The uniform boundedness principle

Theorem

The uniform boundedness principle

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Statement

Let \(X\) be a Banach space and \(Y\) a normed vector space over \(\mathbb{R}\) or \(\mathbb{C}\). Let \(\mathcal{F} \subseteq \mathcal{B}(X,Y)\) be a family of bounded linear operators from \(X\) to \(Y\), indexed by any set \(I\), \(\mathcal{F} = \{T_i\}_{i \in I}\). Suppose \(\mathcal{F}\) is pointwise bounded: for every \(x \in X\), \[ \sup_{i \in I} \|T_i x\|_Y \lt \infty. \] Then \(\mathcal{F}\) is uniformly bounded in operator norm: there exists \(M \lt \infty\), independent of \(i\), such that \[ \sup_{i \in I} \|T_i\|_{\mathcal{B}(X,Y)} \le M. \]

Why it matters

The uniform boundedness principle (UBP), also called the Banach–Steinhaus theorem, is one of the three pillars of linear functional analysis alongside the open mapping theorem and the Hahn–Banach theorem. Its power lies in converting a pointwise (i.e. one-value-at-a-time) hypothesis into a uniform (whole-family) conclusion, entirely for free, using only completeness of the domain. This is a genuinely surprising phenomenon: pointwise bounds on infinitely many linear maps say nothing a priori about how those bounds interact across different points of \(X\), yet completeness forces them to combine into a single global bound.

In practice the theorem is the standard tool for proving that a naively-defined limiting operation is automatically continuous, or conversely for proving non-convergence results by exhibiting unbounded pointwise behaviour. It underlies the convergence theory of Fourier series (existence of continuous functions whose Fourier series diverge at a point), the theory of weak convergence, and stability arguments in numerical analysis (e.g. the Lax equivalence theorem for finite-difference schemes).

Hypotheses

Completeness of \(X\) (Banach space, not merely normed). Let \(X = c_{00}\), the space of finitely-supported sequences, with the \(\sup\)-norm (a normed but incomplete space). Define \(T_n : c_{00} \to \mathbb{R}\) by \(T_n(x) = n x_n\). Each \(T_n\) is bounded on \(c_{00}\), and for every fixed \(x \in c_{00}\) only finitely many coordinates are nonzero so \(\sup_n |T_n x| \lt \infty\) (pointwise bounded). But \(\|T_n\| = n \to \infty\), so uniform boundedness fails. Completeness of the domain is essential.

Linearity of each \(T_i\). Without linearity the Baire category argument below (which uses that each set \(\{x : \|T_i x\| \le n \ \forall i\}\) is closed because each \(T_i\) is continuous, and that these sets can be intersected in a way compatible with the vector space structure to produce a genuine ball) has no traction; pointwise-bounded families of merely continuous nonlinear maps \(X \to Y\) need not be equicontinuous or uniformly bounded on any ball. E.g. \(f_n(x) = \min(n, 1/|x|)\) on \(\mathbb{R}\setminus\{0\}\), extended by \(f_n(0)=n\), is pointwise bounded at each fixed \(x\ne 0\) only in a trivial sense but the construction of a uniform bound genuinely uses linearity of \(T_i\) to transfer a bound on a ball to a bound on all of \(X\) via scaling; nonlinear analogues fail routinely.

\(Y\) is merely a normed space (no completeness needed on the codomain). This hypothesis is not needed for the conclusion, only for the domain; stating it here to forestall a common confusion—some students hunt for a counterexample with incomplete \(Y\) and are surprised none exists, because the Baire category argument is applied to \(X\), not \(Y\).

Boundedness (continuity) of each \(T_i\) individually. If the \(T_i\) are merely linear (not assumed continuous), pointwise boundedness of a family of everywhere-defined linear maps on a Banach space in fact forces continuity of each map too (this is a consequence of the theorem's proof, not an extra assumption we can drop); but if one works with linear maps defined only on a dense subspace or allows discontinuous linear functionals (which exist assuming the Axiom of Choice, via a Hamel basis), the sets \(\{x : \|T_i x\| \le n\}\) need not be closed, and the Baire category argument collapses entirely.

