The Baire Category Theorem
Three of the cornerstones of functional analysis — the open mapping theorem, the
closed graph theorem, and the principle of uniform boundedness — share a
single engine. 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: they 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 — which 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:
\(\operatorname{int}(\overline{E}) = \varnothing\). Equivalently, \(\overline{E}\) contains no ball; equivalently,
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
\(\bigcap_{n=1}^{\infty} G_n\) 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 inside 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 \(\overline{B}(x_n, r_n)\) non-empty and open ball \(B(x_n, r_n)\) available. 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
\[
0 \;\lt\; r_{n+1} \;\leq\; \tfrac{1}{2} r_n, \qquad r_{n+1} \;\leq\; \tfrac{1}{n+1},
\]
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 \(r_N \lt \varepsilon\)
(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: if \(\mathcal{X} = \bigcup_n E_n\) 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 forces one of them to have non-empty interior; and the interior — a whole ball on which the
bound holds — is then spread across the space by linearity. 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: \(A(G)\) 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 — 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 \(r \gt 0\):
the closure of the image of any origin-ball contains an origin-ball in \(\mathcal{Y}\).
Proof
Because \(A\) is surjective and every \(y \in \mathcal{Y}\) has finite norm, every \(y\) lies in some
\(A(B(k r/2))\) for \(k\) large, so
\[
\mathcal{Y} \;=\; \bigcup_{k=1}^{\infty} A\bigl(B(k r/2)\bigr) \;\subseteq\;
\bigcup_{k=1}^{\infty} \overline{A\bigl(B(k r/2)\bigr)} \;=\;
\bigcup_{k=1}^{\infty} k\,\overline{A\bigl(B(r/2)\bigr)} ,
\]
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 points within that
radius of \(0\) — in particular \(y_n - A(x_n)\) for \(A(x_n)\) chosen close enough to \(y_n\) —
belong to it. 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
\[
\|x\| \;\leq\; \sum_{n=1}^{\infty} \|x_n\| \;\lt\; r, \qquad \text{so } x \in B(r) .
\]
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 — 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: \(A(G)\)
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; 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, \(A x_n \to A x\), forcing \(y = A x\), so 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 \(\mathcal{G}\),
\[
P : \mathcal{G} \to \mathcal{X}, \quad P(x, Ax) = x, \qquad Q : \mathcal{G} \to \mathcal{Y}, \quad Q(x, Ax) = Ax .
\]
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; 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 — assuming the limit exists and pinning 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 — 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; as their union is the whole space, at least one — 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, scale: if \(\|w\| \leq 1\), then \(\|r w\| \leq r\), so \(\|A(r w)\| \leq
2N\), giving \(\|A w\| \leq 2N / r\). 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 — a dual space is always a Banach space, regardless of
whether \(\mathcal{X}\) is — so the principle of uniform boundedness 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
\(\sup_n \|J(x_n)\|_{\mathcal{X}^{**}} \lt \infty\). Since \(J\) is an isometry — a fact resting on the
norming functional corollary
of Hahn-Banach — we have \(\|J(x_n)\|_{\mathcal{X}^{**}} = \|x_n\|\), 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 — the structural content of the Baire category theorem proved at the outset.