclusion, one exhibits the whole space as a countable union of closed
“level sets” defined by a bound. Baire then forces one of them to have non-empty
interior, and that interior is a whole ball on which the bound holds. Linearity spreads it across
the space. The next three sections are three instances of
exactly this move.
The Open Mapping Theorem
A continuous linear map between Banach spaces need not send open sets to open sets. That is the
generic expectation for continuous maps. The open mapping theorem says that for surjective bounded linear maps
between Banach spaces, it always does. The consequences are immediate and far-reaching. A
continuous linear bijection between Banach
spaces automatically has continuous inverse, and a linear map is continuous as soon as its graph
is closed. Both are taken up in the next section. Here we prove the theorem itself. Its
inverse-mapping corollary is what the partition of the spectrum into point, continuous, and residual parts silently relied on to
rule out a bounded inverse outside the resolvent set.
Theorem (Open Mapping Theorem)
Let \(\mathcal{X}\) and \(\mathcal{Y}\) be Banach spaces and let \(A : \mathcal{X} \to \mathcal{Y}\) be a
continuous linear surjection. Then \(A\) is an open map, so that \(A(G)\) is open in \(\mathcal{Y}\)
whenever \(G\) is open in \(\mathcal{X}\).
Throughout, write \(B(r) = \{ x \in \mathcal{X} : \|x\| \lt r \}\) for the open ball of radius \(r\) about the
origin in \(\mathcal{X}\), and recall that \(A\) is linear, so \(A(B(r)) = r\,A(B(1))\) and balls
scale. The proof isolates a single quantitative fact, namely that the closure of the image of a ball
contains a ball about the origin, and then upgrades it to the open-mapping statement. The first step is where Baire is used.
Step 1: The Closed Image Contains a Ball
Lemma (Closure Absorbs a Ball)
Under the hypotheses of the theorem, \(0 \in \operatorname{int}\,\overline{A(B(r))}\) for
every \(r \gt 0\). In other words, the
closure of the image of any origin-ball contains an origin-ball in \(\mathcal{Y}\).
Proof
Because \(A\) is surjective, every \(y \in \mathcal{Y}\) is \(A(x)\) for some \(x \in
\mathcal{X}\). Since \(x\) has finite norm, \(\|x\| \lt k r/2\) for \(k\) large, so \(y \in
A(B(k r/2))\). Hence
\[
\mathcal{Y} = \bigcup_{k=1}^{\infty} A\bigl(B(k r/2)\bigr) \subseteq
\bigcup_{k=1}^{\infty} \overline{A\bigl(B(k r/2)\bigr)} =
\bigcup_{k=1}^{\infty} k\,\overline{A\bigl(B(r/2)\bigr)} ,
\]
the last equality by linearity, \(A(B(k r/2)) = k\,A(B(r/2))\), and continuity of scalar multiplication, which
commutes with closure. The space \(\mathcal{Y}\) is a non-empty complete metric space, so the
Baire category theorem forbids it
from being a countable union of nowhere dense sets. Hence some set
\(k\,\overline{A(B(r/2))}\) is not nowhere dense. Since scaling by the fixed
nonzero \(k\) is a homeomorphism, \(\overline{A(B(r/2))}\) itself has non-empty interior. Let
\[
V = \operatorname{int}\,\overline{A\bigl(B(r/2)\bigr)} \neq \varnothing,
\]
and fix \(y_0 \in V\) with an open ball \(\{ y : \|y - y_0\| \lt s \} \subseteq V \subseteq
\overline{A(B(r/2))}\) for some \(s \gt 0\).
