The Statement and Its Quantifiers
The Uniform Boundedness Principle, also called the Banach–Steinhaus Theorem, turns bounds that may depend on the input vector into a common bound on operator norms. Earlier in this course, the principle was used to study pointwise limits and families whose values become unbounded. Here the focus is its precise statement: which spaces and operators it concerns, what its two boundedness conditions mean, and what the conclusion does—and does not—say.
Let \(X\) be a normed space, let \(Y\) be a normed space, and let \(\mathcal{T}\) be a family of bounded linear operators from \(X\) to \(Y\). Pointwise boundedness requires that each fixed \(x\in X\) have its own finite bound for all the operators. Uniform boundedness instead requires one finite bound for the operator norms. The distinction is captured by the order of the quantifiers: the pointwise bound may depend on \(x\), while the operator-norm bound may not depend on \(T\in\mathcal{T}\).
The hypothesis does not require a single constant \(M\) such that \(\|Tx\|\leq M\) for every \(T\) and every \(x\in X\). Such a bound over the whole space would usually be impossible for any nonzero linear operator, since scaling \(x\) scales \(Tx\). The pointwise condition says that for each fixed input there is a finite bound; the conclusion says there is a common constant \(C\) such that \(\|Tx\|\leq C\|x\|\) for every \(T\in\mathcal{T}\) and every \(x\in X\).
The assumption that \(X\) is Banach is essential to the stated principle. The target \(Y\), by contrast, need only be normed; it need not be complete. Boundedness and linearity of each \(T\) are also part of the setup. The principle does not apply to arbitrary functions merely because their values are bounded at each point.
Reading the Conclusion Correctly
A uniform operator-norm bound controls every input in proportion to its norm. It does not claim that the operators are close to one another, that they converge, or that their values have a common bound independent of the size of the input. A useful way to read the conclusion is as a single estimate valid simultaneously for the entire family:
The operator norm is defined by \(\|T\|=\sup_{\|x\|\leq1}\|Tx\|\). Thus, a bound \(\|T\|\leq C\) immediately gives \(\|Tx\|\leq C\|x\|\): for \(x\ne0\), apply the definition to \(x/\|x\|\), and for \(x=0\), use linearity. Conversely, if the latter estimate holds for every \(x\), then taking the supremum over the unit ball gives \(\|T\|\leq C\). This is why the conclusion can be expressed either as a bound on operator norms or as one simultaneous estimate for all inputs.
Worked Example: Coordinate Functionals on \(\ell^2\)
Let \(X=\ell^2\), and for each positive integer \(n\) define \(T_n(x)=x_n\), viewed as an element of \(\mathbb{R}\). Each \(T_n\) is linear. For \(x=(x_k)_{k\geq1}\in\ell^2\), \[ |T_n(x)|^2=|x_n|^2\leq\sum_{k=1}^{\infty}|x_k|^2=\|x\|_2^2, \] so \(\|T_n(x)\|\leq\|x\|_2\). In particular, each \(T_n\) is bounded. The sequence \(e_n\), with \(n\)th entry \(1\) and all other entries \(0\), has \(\|e_n\|_2=1\) and \(T_n(e_n)=1\). Therefore \(\|T_n\|=1\).
For every fixed \(x\), the pointwise bound can be taken to be \(M_x=\|x\|_2\), since \(|T_n(x)|\leq\|x\|_2\) for every \(n\). The Uniform Boundedness Principle applies because \(\ell^2\) is Banach, and concludes that the operator norms are uniformly bounded. In this example they are all exactly \(1\). Notice that the pointwise bound is naturally allowed to vary with \(x\); it is not a single bound on all vectors in \(\ell^2\).
Worked Example: A Family of Functionals on a Finite-Dimensional Space
On \(\mathbb{R}^2\) with its Euclidean norm, define \(S_\theta(x,y)=x\cos\theta+y\sin\theta\) for \(0\leq\theta\leq2\pi\). Each \(S_\theta\) is a linear functional. By the Cauchy–Schwarz inequality, \[ |S_\theta(x,y)|\leq\sqrt{x^2+y^2}\sqrt{\cos^2\theta+\sin^2\theta} =\sqrt{x^2+y^2}. \] The identity \(\cos^2\theta+\sin^2\theta=1\) verifies the final equality, so \(\|S_\theta\|\leq1\). For each \(\theta\), the unit vector \((\cos\theta,\sin\theta)\) satisfies \[ S_\theta(\cos\theta,\sin\theta)=\cos^2\theta+\sin^2\theta=1. \] Consequently \(\|S_\theta\|=1\) for every \(\theta\).
For each fixed \((x,y)\), the estimate above gives a finite bound on all the values \(S_\theta(x,y)\), so the family is pointwise bounded. The Banach-space principle applies, but this example also belongs to the finite-dimensional setting considered below, where a direct argument is available. It illustrates that the theorem concerns families of any size; the index set need not be countable.
A Local Bound Gives a Norm Estimate
The global conclusion of the Uniform Boundedness Principle is often reached by first finding a region on which all members of a family have a common bound. The following elementary estimate explains how a bound on a ball controls operator norms. It is useful independently of the principle: whenever a common bound on such a ball is already known, the estimate quantifies its consequence.
