From an Abstract Principle to Concrete Consequences
The Uniform Boundedness Principle converts pointwise control into a bound that is uniform over a family of operators. Its applications often begin with scalar quantities: inner products, evaluations, or difference quotients. The key is to regard each quantity as a bounded linear operator on a complete normed space. If the operator norms grow without bound, the principle also gives a useful contrapositive: some single input must make the outputs unbounded.
We will use the Uniform Boundedness Principle in the form established earlier in this course. If \(X\) is a Banach space and \((T_n)\) is a sequence of bounded linear operators from \(X\) to a normed space, then pointwise boundedness, \(\sup_n\|T_nx\|<\infty\) for every \(x\in X\), implies \(\sup_n\|T_n\|<\infty\). Consequently, if the operator norms are unbounded, there must be at least one \(x\in X\) for which \(\sup_n\|T_nx\|=\infty\). The latter implication is not a way to specify the input in advance; it guarantees that such an input exists.
The first application concerns sequences in a Hilbert space. A sequence can be tested by taking its inner product with each fixed vector. If all these scalar sequences are bounded, the vectors themselves cannot have arbitrarily large norms.
Scalar Tests Bound a Sequence in a Hilbert Space
Proof. For each \(n\), define \(T_n:H\to\mathbb{R}\) by \(T_n(y)=\langle y,x_n\rangle\). The inner product is linear in \(y\), so each \(T_n\) is linear. The Cauchy–Schwarz inequality gives \(|T_n(y)|\leq\|y\|\|x_n\|\), which shows that \(T_n\) is bounded.
Its operator norm is exactly \(\|x_n\|\). The Cauchy–Schwarz inequality gives \(\|T_n\|\leq\|x_n\|\). If \(x_n\neq0\), choose \(y=x_n/\|x_n\|\). Then \(\|y\|=1\) and \(|T_n(y)|=|\langle x_n/\|x_n\|,x_n\rangle|=\|x_n\|\). Thus \(\|T_n\|\geq\|x_n\|\). If \(x_n=0\), both norms are zero. Hence \(\|T_n\|=\|x_n\|\) in every case.
By hypothesis, for each fixed \(y\in H\), \(\sup_n|T_n(y)|<\infty\). The family \((T_n)\) is therefore pointwise bounded. Since \(H\) is complete, the Uniform Boundedness Principle applies and gives \(\sup_n\|T_n\|<\infty\). The norm identity just proved yields \(\sup_n\|x_n\|<\infty\). \(\square\)
This is often expressed by saying that a weakly bounded sequence in a Hilbert space is norm bounded: each fixed vector supplies a scalar test, and those tests are bounded uniformly along the sequence. In particular, if \(x_n\) converges weakly to \(x\), then for every \(y\), the scalars \(\langle y,x_n\rangle\) converge to \(\langle y,x\rangle\), and so are bounded. The theorem then implies that \((x_n)\) is norm bounded. It does not say that weak convergence implies norm convergence.
Worked Example: The Standard Unit Vectors in \(\ell^2\)
Let \(e_n=(0,\ldots,0,1,0,\ldots)\), with the \(1\) in coordinate \(n\), in the real Hilbert space \(\ell^2\). For each \(y=(y_k)\in\ell^2\), \(\langle y,e_n\rangle=y_n\). Since \(\sum_{k=1}^{\infty}|y_k|^2<\infty\), its terms satisfy \(y_n\to0\), so \((\langle y,e_n\rangle)\) is bounded for every fixed \(y\). The theorem guarantees norm boundedness; directly, \(\|e_n\|_2=1\) for every \(n\).
This example also shows why the conclusion should not be mistaken for convergence in norm. For distinct \(m,n\), the vectors \(e_m-e_n\) have two nonzero coordinates, each of absolute value \(1\), so \(\|e_m-e_n\|_2=\sqrt{2}\). Thus the sequence is not Cauchy in norm, even though its inner products with every fixed \(y\in\ell^2\) tend to zero.
Growing Difference-Quotient Norms Force Irregularity
A second application uses continuous functions on a compact interval. For each positive integer \(n\), a difference quotient is a bounded linear functional on \(C([0,1])\), equipped with the supremum norm. Its value at a fixed function measures a slope over a short interval. If these functionals have unbounded norms, uniform boundedness says there is a continuous function for which the corresponding slopes are unbounded.
Proof. The space \(C([0,1])\), with norm \(\|f\|_\infty=\sup_{t\in[0,1]}|f(t)|\), is a Banach space. Define \(T_n(f)=n(f(1/n)-f(0))\). Each \(T_n\) is linear, and \[ |T_n(f)|\leq n\bigl(|f(1/n)|+|f(0)|\bigr)\leq2n\|f\|_\infty. \] Thus \(T_n\) is bounded and \(\|T_n\|\leq2n\).