We claim \(\{ y : \|y\| \lt s \} \subseteq \overline{A(B(r))}\), which gives
\(0 \in \operatorname{int}\,\overline{A(B(r))}\). Fix \(y\) with \(\|y\| \lt s\). Both \(y_0\) and \(y_0 + y\)
lie in the ball \(\{ \cdot : \|\cdot - y_0\| \lt s \} \subseteq \overline{A(B(r/2))}\), so there are sequences
\(\{x_n\}, \{z_n\} \subseteq B(r/2)\) with \(A(x_n) \to y_0\) and \(A(z_n) \to y_0 + y\). Then
\(z_n - x_n \in B(r)\) (since \(\|z_n - x_n\| \leq \|z_n\| + \|x_n\| \lt r\)) and
\(A(z_n - x_n) = A(z_n) - A(x_n) \to (y_0 + y) - y_0 = y\). Thus \(y \in \overline{A(B(r))}\), proving the
claim.
Step 2: Removing the Closure
The lemma controls \(\overline{A(B(r))}\), but openness requires control of \(A(B(r))\) itself.
The closure is removed by an iteration. A target in the closure of \(A(B(r/2))\) is hit exactly by
a convergent series of successive corrections, each drawn from a ball half the size of the last.
Completeness of \(\mathcal{X}\) sums the
series.
Lemma (Closure Removed)
Under the hypotheses of the theorem, \(\overline{A(B(r/2))} \subseteq A(B(r))\) for every \(r \gt 0\).
Proof
Applying the previous lemma with the radius \(2^{-n} r\) in place of \(r\) gives, for each \(n \geq 1\),
\[
0 \in \operatorname{int}\,\overline{A\bigl(B(2^{-n} r)\bigr)} ,
\]
so each closure \(\overline{A(B(2^{-n} r))}\) contains an origin-ball. Equivalently, any point
arbitrarily close to a point of \(\overline{A(B(2^{-n} r))}\) can be approximated from
\(A(B(2^{-n} r))\) within any prescribed tolerance. Fix
\(y_1 \in \overline{A(B(r/2))} = \overline{A(B(2^{-1} r))}\). We build sequences
\(\{x_n\} \subseteq \mathcal{X}\) and \(\{y_n\} \subseteq \mathcal{Y}\) with
\[
\begin{align*}
&\text{(i)} \quad x_n \in B(2^{-n} r), \\\\
&\text{(ii)} \quad y_n \in \overline{A\bigl(B(2^{-n} r)\bigr)}, \\\\
&\text{(iii)} \quad y_{n+1} = y_n - A(x_n).
\end{align*}
\]
Given \(y_n \in \overline{A(B(2^{-n} r))}\), choose \(x_n \in B(2^{-n} r)\) with \(A(x_n)\) so close to
\(y_n\) that the remainder \(y_{n+1} := y_n - A(x_n)\) lies in \(\overline{A(B(2^{-(n+1)}
r))}\). This is possible because the latter closure contains an origin-ball of some positive
radius, so every point within that radius of \(0\) belongs to it, and \(y_n - A(x_n)\) is such
a point once \(A(x_n)\) is chosen close enough to \(y_n\). Starting from the given \(y_1\)
and iterating produces the three properties.
By (i), \(\|x_n\| \lt 2^{-n} r\), so \(\sum_{n=1}^{\infty} \|x_n\| \lt \sum_{n=1}^{\infty} 2^{-n} r = r\), and
the partial sums of \(\sum x_n\) are Cauchy in \(\mathcal{X}\). Here completeness of
\(\mathcal{X}\) enters. The series converges to some \(x
= \sum_{n=1}^{\infty} x_n\) with
\[
\|x\| \leq \sum_{n=1}^{\infty} \|x_n\| \lt r, \quad \text{so } x \in B(r) .
\]
By continuity of \(A\) and the telescoping identity from (iii),
\[
\begin{align*}
A(x) &= \sum_{k=1}^{\infty} A(x_k) = \lim_{n \to \infty} \sum_{k=1}^{n} A(x_k) \\\\
&= \lim_{n \to \infty} \sum_{k=1}^{n} (y_k - y_{k+1}) = \lim_{n \to \infty} (y_1 - y_{n+1}) = y_1 ,
\end{align*}
\]
where the final limit uses (ii): \(\|y_{n+1}\| \leq \|A\|\, 2^{-(n+1)} r \to 0\), since every element of
\(\overline{A(B(\rho))}\) has norm at most \(\|A\|\,\rho\). Therefore \(y_1 = A(x) \in A(B(r))\). As \(y_1\)
was an arbitrary point of \(\overline{A(B(r/2))}\), the inclusion follows.
