The uniform boundedness principle (Banach-Steinhaus)

The hard direction is (1) $\implies$ (2). The key strategy of the proof is to invoke the Baire category theorem Theorem 1. We invoke the theorem by constructing a covering for the complete space $X$. In detail, define the following subsets of $X$:

Since by assumption we have that $\sup_{\alpha \in A} || T_\alpha x || < \infty$ this implies that the $E_j$ indeed cover $X$ i.e.

Since $X$ is complete, Theorem 1 says that at least one of the $E_j$ is not nowhere dense i.e. has non-empty interior. Let’s denote that set by $E_n$. Then there exists $B_r(y) \subset E_n$. Let $z \in B_r(y)$, then $z \in E_n$ and $||y - z|| < r$ and we can write $z = y + x = y + (z - y)$ then we have $||x|| < r$, and $y + x \in E_n$.

for some $R > 0$. In particular, for $||x|| = r$ we have

Since for arbitrary $x \in X$ we can consider $\tilde{x} = r \frac{x}{||x||}$ with $||\tilde{x}|| = r$ and we have

All that remains now is to compute the operator norm from its definition, to find that

Let $\{T_\alpha\}$ be a family of linear maps $\mathbb{R}^n \to Y$, pointwise bounded. Pick a basis $e_1, \dots, e_n$. Pointwise boundedness gives finite constants $M_i = \sup_\alpha \|T_\alpha e_i\| < \infty$ for each $i$. For any $x = \sum c_i e_i$ with $\|x\| \leq 1$, equivalence of norms gives $|c_i| \leq C$ for some universal constant, so:

Done. $\sup_\alpha \|T_\alpha\| \leq C(M_1 + \cdots + M_n)$. The argument is just the triangle inequality applied to finitely many basis vectors. Pointwise boundedness on $n$ basis vectors immediately gives a uniform bound, because every point is a finite linear combination of them with controlled coefficients. In infinite dimensions this sum becomes an infinite series, which has no reason to converge --- the finite-dimensional trick breaks down completely.

Let $X = c_{00}$ be the space of real sequences with only finitely many nonzero terms, equipped with the sup-norm $\|x\|_\infty = \max_k |x_k|$.

Why $c_{00}$ is not complete. Consider the sequence of elements

This is Cauchy in the sup-norm: for $m > n$, $\|x^{(m)} - x^{(n)}\|_\infty = \max_{k=n+1}^{m} \frac{1}{k} = \frac{1}{n+1} \to 0.$ But the limit $x = (1, 1/2, 1/3, \ldots)$ has infinitely many nonzero terms, so $x \notin c_{00}$. The completion of $c_{00}$ in the sup-norm is $c_0$, the space of sequences converging to zero.

The operators. Define $T_n : c_{00} \to \mathbb{R}$ by $T_n(x) = n \cdot x_n$. Each $T_n$ is a bounded linear functional with $\|T_n\|_{\mathrm{op}} = n$.

Pointwise bounded: For any fixed $x \in c_{00}$, there exists $N$ such that $x_k = 0$ for $k > N$. Then $T_n(x) = n \cdot x_n = 0$ for all $n > N$, so $\sup_n |T_n(x)| = \max_{1 \leq n \leq N} n|x_n| < \infty.$

Not uniformly bounded: But $\|T_n\|_{\mathrm{op}} = n \to \infty$.

So Banach-Steinhaus fails on the incomplete space $c_{00}$: the family is pointwise bounded but uniformly unbounded.

Where is the witness?¶

The contrapositive of Banach-Steinhaus promises: if the family is uniformly unbounded, there exists a single $x$ with $\sup_n n|x_n| = \infty$. Since $\|T_n\|_{\mathrm{op}} = n \to \infty$, Banach-Steinhaus applied on the complete space $c_0$ tells us (by contrapositive) there must exist an $x \in c_0$ where $\sup_n |T_n(x)| = \infty$. We can find it explicitly: take

This is in $c_0$ (since $1/\sqrt{n} \to 0$) but not in $c_{00}$. Then $T_n(x) = n \cdot \frac{1}{\sqrt{n}} = \sqrt{n} \to \infty.$

But this witness is not in $c_{00}$. It has infinite support. The witness lives in the completion $c_0$, not in the finitely supported subspace. The incomplete space has a hole exactly where it needs to be.

The catastrophe was always there, you just couldn’t see it because $c_{00}$ is missing the elements that expose the problem. The incomplete space gives a false sense of security: everything you can test looks fine, but the moment someone hands you an input from the completion, it blows up. This is the applied mathematics moral: you build your numerical scheme on finitely supported data (effectively $c_{00}$), your convergence tests all pass, and then in production someone feeds in an input from $c_0$ and the scheme explodes. Completeness isn’t an abstraction, it’s the guarantee that your test cases are representative of the full space.

What goes wrong with the Baire argument? We would write $c_{00} = \bigcup_j F_j$ where $F_j = \{x : |T_n(x)| \leq j \text{ for all } n\}$. These $F_j$ do cover $c_{00}$, but since $c_{00}$ is not complete, it is not a Baire space. Each $F_j$ is nowhere dense: given any $x \in F_j$ and $\varepsilon > 0$, choose $n > j/\varepsilon$ and perturb the $n$-th coordinate by $\varepsilon/2$ to find a point $y \in B_\varepsilon(x) \setminus F_j$. No $F_j$ contains a ball, so the Baire argument has no traction. Completeness would forbid this: in a Banach space, at least one $F_j$ must have nonempty interior, and linearity propagates that local bound to a global one.

The partial sum operators $S_n : C[0,1] \to C[0,1]$ satisfy $\|S_n\| \to \infty$ (the Lebesgue constants grow like $\log n$). Banach-Steinhaus immediately gives: there exists a specific continuous function whose Fourier partial sums diverge at a point. Not a pathological distribution, not something outside $C[0,1]$, but an actual continuous function. This is the du Bois-Reymond theorem (1876), and the Banach-Steinhaus proof is essentially one line once you know the norms grow. Moreover, such functions are generic: the typical continuous function has a divergent Fourier series.

If an approximation scheme has operator norms growing to infinity, Banach-Steinhaus guarantees a concrete input where convergence fails, and generically most inputs fail. The Runge phenomenon (polynomial interpolation blowing up at equispaced nodes) can be understood this way: the interpolation operators are unbounded, so a witness function exists where interpolation diverges.

The practical message is: if you are designing a numerical scheme and you cannot prove uniform boundedness of the operator norms, you should not think “maybe it works for the inputs I care about.” Banach-Steinhaus says the failure is not hiding in some exotic corner, it is dense. You will encounter it.