To verify the reverse inequality, define \(g_n\) to be linear on \([0,1/n]\), with \(g_n(0)=-1\) and \(g_n(1/n)=1\), and constant equal to \(1\) on \([1/n,1]\). This is continuous and \(\|g_n\|_\infty=1\). Also \(T_n(g_n)=n(1-(-1))=2n\), so \(\|T_n\|\geq2n\). Therefore \(\|T_n\|=2n\), and the operator norms are unbounded.
The contrapositive of the Uniform Boundedness Principle now gives an \(f\in C([0,1])\) such that \(\sup_n|T_n(f)|=\infty\). Since \(T_n(f)=n(f(1/n)-f(0))\), these are the difference quotients at \(0\) along the positive increments \(1/n\). If the finite right derivative \(f'_+(0)\) existed, then the difference quotients would tend to \(f'_+(0)\) as \(n\to\infty\), and therefore would be bounded. This contradicts their unboundedness. Hence this \(f\) is not right-differentiable at \(0\). \(\square\)
The theorem asserts existence, not an explicit formula for the function. That is a characteristic feature of this application: unbounded operator norms prove that at least one input fails a desired regularity property, even when the argument does not identify that input. The conclusion is limited to failure of the right derivative at the endpoint; it does not assert that the function is nowhere differentiable.
Worked Example: A Particular Function with Unbounded Difference Quotients
The existence theorem does not construct its function, but a simple example illustrates the same behavior directly. Set \(f(t)=\sqrt{t}\) on \([0,1]\). This function is continuous and \(f(0)=0\). At the points \(1/n\), \[ n\bigl(f(1/n)-f(0)\bigr)=n\left(\frac{1}{\sqrt{n}}-0\right)=\sqrt{n}. \] Since \(\sqrt{n}\) is unbounded, these difference quotients are unbounded. This verifies directly that \(f\) has no finite right derivative at \(0\).
The example confirms the kind of behavior guaranteed by the theorem, but it is not a substitute for its proof: the operator argument guarantees such a function without assuming that a convenient formula is known.
Weighted Evaluations and the Pointwise-Boundedness Test
A third useful setting is a family of weighted evaluations. For chosen points \(t_n\in[0,1]\) and real weights \(a_n\), define \(S_n(f)=a_nf(t_n)\) on \(C([0,1])\). These are bounded linear functionals. The inequality \(|S_n(f)|\leq|a_n|\|f\|_\infty\) gives \(\|S_n\|\leq|a_n|\); testing on the constant function \(1\), whose norm is \(1\), gives \(\|S_n\|\geq|a_n|\). Thus \(\|S_n\|=|a_n|\). When the weights are unbounded, the Uniform Boundedness Principle ensures that some continuous function has unbounded weighted values. The next example verifies this with a specific choice.
Worked Example: Weighted Samples Near Zero
Let \(t_n=1/n\), \(a_n=n\), and \(S_n(f)=nf(1/n)\). As just shown, \(\|S_n\|=n\), so the operator norms are unbounded. In this case an input displaying unbounded outputs is \(f(t)=\sqrt{t}\): it satisfies \(S_n(f)=n\sqrt{1/n}=\sqrt{n}\), which is unbounded.
By contrast, for the constant function \(f(t)=1\), \(S_n(f)=n\), which is also unbounded. The particular input is therefore not unique. What the general principle adds is a guarantee of an input whenever the norms grow, even for weights and sample points where direct construction is difficult.
What These Applications Do—and Do Not—Say
Each application begins with a complete domain and a family of bounded linear maps. The hypotheses must be checked on the whole domain: knowing that a family is bounded on a dense subset, for example, is not by itself the pointwise-boundedness condition in the Uniform Boundedness Principle. In the Hilbert-space application, the fixed test vector is \(y\); in the differentiation application, the input is a continuous function \(f\). Keeping the roles of operator and input distinct helps prevent a common mistake: bounds that hold separately for different inputs need not be uniform over all inputs.
The principle also does not usually provide an explicit estimate for the resulting input, nor does it identify where a function is irregular beyond what the chosen operators test. Its strength is structural. A sequence of scalar tests that is bounded for each fixed input must have uniformly bounded operator norms; if those norms instead grow, pointwise boundedness fails somewhere. Inner products turn that alternative into a bound on Hilbert-space vectors, while difference quotients turn it into a continuous function with an unbounded sequence of slopes.
Check Your Understanding
Use the operator-norm viewpoint to answer the following questions.
- In the Hilbert-space theorem, why does the operator \(T_n(y)=\langle y,x_n\rangle\) have norm exactly \(\|x_n\|\), including when \(x_n=0\)?
- Why does weak convergence of a sequence imply that its scalar inner-product tests are bounded?
- For \(T_n(f)=n(f(1/n)-f(0))\), which test function proves the lower bound \(\|T_n\|\geq2n\)?
- Why would a finite right derivative at \(0\) contradict unboundedness of the difference quotients along \(1/n\)?
- For \(S_n(f)=nf(1/n)\) and \(f(t)=\sqrt{t}\), calculate \(S_n(f)\) and explain why these values are unbounded.