Step 3: From the Origin to Every Open Set
Proof of the Theorem
Combining the two lemmas, for every \(r \gt 0\),
\[
0 \in \operatorname{int}\,\overline{A\bigl(B(r/2)\bigr)} \subseteq
\operatorname{int}\, A(B(r)) ,
\]
so \(A(B(r))\) contains an origin-ball for every \(r \gt 0\). Now let \(G \subseteq \mathcal{X}\) be open and
let \(y \in A(G)\), say \(y = A(x)\) with \(x \in G\). Pick \(\rho \gt 0\) with \(B(x; \rho) := x + B(\rho)
\subseteq G\). Since \(A(B(\rho))\) contains an origin-ball \(\{ \cdot : \|\cdot\| \lt \sigma \}\) for some
\(\sigma \gt 0\), translation by \(y = A(x)\) gives
\[
\{ w : \|w - y\| \lt \sigma \} = y + \{ \cdot : \|\cdot\| \lt \sigma \} \subseteq
A(x) + A(B(\rho)) = A(x + B(\rho)) \subseteq A(G) .
\]
Thus every point of \(A(G)\) is interior to \(A(G)\), so \(A(G)\) is open. \(A\) is an open map.
Inverse Mapping & Closed Graph
Two corollaries of the open mapping theorem govern when linearity forces continuity. The first
removes a verification that one might expect to be necessary. For a continuous linear
bijection between Banach spaces, the inverse is automatically continuous, and one never has to
check it. The second replaces the direct test of continuity by a
test on the graph, often far easier to apply.
The Inverse Mapping Theorem
Theorem (Inverse Mapping Theorem)
Let \(\mathcal{X}\) and \(\mathcal{Y}\) be Banach spaces and let \(A : \mathcal{X} \to \mathcal{Y}\) be a
bounded linear bijection. Then the inverse \(A^{-1} : \mathcal{Y} \to \mathcal{X}\) is bounded.
Equivalently, a continuous linear bijection between Banach spaces is a homeomorphism.
Proof
Being bijective, \(A\) is in particular a continuous linear surjection, so by the
open mapping theorem it is an
open map, and \(A(G)\) is open for every open \(G \subseteq \mathcal{X}\). The inverse
\(A^{-1}\) is a function because \(A\) is a
bijection, and its continuity is exactly the requirement that preimages of open sets under \(A^{-1}\) be open.
But the preimage of \(G\) under \(A^{-1}\) is \((A^{-1})^{-1}(G) = A(G)\), which is open.
Hence \(A^{-1}\) is continuous, and being linear and continuous it is bounded.
This is the resolution promised when the spectrum was partitioned. At a point \(\lambda\) where
\(T - \lambda I\) is a continuous linear bijection of \(\mathcal{X}\) onto itself, the inverse
mapping theorem makes \((T - \lambda I)^{-1}\) automatically bounded, placing \(\lambda\) in the
resolvent set. The unboundedness of the inverse on the continuous spectrum is therefore not an
oversight but a structural necessity. If the inverse were bounded, \(\lambda\) would not be in the
spectrum at all.
The Closed Graph Theorem
The graph of a linear map \(A : \mathcal{X} \to \mathcal{Y}\) is the set
\[
\operatorname{gra} A = \{\, (x, Ax) : x \in \mathcal{X} \,\} \subseteq \mathcal{X} \oplus \mathcal{Y} .
\]
We equip the direct sum \(\mathcal{X} \oplus \mathcal{Y}\) with the norm \(\|(x, y)\| = \|x\|_{\mathcal{X}} +
\|y\|_{\mathcal{Y}}\). Under this norm convergence means convergence in each coordinate, and the
direct sum is complete because a sequence Cauchy in the sum is Cauchy in each coordinate, with
limits supplied by completeness of \(\mathcal{X}\) and \(\mathcal{Y}\). Thus \(\mathcal{X} \oplus \mathcal{Y}\) is a Banach space.
