From Category to Uniform Bounds
The Baire Category Theorem is more than a statement about intersections of open dense sets. It also converts countably many pointwise conditions into a conclusion that holds throughout some open region. In this tutorial, we develop two applications of that idea: a local boundedness result for continuous functions and the Uniform Boundedness Principle for operators on a Banach space.
The common structure is simple. A pointwise bound says that every point belongs to at least one of a countable collection of sets on which a uniform bound holds. If those sets are closed and cover a complete metric space, the Baire Category Theorem prevents all of them from having empty interior. At least one bound must therefore hold on an open set. For operators, linearity then turns that open set into a bound on the operator norms over the entire unit ball.
Recall that a Banach space is a normed vector space that is complete in the metric induced by its norm. A bounded linear operator \(T:X\to Y\) between normed spaces has operator norm \(\|T\|=\sup\{\|Tx\|_Y:\|x\|_X\leq1\}\). The operator is bounded precisely when this supremum is finite.
Proof. For each positive integer \(m\), define \(E_m=\{x\in X: |f_n(x)|\leq m\text{ for every }n\}\). For each fixed \(n\), the set \(\{x:|f_n(x)|\leq m\}\) is closed, because \(f_n\) is continuous and the interval \([-m,m]\) is closed. Thus \(E_m=\bigcap_{n=1}^{\infty}\{x:|f_n(x)|\leq m\}\) is closed. The pointwise boundedness assumption says that for each \(x\), some positive integer \(m\) is at least \(\sup_n|f_n(x)|\). Hence \(X=\bigcup_{m=1}^{\infty}E_m\).
If every \(E_m\) had empty interior, then, being closed, each would be nowhere dense. The Baire Category Theorem would imply that their union cannot contain the nonempty open set \(X\), contradicting \(X=\bigcup_mE_m\). Therefore some \(E_M\) has nonempty interior. Let \(U\) be a nonempty open subset of \(E_M\). By its definition, \(|f_n(x)|\leq M\) for all \(x\in U\) and all \(n\). This proves the result. \(\square\)
The conclusion is deliberately local: a common bound is obtained on some nonempty open set, not necessarily on all of \(X\). The proof uses continuity to make the sets \(E_m\) closed, and completeness to apply the Baire Category Theorem. The next result uses the same closed-set idea, together with the algebraic structure of linear operators, to obtain a global bound on operator norms.
The Uniform Boundedness Principle
Proof. If \(X=\{0\}\), every linear operator from \(X\) is zero, so the conclusion holds. Suppose \(X\neq\{0\}\). For each positive integer \(m\), set \(E_m=\{x\in X:\|Tx\|_Y\leq m\text{ for every }T\in\mathcal{T}\}\). For each \(T\), the function \(x\mapsto\|Tx\|_Y\) is continuous: boundedness of \(T\) gives \(\bigl|\|Tx\|_Y-\|Tz\|_Y\bigr|\leq\|T(x-z)\|_Y\leq\|T\|\|x-z\|_X\). Consequently, \(\{x:\|Tx\|_Y\leq m\}\) is closed, and \(E_m\), the intersection of these closed sets over \(T\in\mathcal{T}\), is closed. Pointwise boundedness implies \(X=\bigcup_{m=1}^{\infty}E_m\).
The space \(X\) is complete, so the Baire Category Theorem shows that some \(E_N\) has nonempty interior. There are \(x_0\in X\) and \(r>0\) such that \(B(x_0,r)\subseteq E_N\). In particular, \(x_0\in E_N\). Let \(u\in X\) satisfy \(\|u\|_X\leq1\), and put \(h=(r/2)u\). Since \(\|h\|_X\leq r/2<r\), both \(x_0+h\) and \(x_0\) belong to \(E_N\). Therefore, for every \(T\in\mathcal{T}\), linearity and the triangle inequality give \(\|Th\|_Y=\|T(x_0+h)-Tx_0\|_Y\leq\|T(x_0+h)\|_Y+\|Tx_0\|_Y\leq2N\). Because \(h=(r/2)u\), homogeneity yields \(\|Tu\|_Y\leq4N/r\). This bound holds for every \(u\) with \(\|u\|_X\leq1\) and every \(T\in\mathcal{T}\). Taking the supremum first over \(u\) and then over \(T\) gives \(\sup_{T\in\mathcal{T}}\|T\|\leq4N/r<\infty\). \(\square\)
The pointwise hypothesis allows the bound to depend on \(x\): the quantity \(\sup_{T\in\mathcal{T}}\|Tx\|_Y\) may vary from point to point. The theorem asserts that, on a Banach space, these individual bounds force one common bound for all the operator norms. Completeness is what rules out the possibility that the closed sets encoding the pointwise bounds all have empty interior.