Proof

For each \(n \in \mathbb{N}\), define \(E_n = \{x \in X : \sup_{i \in I} \|T_i x\|_Y \le n\} = \bigcap_{i \in I} \{x \in X : \|T_i x\|_Y \le n\}.\)

Each map \(x \mapsto \|T_i x\|_Y\) is continuous, being the composition of the continuous linear map \(T_i\) with the continuous norm on \(Y\); hence each set \(\{x : \|T_i x\|_Y \le n\}\) is closed, being the preimage of the closed set \([0,n]\). An arbitrary intersection of closed sets is closed, so each \(E_n\) is closed. A

\(X = \bigcup_{n=1}^{\infty} E_n.\)

By the pointwise boundedness hypothesis, for every \(x \in X\), \(\sup_{i} \|T_i x\|_Y \lt \infty\), so this supremum is at most some natural number \(n\), i.e. \(x \in E_n\) for that \(n\). Hence every point of \(X\) lies in some \(E_n\). A

Some \(E_{n_0}\) has nonempty interior.

This is the crux of the proof and uses the Baire Category Theorem: a complete metric space is not a countable union of nowhere-dense closed sets. Since \(X\) is a Banach space it is a complete metric space (metric induced by the norm). Steps 1–2 exhibit \(X\) as a countable union of closed sets \(E_n\). If every \(E_n\) were nowhere dense (i.e. had empty interior, since they are already closed), Baire's theorem would be contradicted by Step 2. Hence at least one \(E_{n_0}\) contains a nonempty open set. B t3

There exist \(x_0 \in X\) and \(r \gt 0\) with \(\overline{B}(x_0, r) \subseteq E_{n_0}\), i.e. \(\|x - x_0\| \le r \implies \sup_i \|T_i x\|_Y \le n_0.\)

\(E_{n_0}\) has nonempty interior (Step 3) and is closed, so it contains the closure of some open ball; shrinking slightly if necessary we obtain a closed ball of some radius \(r\gt 0\) centred at some \(x_0\), entirely inside \(E_{n_0}\), by definition of interior point in a normed space. A

For \(\|h\| \le r\) and every \(i \in I\): \(\|T_i h\|_Y \le \|T_i(x_0+h)\|_Y + \|T_i x_0\|_Y \le n_0 + n_0 = 2n_0.\)

Write \(h = (x_0+h) - x_0\) and use linearity of \(T_i\): \(T_i h = T_i(x_0+h) - T_i x_0\), then the triangle inequality in \(Y\). Both \(x_0+h\) and \(x_0\) lie in \(\overline{B}(x_0,r) \subseteq E_{n_0}\) by Step 4 (note \(x_0 \in \overline B(x_0,r)\) trivially, taking \(h=0\)), so each term is bounded by \(n_0\) by definition of \(E_{n_0}\). B

For every \(z \in X\) with \(z \ne 0\), setting \(h = \dfrac{r}{\|z\|}z\) (so \(\|h\|=r\)) gives \(\|T_i z\|_Y = \dfrac{\|z\|}{r}\|T_i h\|_Y \le \dfrac{\|z\|}{r}\cdot 2n_0 = \dfrac{2n_0}{r}\|z\|_X.\)

Rescaling: linearity gives \(T_i h = \frac{r}{\|z\|} T_i z\), so \(\|T_i z\|_Y = \frac{\|z\|}{r}\|T_i h\|_Y\); apply the bound from Step 5 to \(\|T_i h\|_Y\) since \(\|h\|=r\le r\). This holds for every \(i \in I\) simultaneously, with the same constant. B

Set \(M = \dfrac{2n_0}{r}\). Then \(\|T_i\|_{\mathcal{B}(X,Y)} = \sup_{\|z\|\le 1} \|T_i z\|_Y \le M\) for every \(i \in I\), and this bound is verified trivially at \(z=0\) too.

Step 6 gives \(\|T_i z\|_Y \le M\|z\|_X\) for all \(z \ne 0\) (and \(z=0\) is immediate since both sides vanish); this is precisely the definition of the operator norm \(\|T_i\|\) being at most \(M\), by the standard equivalent characterisation \(\|T_i\| = \inf\{c \ge 0 : \|T_i z\| \le c\|z\| \ \forall z\}\). The bound \(M\) does not depend on \(i\), completing the proof. A

Result

\(X\) Banach, \(\{T_i\}_{i\in I} \subseteq \mathcal{B}(X,Y)\), \(\ \sup_i \|T_i x\| \lt \infty\ \ \forall x \in X\ \implies\ \sup_i \|T_i\| \lt \infty.\)

Reading. If a family of bounded linear operators on a complete normed space never blows up at any single input, then it cannot blow up in operator norm either — the family is trapped inside a single ball of \(\mathcal{B}(X,Y)\), even if \(I\) is uncountable.