Continuity of \(A\) always forces \(\operatorname{gra} A\) to be closed. If \((x_n, A x_n) \to (x,
y)\) then \(x_n \to x\) and, by continuity, \(A x_n \to A x\), so \(y = A x\) and the limit lies
in the graph. The closed graph theorem is the converse, and it is the deeper statement. Closedness
of the graph is a weaker-looking hypothesis than continuity, yet between Banach spaces it is
equivalent to it.
Theorem (Closed Graph Theorem)
Let \(\mathcal{X}\) and \(\mathcal{Y}\) be Banach spaces and let \(A : \mathcal{X} \to \mathcal{Y}\) be linear.
If \(\operatorname{gra} A\) is closed in \(\mathcal{X} \oplus \mathcal{Y}\), then \(A\) is continuous.
Proof
Let \(\mathcal{G} = \operatorname{gra} A\). As a closed linear subspace of the Banach space \(\mathcal{X}
\oplus \mathcal{Y}\), \(\mathcal{G}\) is itself a Banach space. Consider the two coordinate projections
restricted to \(\mathcal{G}\),
\[
P : \mathcal{G} \to \mathcal{X}, \quad P(x, Ax) = x, \quad Q : \mathcal{G} \to \mathcal{Y}, \quad Q(x, Ax) = Ax .
\]
Both are linear and bounded, since \(\|P(x, Ax)\| = \|x\| \leq \|(x, Ax)\|\) and likewise \(\|Q(x, Ax)\| =
\|Ax\| \leq \|(x, Ax)\|\). The map \(P\) is a bijection of \(\mathcal{G}\) onto \(\mathcal{X}\). It is surjective because
every \(x \in \mathcal{X}\) yields the point \((x, Ax) \in \mathcal{G}\), and injective
because \(P(x, Ax) = 0\) gives \(x = 0\), hence \((x, Ax) = (0, 0)\). By the
inverse mapping theorem, \(P^{-1} :
\mathcal{X} \to \mathcal{G}\), \(P^{-1}(x) = (x, Ax)\), is bounded. Therefore \(A = Q \circ P^{-1}\) is a
composition of bounded maps,
\[
x \xrightarrow{P^{-1}} (x, Ax) \xrightarrow{Q} Ax ,
\]
and is bounded, hence continuous.
The Sequential Criterion
In practice one rarely verifies closedness of the graph as a topological condition. Instead one
uses a sequential test that isolates exactly what closedness adds beyond mere convergence. To show
\(A\) continuous it would suffice to show that \(x_n \to 0\) implies \(A x_n \to 0\). The
difficulty is that \(\{A x_n\}\) need not converge at all. The
following criterion shows that closedness of the graph lets one assume \(\{A x_n\}\) converges and only
check that its limit is forced to be \(0\).
Proposition (Sequential Criterion for a Closed Graph)
Let \(\mathcal{X}\) and \(\mathcal{Y}\) be normed spaces and \(A : \mathcal{X} \to \mathcal{Y}\) linear. Then
\(\operatorname{gra} A\) is closed if and only if the following holds. Whenever \(x_n \to 0\)
in \(\mathcal{X}\) and \(A x_n \to y\) in \(\mathcal{Y}\), it must be that \(y = 0\).
Proof
Closed graph implies the criterion.
Suppose \(\operatorname{gra} A\) is closed and \(x_n \to 0\), \(A x_n \to y\). Then \((x_n, A x_n) \to (0, y)\)
in \(\mathcal{X} \oplus \mathcal{Y}\), and the points \((x_n, A x_n)\) lie in \(\operatorname{gra} A\). A closed
set contains the limits of its sequences, so \((0, y) \in \operatorname{gra} A\). By
definition of the graph this means \(y = A(0) = 0\).