Proof. Fix \(T\in\mathcal{T}\). Since \(x_0\) is in the ball, \(\|Tx_0\|\leq M\). For any \(h\) with \(\|h\|<r\), the point \(x_0+h\) is also in the ball, so \(\|T(x_0+h)\|\leq M\). By linearity and the triangle inequality, \[ \|Th\|=\|T(x_0+h)-Tx_0\| \leq\|T(x_0+h)\|+\|Tx_0\| \leq2M. \] Now take any \(u\in X\) with \(\|u\|\leq1\), and choose a real number \(\rho\) with \(0<\rho<r\). Then \(\|\rho u\|\leq\rho<r\), so the inequality just proved gives \(\rho\|Tu\|=\|T(\rho u)\|\leq2M\). Hence \(\|Tu\|\leq2M/\rho\). Taking the supremum over all such \(u\) yields \(\|T\|\leq2M/\rho\). This holds for every \(\rho<r\); letting \(\rho\) increase to \(r\) gives \(\|T\|\leq2M/r\). The bound is independent of \(T\), so taking the supremum over the family proves the proposition. \(\square\)
Worked Example: Applying the Ball Estimate
Suppose a family \(\mathcal{T}\) of bounded linear operators satisfies \(\|Tx\|\leq5\) for every \(T\in\mathcal{T}\) and every \(x\) with \(\|x-x_0\|<2\), for some \(x_0\in X\). In the proposition, \(M=5\) and \(r=2\). Therefore \[ \sup_{T\in\mathcal{T}}\|T\|\leq\frac{2M}{r}=\frac{2\cdot5}{2}=5. \] In particular, every \(T\) obeys \(\|Tx\|\leq5\|x\|\) for all \(x\in X\). The estimate depends on having the common bound throughout a ball, not merely at its center. A bound at \(x_0\) alone gives no control over how rapidly the operators can change away from that point.
Why Finite-Dimensional Domains Are Different
Completeness is the hypothesis that enables the general Uniform Boundedness Principle, but finite-dimensional normed spaces have an additional structure: coordinates in a fixed basis are controlled by the norm. This lets us prove uniform boundedness directly, without assuming the domain is complete. In fact, every finite-dimensional normed space is complete, but the argument below uses its finite basis and does not need to invoke completeness or the general principle.
Proof. If \(X=\{0\}\), every linear operator from \(X\) is zero, so the conclusion holds. Otherwise choose a basis \(e_1,\ldots,e_d\) for \(X\). For each \(i\), pointwise boundedness at \(e_i\) gives a finite number \(M_i=\sup_{T\in\mathcal{T}}\|Te_i\|\).
Write \(x=\sum_{i=1}^d a_i e_i\). The coordinate maps \(x\mapsto a_i\) are continuous on a finite-dimensional normed space. Equivalently, since the unit sphere is compact, each coordinate map has a finite maximum in absolute value there. Thus there is a finite constant \(C_i\) such that \(|a_i|\leq C_i\|x\|\) for every \(x\in X\). For every \(T\in\mathcal{T}\), linearity and the triangle inequality now give \[ \|Tx\|=\left\|\sum_{i=1}^d a_iTe_i\right\| \leq\sum_{i=1}^d |a_i|\|Te_i\| \leq\sum_{i=1}^d C_iM_i\|x\| =\left(\sum_{i=1}^d C_iM_i\right)\|x\|. \] The constant \(\sum_{i=1}^d C_iM_i\) is finite and independent of \(T\) and \(x\). Taking the supremum over \(\|x\|\leq1\) shows that every \(\|T\|\) is at most this constant. Therefore \(\sup_{T\in\mathcal{T}}\|T\|<\infty\). \(\square\)
The proof shows exactly what finite dimension contributes: it is enough to bound the family at the finitely many basis vectors, because every other vector is a finite linear combination of them with controlled coefficients. In an infinite-dimensional space, there is no finite basis that reduces pointwise boundedness to finitely many checks. The Banach-space principle provides a different mechanism in that setting.
When the Hypothesis Fails
The principle is an implication, not a claim that every family of bounded linear operators has bounded operator norms. Pointwise boundedness must be checked. A simple family on a Banach space shows what can happen when it fails.
Worked Example: Unbounded Norms with Pointwise Blow-Up
Let \(X=C([0,1])\), equipped with the supremum norm, and define \(T_n(f)=n f(0)\). Each \(T_n\) is linear, and \[ |T_n(f)|=n|f(0)|\leq n\|f\|_\infty. \] Thus \(T_n\) is bounded and \(\|T_n\|\leq n\). For the constant function \(\mathbf{1}(t)=1\), we have \(\|\mathbf{1}\|_\infty=1\) and \(|T_n(\mathbf{1})|=n\), so \(\|T_n\|\geq n\). Hence \(\|T_n\|=n\), and the operator norms are unbounded.
At the same time, for every \(f\) with \(f(0)\ne0\), \(|T_n(f)|=n|f(0)|\) tends to infinity. In particular, the family is not pointwise bounded. This is precisely why the Uniform Boundedness Principle makes no assertion that these norms must be bounded.
A reliable application begins by checking the setting: the domain must be Banach, the members must be bounded linear operators, and the family must be pointwise bounded. Once those conditions are established, the conclusion is a common bound on operator norms—not convergence, and not a bound independent of the size of the input. When a common bound is known on a whole ball, the local estimate above makes that conclusion quantitative; in finite dimensions, the basis argument gives uniform boundedness directly.
Check Your Understanding
Use the statement and results in this tutorial to answer the following questions.
- In pointwise boundedness, which quantity is allowed to depend on the input vector?
- Why does a bound on operator norms imply a common estimate of the form \(\|Tx\|\leq C\|x\|\)?
- In the local ball estimate, why is it useful to compare \(T(x_0+h)\) with \(Tx_0\)?
- Where does finite dimensionality enter the direct proof for pointwise bounded families?
- For the family \(T_n(f)=nf(0)\), what input shows that pointwise boundedness fails?