Scope. Requires \(X\) complete (Banach); \(Y\) need only be normed. Applies to any index set \(I\), not just sequences. It is an existence statement for a bound \(M\); the proof (via Baire category) is non-constructive and gives no explicit value or rate for \(M\) in terms of the pointwise bounds. It says nothing if \(X\) is merely normed but not complete (see Fails without).

Corollaries & converses

Banach–Steinhaus convergence corollary. If \(T_n \in \mathcal{B}(X,Y)\) with \(Y\) a Banach space, and \(T_n x \to Tx\) for every \(x \in X\), then \(\sup_n \|T_n\| \lt \infty\) (by UBP, since a convergent sequence in \(Y\) is bounded, giving pointwise boundedness) and moreover \(T\) itself is linear and bounded, with \(\|T\| \le \liminf_n \|T_n\|\).
Non-convergence criterion. If \(\sup_n \|T_n\| = \infty\) for a sequence of bounded operators on a Banach space, then \(\{T_n x\}\) fails to converge (indeed fails to be bounded) for at least one \(x\) — in fact the set of such "bad" \(x\) is residual (comeagre) in \(X\), a strengthening obtained from the same Baire category argument (the Banach–Steinhaus "resonance" theorem).
Weak boundedness corollary. A weakly bounded subset \(S\) of a normed space \(X\) (i.e. \(\sup_{x\in S}|f(x)|\lt\infty\) for every \(f \in X^*\)) is norm-bounded. This is UBP applied to the canonical embeddings \(J(x) \in X^{**} = \mathcal{B}(X^*, \mathbb{K})\) with the complete space now being \(X^*\) (always Banach), not \(X\).
Converse fails. The converse "uniformly bounded \(\implies\) pointwise bounded" is trivially true (uniform boundedness is the stronger statement, since \(\|T_ix\|\le\|T_i\|\|x\|\)) and is not the content of the theorem; the interesting logical direction, pointwise \(\implies\) uniform, does not reverse into an "if and only if" needing separate proof — there is nothing more to state. What genuinely fails to converse is completeness: uniform boundedness of a pointwise-bounded family can occur even when \(X\) is incomplete for special families, so the theorem's hypothesis is sufficient but not in general necessary for any particular family.

Fails without

Completeness of \(X\) dropped: on \(X = c_{00}\) (finite sequences, sup norm), \(T_n(x) = n x_n\) is pointwise bounded (each \(x\) has finite support so only finitely many \(T_n\) act nontrivially on it) yet \(\|T_n\| = n \to \infty\); no uniform bound exists. The completion of \(c_{00}\) is \(c_0\), on which the analogous statement would force \(T_n\) to not even be pointwise bounded at, e.g., \(x=(1,1,1,\dots)\notin c_0\) — illustrating exactly how incompleteness lets pathological unbounded points "hide" outside the space.
Linearity dropped: on \(X=\mathbb{R}\) (complete), let \(f_n(x) = nx\) for \(x \in [0,1/n]\) and \(f_n(x) = 1\) for \(x\gt 1/n\), extended evenly; these are continuous but only piecewise linear "kinks," and one can engineer pointwise-bounded but norm-blind-up families once linearity (needed for the rescaling in Step 6) is abandoned; more sharply, replace \(T_i\) by non-linear continuous \(g_n(x) = \max(0, n - n^2|x - 1/n|)\) (a spike of height \(n\) near \(1/n\)): at any fixed \(x\), \(g_n(x)=0\) for all large \(n\), so pointwise bounded (indeed pointwise convergent to 0), yet \(\sup_x g_n(x) = n \to \infty\), so no uniform bound on the family exists, precisely because there is no linearity to convert the ball-bound of Step 4 into a global bound via rescaling.
Individual boundedness (continuity) of each \(T_i\) dropped: using a Hamel basis of \(X=\ell^2\) one can construct a single discontinuous linear functional \(f\); then even the one-element family \(\{f\}\) is "pointwise bounded" only trivially (a family of size one is always pointwise bounded, at each \(x\) the value \(f(x)\) is just some fixed finite number) but is not in \(\mathcal{B}(X,\mathbb{K})\) and the notion of "\(\sup_i \|T_i\|\)" is not even finite for that single map. The theorem's proof needs continuity of each \(T_i\) to make the sets \(E_n\) closed in Step 1; without it the Baire argument does not apply.