Worked Applications
Worked Example: Coordinate Functionals on a Sequence Space
Let \(X=\ell^2\), the Banach space of square-summable real sequences with norm \(\|x\|_2=(\sum_{k=1}^{\infty}|x_k|^2)^{1/2}\). For each positive integer \(n\), define \(T_n:\ell^2\to\mathbb{R}\) by \(T_n(x)=x_n\). This is linear, and \(|T_n(x)|=|x_n|\leq\|x\|_2\), so \(T_n\) is bounded and \(\|T_n\|\leq1\).
For the sequence \(e_n\), whose \(n\)-th entry is \(1\) and whose other entries are \(0\), we have \(\|e_n\|_2=1\) and \(T_n(e_n)=1\). Thus \(\|T_n\|\geq1\), and hence \(\|T_n\|=1\). For any fixed \(x\in\ell^2\), the estimate \(|T_n(x)|=|x_n|\leq\|x\|_2\) holds for every \(n\), so the family is pointwise bounded. The Uniform Boundedness Principle applies, and its conclusion \(\sup_n\|T_n\|=1<\infty\) agrees with the direct calculation.
Worked Example: Why Completeness Matters
Let \(c_{00}\) be the vector space of real sequences with only finitely many nonzero entries, equipped with the norm inherited from \(\ell^2\). Define \(T_n:c_{00}\to\mathbb{R}\) by \(T_n(x)=n x_n\). Each \(T_n\) is linear, and \(|T_n(x)|=n|x_n|\leq n\|x\|_2\), so it is bounded and \(\|T_n\|\leq n\). Taking \(x=e_n\), for which \(\|e_n\|_2=1\), gives \(T_n(e_n)=n\). Therefore \(\|T_n\|=n\), and the operator norms are unbounded.
Nevertheless, this family is pointwise bounded. If \(x\in c_{00}\), there is an integer \(K\) such that \(x_n=0\) for every \(n>K\). Thus \(\sup_{n\geq1}|T_n(x)|=\max_{1\leq n\leq K}n|x_n|<\infty\) when \(K\geq1\); if \(x=0\), the supremum is \(0\). The apparent contradiction with the Uniform Boundedness Principle disappears because \(c_{00}\) is not complete in the inherited norm. For instance, the finite truncations of the sequence \((1/n)_{n\geq1}\) are Cauchy in \(\ell^2\), but their limit is not in \(c_{00}\). This example shows why the Banach-space hypothesis cannot simply be omitted.
Worked Example: Pointwise Convergence Without Uniform Boundedness
On \(X=[0,1]\), with its usual metric, let \(c_n=2^{-n}\), \(w_n=2^{-n}/4\), and define \(f_n(x)=n\max(0,1-|x-c_n|/w_n)\). Each \(f_n\) is continuous, is zero outside \(S_n=[(3/4)2^{-n},(5/4)2^{-n}]\), and has maximum \(f_n(c_n)=n\). In particular, \(\sup_{x\in[0,1]}|f_n(x)|=n\).
The intervals \(S_n\) are pairwise disjoint: the upper endpoint of \(S_{n+1}\) is \((5/8)2^{-n}\), which is less than the lower endpoint \((3/4)2^{-n}\) of \(S_n\). Hence any fixed \(x>0\) belongs to at most one of these intervals, and \(x=0\) belongs to none. It follows that \(\sup_n|f_n(x)|<\infty\) for every \(x\). In fact, \(f_n(x)\to0\) at every \(x\), since at most one term can be nonzero at that point.
The sequence is not uniformly bounded on all of \(X\), because its supremum norms are \(n\). This does not conflict with the local uniform boundedness theorem. For example, every \(f_n\) is zero on the nonempty relatively open set \((5/8,1]\), so the family is uniformly bounded there. More generally, the theorem guarantees some nonempty open region of uniform boundedness; it does not claim a common bound on the entire space.
What the Applications Do—and Do Not—Say
The two results illustrate different strengths of the same category argument. For continuous functions on a complete metric space, pointwise boundedness guarantees a common bound on at least one nonempty open set. For bounded linear operators on a Banach space, the additional linear structure upgrades that local conclusion to a uniform bound on the operator norms, which controls every vector in the unit ball.
A common pitfall is to treat pointwise boundedness as if it already meant uniform boundedness. It does not: the bound may depend on the point, as the continuous spike example demonstrates. Another is to omit completeness. In each proof, pointwise boundedness gives a countable closed cover. Baire category supplies interior to one of the closed sets only because the underlying space is complete (or, more generally, is a Baire space). Without that hypothesis, the conclusion can fail.
Check Your Understanding
Use the closed-set covers and their hypotheses to answer the following questions.
- Why is each set \(E_m\) in the local uniform boundedness proof closed?
- Where does completeness enter the proof of the Uniform Boundedness Principle?
- Why does a bound on \(T(x_0+h)\) and \(T(x_0)\) give a bound on \(T(u)\) for \(\|u\|_X\leq1\)?
- For the coordinate functionals on \(\ell^2\), which vector shows that \(\|T_n\|\geq1\)?
- How can the operators on \(c_{00}\) be pointwise bounded even though their norms tend to infinity?
- What does the local uniform boundedness theorem guarantee for the continuous spike sequence, and what does it not guarantee?