The Uniform Boundedness Principle

From Category to Operator Norms

The previous tutorials developed the Baire Category Theorem and the language of meager and comeager sets. These ideas have a striking consequence for families of linear operators: on a Banach space, bounds that hold separately at each point force one bound that works for every operator. The Uniform Boundedness Principle, also called the Banach–Steinhaus Theorem, was stated earlier in this course. Here we examine how to use it, what it implies for convergent sequences of operators, and what category says when the operator norms are not bounded.

Let \(X\) and \(Y\) be normed spaces. For a bounded linear operator \(T:X\to Y\), its operator norm is \(\|T\|=\sup_{\|x\|\leq 1}\|Tx\|\). A family \(\mathcal{T}\) is pointwise bounded if, for each fixed \(x\in X\), the values \(\|Tx\|\) have a finite common bound as \(T\) ranges over \(\mathcal{T}\). This bound is allowed to depend on \(x\). Uniform boundedness is the stronger-looking conclusion that one finite constant bounds all the operator norms.

The key distinction is the order of the quantifiers. Pointwise boundedness says that for every \(x\), there is a bound \(M_x\) that works for all operators at that \(x\). It does not say that the same \(M_x\) works for every \(x\). The principle says that, provided the domain is complete, this collection of point-dependent bounds nevertheless implies a common bound on the operator norms.

Consequences for Pointwise-Convergent Sequences

A sequence \(T_n:X\to Y\) is pointwise convergent if, for each \(x\in X\), the sequence \(T_nx\) converges in \(Y\). A convergent sequence is bounded, so pointwise convergence supplies the pointwise boundedness required by the Uniform Boundedness Principle. This gives a uniform bound on \(\|T_n\|\). The pointwise limit also inherits linearity and boundedness, as the following result makes precise.

Proof. For each \(x\in X\), the sequence \(T_nx\) converges and is therefore bounded. The family \(\{T_n:n\geq1\}\) is pointwise bounded. Since \(X\) is Banach, the Uniform Boundedness Principle gives a finite constant \(C=\sup_n\|T_n\|\).

For \(x,z\in X\) and scalars \(a,b\), linearity of every \(T_n\) gives \(T_n(ax+bz)=aT_nx+bT_nz\). Taking limits in \(Y\), using the assumed convergence at \(ax+bz\), \(x\), and \(z\), yields \(T(ax+bz)=aTx+bTz\). Thus \(T\) is linear. Moreover, for every \(n\), \(\|T_nx\|\leq C\|x\|\). Continuity of the norm under convergence in \(Y\) gives \(\|Tx\|=\lim_{n\to\infty}\|T_nx\|\leq C\|x\|\). Hence \(T\) is bounded and \(\|T\|\leq C=\sup_n\|T_n\|\). \(\square\)

The theorem gives a uniform upper bound for the norms in this example, as it must. It does not say that the norms tend to zero, or even that the operators approach their pointwise limit uniformly on the unit ball. Those are stronger conclusions and require additional hypotheses.

What Happens When the Norms Are Unbounded?

There is a useful category-theoretic strengthening of the usual conclusion. Instead of merely saying that pointwise boundedness implies bounded operator norms, we can describe what happens if the operator norms are unbounded: the points where the operator values are unbounded form a dense \(G_\delta\) subset of the Banach space. Here a \(G_\delta\) is a countable intersection of open sets.

Proof. For each positive integer \(m\), define \(E_m=\{x\in X:\|T_nx\|\leq m\text{ for every }n\}\). Each \(T_n\) is continuous, so \(\{x:\|T_nx\|\leq m\}\) is closed. Since \(E_m=\bigcap_{n=1}^{\infty}\{x:\|T_nx\|\leq m\}\), the set \(E_m\) is closed.

We show that \(E_m\) has empty interior. Suppose instead that it contains an open ball \(B(x_0,r)\), where \(r>0\). Because \(x_0\in E_m\), we have \(\|T_nx_0\|\leq m\) for every \(n\). If \(\|h\|<r\), then \(x_0+h\in E_m\), so \(\|T_n(x_0+h)\|\leq m\) as well. Linearity and the triangle inequality give, for every \(n\), \(\|T_nh\|\leq\|T_n(x_0+h)\|+\|T_nx_0\|\leq2m\).

For any \(u\) with \(\|u\|\leq1\), take \(h=(r/2)u\). Then \(\|h\|\leq r/2<r\), so \((r/2)\|T_nu\|=\|T_nh\|\leq2m\), and consequently \(\|T_nu\|\leq4m/r\). Taking the supremum over such \(u\) gives \(\|T_n\|\leq4m/r\) for every \(n\). This contradicts \(\sup_n\|T_n\|=\infty\). Thus \(E_m\) has empty interior. Since it is closed, it is nowhere dense.

By the Baire Category Theorem, the intersection \(\bigcap_{m=1}^{\infty}(X\setminus E_m)\) is dense; it is a \(G_\delta\) because every \(X\setminus E_m\) is open. A point \(x\) belongs to this intersection exactly when, for every positive integer \(m\), some \(n\) satisfies \(\|T_nx\|>m\). This is equivalent to \(\sup_n\|T_nx\|=\infty\), proving the claim. \(\square\)

Using the Principle Carefully

The Uniform Boundedness Principle is useful whenever a family of operators is easier to control at each fixed input than on the entire unit ball. A typical argument first verifies that for every \(x\), the numbers \(\|T_nx\|\) remain bounded, perhaps because \(T_nx\) converges. The principle then supplies a common bound on \(\|T_n\|\). The dense blow-up result gives a complementary diagnostic: if those norms are unbounded on a Banach space, pointwise boundedness must fail, and the failure occurs on a dense \(G_\delta\) set.

Two cautions are important. First, the common bound on operator norms does not imply operator-norm convergence, as the coordinate-functional example demonstrates. Second, completeness belongs to the domain space \(X\), not necessarily to the target \(Y\). It is the Baire property of the domain that drives the principle. In applications, identify the space on which the operators act and check that it is Banach before invoking the result.