Common errors

Believing UBP gives an explicit formula or algorithm for the bound \(M\) in terms of the pointwise bounds \(\sup_i\|T_ix\|\); the proof is non-constructive (existence of \(n_0\) via Baire category, no way to compute it) and no such formula exists in general.
Forgetting that completeness must hold for the domain \(X\), not the codomain \(Y\); students hunt fruitlessly for counterexamples with \(Y\) incomplete, when in fact \(Y\) plays no role in the Baire category argument at all.
Applying UBP to a family that is only pointwise bounded on a proper dense subspace of \(X\), rather than on all of \(X\); the union \(X=\bigcup E_n\) in Step 2 genuinely requires every point of \(X\), not merely a dense subset, to appear in some \(E_n\) (though there are refined versions using an equicontinuity-plus-density argument, they are separate results, not a direct application of the theorem as stated).
Confusing "pointwise bounded" (\(\sup_i \|T_i x\|\lt\infty\) for each fixed \(x\)) with "pointwise convergent" (\(T_i x\) converges as \(i\to\infty\)); convergence implies boundedness but the theorem only needs the weaker hypothesis, and misstating the hypothesis as convergence over-restricts the theorem's applicability.
Trying to prove UBP by a "naive supremum swap" \(\sup_i \sup_{\|x\|\le1} \|T_ix\| = \sup_{\|x\|\le1}\sup_i\|T_ix\|\) and asserting the right side is finite because each inner sup is — this begs the question entirely: finiteness of \(\sup_{\|x\|\le 1}\sup_i \|T_ix\|\) is exactly \(\sup_i\|T_i\|<\infty\), the very conclusion being sought, and no elementary swap of suprema can produce it without Baire category or an equivalent completeness argument.

Discussion

The theorem was proved by Stefan Banach and Hugo Steinhaus in 1927, building on earlier special cases studied by Hans Hahn and by Eduard Helly, and on the general Baire category machinery developed by René-Louis Baire in his 1899 thesis for the real line. The Baire category theorem itself says that a complete metric space cannot be written as a countable union of nowhere-dense sets; the genius of the Banach–Steinhaus argument is recognising that the "bad set" for a pointwise-bounded family, sliced into levels \(E_n\), is exactly amenable to that dichotomy, and that once you have one ball on which the family is uniformly bounded, linearity transports that bound to the whole space by rescaling.

UBP sits alongside the open mapping theorem and the closed graph theorem as one of the three classical "Baire category consequences" in Banach space theory; all three ultimately rest on the same completeness-driven category argument, and each can in fact be derived from either of the others with some work, though the direct proofs (as given here for UBP) are the standard route in most treatments.

A particularly striking classical application is the existence of a continuous \(2\pi\)-periodic function whose Fourier series diverges at a point. Let \(S_n(f)\) denote the partial-sum operator on \(C(\mathbb{T})\) (continuous periodic functions, sup norm, a Banach space) evaluated at \(0\); one computes \(\|S_n\| \asymp \log n \to \infty\) (the Lebesgue constants), and UBP's contrapositive (via the corollary above) then guarantees a continuous \(f\) with \(S_n(f)(0)\) unbounded, without ever constructing \(f\) explicitly — a beautiful instance of a "soft" existence proof from Baire category succeeding where explicit construction is painful.

A subtler point for advanced students: the "resonance" strengthening (mentioned under Corollaries) shows that whenever a sequence of bounded operators fails to be uniformly bounded, the set of points at which the sequence is even bounded is not just non-everything but meagre (a countable union of nowhere-dense sets) — so "most" points, in the Baire category sense, exhibit the blow-up. This is considerably stronger than merely asserting one bad point exists, and is often what is actually used (e.g. in the Fourier divergence example, to get a generic rather than merely existential counterexample). Common misconception: that UBP requires \(I\) to be countable, or that "uniformly bounded" means "bounded by the same operator norm value across a sequence indexed by \(\mathbb{N}\)" — the theorem holds for an arbitrary index set \(I\), countable or not, since the Baire category argument never uses countability of \(I\) itself, only countability of the levels \(n \in \mathbb{N}\) used to slice \(X\).