The criterion implies closed graph.
Suppose the criterion holds and \((x_n, A x_n) \to (x, y)\) in \(\mathcal{X} \oplus
\mathcal{Y}\). We show \((x, y) \in \operatorname{gra} A\), i.e. \(y = A
x\). Coordinatewise, \(x_n \to x\) and \(A x_n \to y\). Set
\(u_n := x_n - x\). Then \(u_n \to 0\), and by linearity \(A u_n = A x_n - A x \to y - A x\). Applying the
criterion to the sequence \(\{u_n\}\), whose images converge to \(y - A x\), forces \(y - A x = 0\), that is
\(y = A x\). Hence the limit lies in the graph, and \(\operatorname{gra} A\) is closed.
This is what makes the closed graph theorem a labor-saving device rather than a restatement.
Closedness is checked through the proposition, which lets one assume the limit exists and pin down
only its value, and the theorem upgrades it to full continuity at no further cost.
The Principle of Uniform Boundedness
The third pillar concerns a family of operators rather than a single one. It converts a
bound that holds separately at each point into a bound that holds uniformly across
the whole family, so that pointwise boundedness becomes norm boundedness. Like the open mapping
theorem, it runs on Baire’s engine. The domain is written as a countable union of closed
level sets, one of which must contain a ball, and linearity then spreads the bound from that ball
across the space.
Proof
For each \(n \geq 1\), define
\[
E_n = \bigl\{\, x \in \mathcal{X} : \|A x\| \leq n \text{ for every } A \in \mathcal{A} \,\bigr\}
= \bigcap_{A \in \mathcal{A}} \bigl\{\, x : \|A x\| \leq n \,\bigr\} .
\]
Each set \(\{ x : \|A x\| \leq n \}\) is closed, being the preimage of the closed interval \([0, n]\) under the
continuous map \(x \mapsto \|A x\|\). An intersection of closed sets is closed, so each
\(E_n\) is closed. Pointwise boundedness says exactly that every \(x\) lies in some \(E_n\).
Given \(x\), the finite number \(\sup_{A} \|A x\|\) is at most some integer \(n\), so \(x \in
E_n\). Hence
\(\mathcal{X} = \bigcup_{n=1}^{\infty} E_n\).
Since \(\mathcal{X}\) is a non-empty complete metric space, the
Baire category theorem forbids it from
being a countable union of nowhere dense sets. The \(E_n\) are closed, so a nowhere dense \(E_n\) would equal its
closure with empty interior. Since their union is the whole space, at least one of them, say
\(E_N\), has non-empty interior. Choose \(x_0 \in \mathcal{X}\) and \(r \gt 0\) with the closed ball
\(\{ x : \|x - x_0\| \leq r \} \subseteq E_N\).
Spreading the bound.
Fix any \(A \in \mathcal{A}\) and any \(z \in \mathcal{X}\) with \(\|z\| \leq r\). Both \(x_0\) and \(x_0 + z\)
lie in the ball, hence in \(E_N\), so \(\|A x_0\| \leq N\) and \(\|A(x_0 + z)\| \leq N\). By linearity and the
triangle inequality,
\[
\|A z\| = \|A(x_0 + z) - A x_0\| \leq \|A(x_0 + z)\| + \|A x_0\| \leq 2N .
\]
For an arbitrary unit-bounded vector we scale. If \(\|w\| \leq 1\), then \(\|r w\| \leq r\),
so \(\|A(r w)\| \leq 2N\) and \(\|A w\| \leq 2N / r\). Taking the supremum over \(\|w\| \leq 1\),
\[
\|A\| \leq \frac{2N}{r} .
\]
The bound \(2N/r\) depends only on \(N\) and \(r\), not on the choice of \(A \in \mathcal{A}\). Therefore
\(\sup_{A \in \mathcal{A}} \|A\| \leq 2N/r \lt \infty\).
The Banach-Steinhaus Theorem
The most frequent use of the principle concerns a sequence of operators that converges pointwise.