Worked examples

Let \(X = \ell^1(\mathbb{N})\) (Banach, with \(\|x\|_1 = \sum_k |x_k|\)) and for each \(n\in\mathbb{N}\) define \(T_n : \ell^1 \to \mathbb{K}\) by \(T_n(x) = \sum_{k=1}^n k\, x_k\). Show \(\{T_n\}\) is pointwise bounded and find an explicit uniform bound, comparing it with the abstract UBP conclusion.

Setting up the example: each \(T_n\) is a finite linear combination of coordinate functionals, hence clearly linear and bounded. A

For fixed \(x \in \ell^1\): \(|T_n(x)| = \left|\sum_{k=1}^n k x_k\right| \le \sum_{k=1}^n k|x_k| \le \sum_{k=1}^{\infty} k|x_k|.\) If additionally \(x\) has, say, only finitely many nonzero terms or \(\sum k|x_k|\lt\infty\), this is a finite bound independent of \(n\); pointwise boundedness holds provided we restrict to \(x\) with \(\sum k|x_k| \lt \infty\) — but for general \(x \in \ell^1\), \(\sum k |x_k|\) may diverge (e.g. \(x_k = 1/k^2\) gives \(\sum k\cdot k^{-2} = \sum 1/k\) divergent), so in fact \(\sup_n |T_n x|\) can be \(\infty\) for such \(x\): this family is not pointwise bounded on all of \(\ell^1\), correctly reflecting that \(\|T_n\|_{(\ell^1)^*} = n \to \infty\) (since \(\|T_n\| = \max_{1\le k\le n} k = n\), attained at the standard basis vector \(e_n\)), so no uniform bound exists — consistent with UBP's contrapositive, not a violation of it.

Direct computation of the operator norm on \(\ell^1\): for functionals of the form \(x\mapsto\sum c_kx_k\), \(\|T\|_{(\ell^1)^*}=\sup_k|c_k|\), a standard fact about the dual of \(\ell^1\) being \(\ell^\infty\). B

\(\sup_n \|T_n\|=\infty \implies\) (by UBP's contrapositive) the family \(\{T_n\}\) cannot be pointwise bounded on all of \(\ell^1\); indeed it fails exactly on those \(x\) with \(\sum k|x_k|=\infty\).

Reading. This example demonstrates the theorem's contrapositive in action: by computing operator norms directly and finding they are unbounded, we can conclude — without searching case by case — that some (indeed generically most, by the resonance corollary) vector in \(\ell^1\) must produce unbounded values, and we identified explicitly which vectors fail.

Let \(X = C[0,1]\) (Banach, sup norm) and let \(x_n = 1/n \in [0,1]\) for \(n\ge 1\). Define \(T_n : X \to \mathbb{R}\) by \(T_n(f) = n\big(f(1/n) - f(0)\big)\), a discretised derivative-at-0 operator. Determine whether \(\{T_n\}\) is pointwise bounded, and if not, deduce a conclusion about differentiability via UBP's contrapositive.

Each \(T_n\) is linear (difference of evaluation functionals scaled by \(n\)) and bounded, with \(\|T_n\| \le n\cdot(1+1) = 2n\) by the triangle inequality and \(\|f\|_\infty\ge|f(1/n)|,|f(0)|\); in fact \(\|T_n\|=2n\) exactly, attained by an \(f\) with \(f(1/n)=1,f(0)=-1,\|f\|_\infty=1\). A

Since \(\|T_n\| = 2n \to \infty\), UBP's contrapositive says \(\{T_n\}\) cannot be pointwise bounded on all of \(C[0,1]\): there exists (in fact a residual set of) \(f \in C[0,1]\) with \(\sup_n |T_n(f)| = \sup_n n|f(1/n)-f(0)| = \infty\).

Applying the corollary "if \(\sup_n\|T_n\|=\infty\) then pointwise boundedness fails somewhere," established above as a direct restatement of the theorem's contrapositive. B

Interpretation: such an \(f\) has right-hand difference quotients at \(0\) that are unbounded as \(n\to\infty\) along \(x_n=1/n\), so \(f\) is not right-differentiable at \(0\) (a fortiori not differentiable at \(0\)).

Direct unwinding of the definition: \(T_n(f) = \frac{f(1/n)-f(0)}{1/n}\) is exactly the difference quotient; if this is unbounded as \(n\to\infty\) the limit defining the right derivative cannot exist (a convergent sequence of reals is bounded). A

There exists a continuous \(f:[0,1]\to\mathbb{R}\) that is not differentiable at \(0\), produced abstractly (no explicit formula needed) via UBP.

Reading. This mirrors the classical existence proof of continuous-but-nowhere-differentiable functions: rather than exhibiting a Weierstrass-type formula, UBP guarantees existence purely from the growth rate \(\|T_n\|\to\infty\) of the associated difference-quotient functionals, and moreover shows the "bad" functions are topologically generic, not rare exceptions.

Problems

Let \(X\) be a Banach space and \(\{f_n\} \subseteq X^*\) (continuous linear functionals) satisfy \(\sup_n |f_n(x)| \lt \infty\) for every \(x \in X\). Prove that \(\{f_n\}\) converges weak-* to some \(f \in X^*\) whenever it also converges pointwise on a dense subset \(D\subseteq X\) with \(\|f_n\|\le M\) for all \(n\) already known independently. State clearly which step uses UBP itself and which uses the density/\(\varepsilon\)-argument.
Solution
First, if \(\|f_n\|\le M\) is already assumed, UBP is not strictly needed to get the bound (it is given); the role of UBP in the general version of this exercise (without the extra assumption) is to derive \(\sup_n\|f_n\|=M'\lt\infty\) directly from the pointwise boundedness hypothesis \(\sup_n|f_n(x)|\lt\infty\) for every \(x\in X\), via the theorem exactly as proved above.

Given the uniform bound \(M\) (from UBP or by hypothesis) and pointwise convergence \(f_n(d)\to L(d)\) on a dense set \(D\), extend to a limit functional on all of \(X\): for \(x\in X\) and \(\varepsilon\gt0\), choose \(d\in D\) with \(\|x-d\|\lt \varepsilon/(3M)\) (density of \(D\)); then for \(m,n\) large enough that \(|f_n(d)-f_m(d)|\lt\varepsilon/3\) (Cauchy on \(D\), from pointwise convergence there), \[|f_n(x)-f_m(x)| \le |f_n(x)-f_n(d)| + |f_n(d)-f_m(d)| + |f_m(d)-f_m(x)| \le M\|x-d\|+\varepsilon/3+M\|x-d\| \lt \varepsilon.\] So \((f_n(x))_n\) is Cauchy in \(\mathbb{K}\) (complete), hence converges to some \(f(x)\); linearity of \(f\) follows from linearity of each \(f_n\) and pointwise limits preserving linear combinations, and \(f\) is bounded with \(\|f\|\le M\) since \(|f(x)|=\lim_n|f_n(x)|\le M\|x\|\). Hence \(f_n \to f\) weak-* and \(f\in X^*\).
Give an example of a family of bounded linear operators \(\{T_i\}_{i\in I}\) on a Banach space \(X\) that is uniformly bounded in norm but where the index set \(I\) is uncountable, illustrating that UBP is not merely a "sequence" theorem.
Solution
Let \(X = \ell^2(\mathbb{R})\) (an uncountable orthonormal-basis Hilbert space, technically requires care but a cleaner concrete choice: let \(X=L^2[0,1]\) and for each \(t\in[0,1]\) let \(T_t: X\to\mathbb{R}\) be... evaluation functionals are not bounded on \(L^2\), so instead use: let \(X = C[0,1]\) and for each \(t \in [0,1]\) (uncountable index set) define \(T_t(f) = f(t)\), the evaluation functional at \(t\). Each \(T_t\) is linear and bounded with \(\|T_t\| = 1\) exactly (attained by constant function \(1\)), so trivially \(\sup_{t\in[0,1]}\|T_t\| = 1 \lt \infty\): a genuinely uncountably-indexed uniformly bounded family. This also trivially satisfies pointwise boundedness (\(\sup_t |f(t)| = \|f\|_\infty \lt \infty\) for every fixed \(f\in C[0,1]\)), consistent with UBP, though here uniform boundedness is checked directly rather than derived from the theorem.
Let \(X=Y=\ell^2\) and let \(T_n\) be the right-shift-then-truncate operator \(T_n(x) = (0,\dots,0,x_1,x_2,\dots)\) shifted by \(n\) places then truncated to have zero entries beyond index \(2n\) (i.e. \(T_n(x)_k = x_{k-n}\) for \(n\lt k\le 2n\), else \(0\)). Show directly (without invoking UBP) that \(\sup_n\|T_n\|\le 1\), and explain why this is consistent with, but not a proof of, UBP.
Solution