Pointwise convergence alone does not obviously produce a bounded limit operator, nor a bound on
the sequence. The principle supplies both.
Theorem (Banach-Steinhaus)
Let \(\mathcal{X}\) and \(\mathcal{Y}\) be Banach spaces and let \(\{A_n\} \subseteq B(\mathcal{X}, \mathcal{Y})\)
be a sequence such that, for every \(x \in \mathcal{X}\), the limit \(A x := \lim_{n \to \infty} A_n x\) exists in
\(\mathcal{Y}\). Then \(A : \mathcal{X} \to \mathcal{Y}\) is a bounded linear operator, and \(\sup_n \|A_n\| \lt
\infty\).
Proof
Linearity of \(A\) is inherited in the limit: \(A(\alpha x + \beta x') = \lim A_n(\alpha x + \beta x') =
\alpha \lim A_n x + \beta \lim A_n x' = \alpha A x + \beta A x'\). For boundedness, observe that for each fixed
\(x\) the sequence \(\{A_n x\}\) converges, hence is bounded in \(\mathcal{Y}\): \(\sup_n \|A_n x\| \lt \infty\).
Thus \(\{A_n\}\) is pointwise bounded, and the
principle of uniform boundedness gives a
constant \(M\) with \(\|A_n\| \leq M\) for all \(n\), proving \(\sup_n \|A_n\| \lt \infty\). Finally, for every
\(x\),
\[
\|A x\| = \lim_{n \to \infty} \|A_n x\| \leq \limsup_{n \to \infty} \|A_n\|\, \|x\| \leq M \|x\| ,
\]
using continuity of the norm for the first equality. Hence \(\|A\| \leq M\) and \(A\) is bounded.
Weak Convergence Forces Norm Boundedness
A consequence in the geometry of Banach spaces shows the principle at work on a problem it is uniquely suited to.
A sequence converging in the weak
topology is tested one functional at a time, which gives no obvious control on its
norms. The principle is exactly what converts that functional-by-functional information into a
uniform bound.
Corollary (Weakly Convergent Sequences Are Bounded)
Let \(\mathcal{X}\) be a normed space and let \(\{x_n\} \subseteq \mathcal{X}\) converge weakly to some
\(x \in \mathcal{X}\). Then \(\sup_n \|x_n\| \lt \infty\).
Proof
View each \(x_n\) through the
canonical embedding
\(J : \mathcal{X} \to \mathcal{X}^{**}\), so that \(J(x_n)\) is the bounded linear functional on \(\mathcal{X}^*\)
given by \(J(x_n)(\varphi) = \varphi(x_n)\). The family \(\{ J(x_n) \}_n \subseteq B(\mathcal{X}^*, \mathbb{F})\)
acts on \(\mathcal{X}^*\), which is complete, since a dual space is always a Banach
space regardless of whether \(\mathcal{X}\) is. The principle of uniform boundedness therefore
applies with \(\mathcal{X}^*\) in the role of the Banach domain. For each fixed
\(\varphi \in \mathcal{X}^*\), weak convergence gives
\(J(x_n)(\varphi) = \varphi(x_n) \to \varphi(x)\), a convergent and therefore bounded scalar
sequence. Thus \(\{ J(x_n) \}_n\) is pointwise bounded. The
principle of uniform boundedness yields
\(\sup_n \|J(x_n)\|_{\mathcal{X}^{**}} \lt \infty\). The map \(J\) is an isometry, a fact resting on the
norming functional corollary
of Hahn-Banach, so \(\|J(x_n)\|_{\mathcal{X}^{**}} = \|x_n\|\) and therefore \(\sup_n \|x_n\|
\lt \infty\).
With this the three Baire-driven pillars are complete. The open mapping and closed graph theorems
govern the automatic continuity of inverses and of
graph-closed maps. The principle of uniform boundedness governs the passage from pointwise to
uniform control of operator families. Each rests on the single fact that a complete space resists
being built from negligible pieces, which is the structural content of the Baire category theorem
proved at the outset.