For any \(x\in\ell^2\), \(\|T_nx\|_2^2 = \sum_{k=n+1}^{2n}|x_{k-n}|^2 = \sum_{j=1}^{n}|x_j|^2 \le \sum_{j=1}^\infty|x_j|^2 = \|x\|_2^2\), so \(\|T_nx\|\le\|x\|\) for every \(x\), giving \(\|T_n\|\le 1\) for every \(n\), hence \(\sup_n\|T_n\|\le1\) directly. This is an example where the uniform bound is verified by hand, exhibiting explicit control that UBP's non-constructive proof (via Baire category) does not provide in general; it is "consistent with" UBP in the trivial sense that a uniformly bounded family is automatically pointwise bounded (the easy direction), but does not showcase the theorem's genuine content, which is the pointwise-to-uniform implication.
Explain precisely why the Baire category theorem, rather than a naive covering or compactness argument, is essential to Step 3 of the proof: could one instead argue "each \(E_n\) is closed and their union is \(X\), so by compactness some finite subcollection covers a large closed ball" to reach the same conclusion?
Solution
No — this approach fails at the outset because \(X\) is typically not compact (closed bounded sets in an infinite-dimensional normed space are never compact, by Riesz's lemma), so there is no compactness argument available to extract a finite subcover; the sets \(E_n\) are also not required to be bounded or "small" individually, only closed, so no covering-number argument applies either. The Baire category theorem is precisely the correct (and essentially only) tool because it makes no compactness assumption at all, relying solely on completeness of the metric \(d(x,y)=\|x-y\|\) on \(X\): completeness guarantees that a nested sequence of closed balls with radii \(\to 0\) has nonempty intersection (via a Cauchy sequence argument), which is exactly what powers the proof that a countable union of closed sets with empty interior cannot cover a complete space. Attempting the proof via compactness is a common confusion for students transitioning from finite-dimensional intuition, where closed-and-bounded does imply compact (Heine–Borel), but this fails completely in the infinite-dimensional Banach space setting relevant here.
Let \(X\) be a Banach space and suppose \(B: X\times X \to \mathbb{K}\) is a bilinear form such that \(B(x,\cdot)\) is a bounded linear functional for each fixed \(x\), and \(B(\cdot,y)\) is a bounded linear functional for each fixed \(y\) (separate continuity). Use UBP to show \(B\) is jointly continuous, i.e. there exists \(C\) with \(|B(x,y)| \le C\|x\|\|y\|\) for all \(x,y\).
Solution
For each \(x\in X\) with \(\|x\|\le1\), define \(T_x : X \to \mathbb{K}\) by \(T_x(y) = B(x,y)\); by hypothesis (separate continuity in the second slot) each \(T_x\) is a bounded linear functional. Consider the family \(\{T_x : \|x\|\le 1\}\), indexed by the unit ball of \(X\) (an uncountable index set in general, which is fine per Problem 2's observation that UBP does not need countability). Fix \(y\in X\); by separate continuity in the first slot, the map \(x\mapsto B(x,y)=T_x(y)\) is a bounded linear functional of \(x\), so in particular it is bounded on the unit ball: \(\sup_{\|x\|\le1}|T_x(y)| = \sup_{\|x\|\le1}|B(x,y)| \lt \infty\) for this fixed \(y\). This holds for every fixed \(y\in X\), i.e. the family \(\{T_x\}_{\|x\|\le1}\) is pointwise bounded on \(X\) (viewing \(y\) as the variable). Since \(X\) is Banach, UBP applies: \(\sup_{\|x\|\le1}\|T_x\| = C \lt \infty\) for some constant \(C\). Then for general \(x,y\ne0\), by linearity/homogeneity of \(B\) in each slot, \(|B(x,y)| = \|x\|\|y\|\,\big|B(x/\|x\|, y/\|y\|)\big| = \|x\|\|y\|\,|T_{x/\|x\|}(y/\|y\|)| \le \|x\|\|y\|\cdot C\|y/\|y\|\| = C\|x\|\|y\|\), using \(\|T_{x/\|x\|}\|\le C\) and \(\|y/\|y\|\|=1\). This is the standard "separate continuity implies joint continuity for bilinear forms on Banach spaces" theorem, a direct and important application of UBP.