Uniform boundedness principle

In mathematics, the uniform boundedness principle or Banach–Steinhaus theorem is one of the fundamental results in functional analysis. Together with the Hahn–Banach theorem and the open mapping theorem, it is considered one of the cornerstones of the field. In its basic form, it asserts that for a family of continuous linear operators (and thus bounded operators) whose domain is a Banach space, pointwise boundedness is equivalent to uniform boundedness in operator norm.

The theorem was first published in 1927 by Stefan Banach and Hugo Steinhaus, but it was also proven independently by Hans Hahn.

Theorem
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The completeness of $$X$$ enables the following short proof, using the Baire category theorem.

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There are also simple proofs not using the Baire theorem.

Corollaries
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The above corollary does claim that $$T_n$$ converges to $$T$$ in operator norm, that is, uniformly on bounded sets. However, since $$\left\{T_n\right\}$$ is bounded in operator norm, and the limit operator $$T$$ is continuous, a standard "$$3\varepsilon$$" estimate shows that $$T_n$$ converges to $$T$$ uniformly on sets.

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Indeed, the elements of $$S$$ define a pointwise bounded family of continuous linear forms on the Banach space $$X := Y',$$ which is the continuous dual space of $$Y.$$ By the uniform boundedness principle, the norms of elements of $$S,$$ as functionals on $$X,$$ that is, norms in the second dual $$Y'',$$ are bounded. But for every $$s \in S,$$ the norm in the second dual coincides with the norm in $$Y,$$ by a consequence of the Hahn–Banach theorem.

Let $$L(X, Y)$$ denote the continuous operators from $$X$$ to $$Y,$$ endowed with the operator norm. If the collection $$F$$ is unbounded in $$L(X, Y),$$ then the uniform boundedness principle implies: $$R = \left \{x \in X \ : \ \sup\nolimits_{T \in F} \|Tx\|_Y = \infty \right\} \neq \varnothing.$$

In fact, $$R$$ is dense in $$X.$$ The complement of $$R$$ in $$X$$ is the countable union of closed sets $\bigcup X_n.$ By the argument used in proving the theorem, each $$X_n$$ is nowhere dense, i.e. the subset $\bigcup X_n$ is. Therefore $$R$$ is the complement of a subset of first category in a Baire space. By definition of a Baire space, such sets (called or ) are dense. Such reasoning leads to the, which can be formulated as follows:

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Example: pointwise convergence of Fourier series
Let $$\mathbb{T}$$ be the circle, and let $$C(\mathbb{T})$$ be the Banach space of continuous functions on $$\mathbb{T},$$ with the uniform norm. Using the uniform boundedness principle, one can show that there exists an element in $$C(\mathbb{T})$$ for which the Fourier series does not converge pointwise.

For $$f \in C(\mathbb{T}),$$ its Fourier series is defined by $$\sum_{k \in \Z} \hat{f}(k) e^{ikx} = \sum_{k \in \Z} \frac{1}{2\pi} \left (\int_0 ^{2 \pi} f(t) e^{-ikt} dt \right) e^{ikx},$$ and the N-th symmetric partial sum is $$S_N(f)(x) = \sum_{k=-N}^N \hat{f}(k) e^{ikx} = \frac{1}{2 \pi} \int_0^{2 \pi} f(t) D_N(x - t) \, dt,$$ where $$D_N$$ is the $$N$$-th Dirichlet kernel. Fix $$x \in \mathbb{T}$$ and consider the convergence of $$\left\{S_N(f)(x)\right\}.$$ The functional $$\varphi_{N,x} : C(\mathbb{T}) \to \Complex$$ defined by $$\varphi_{N, x}(f) = S_N(f)(x), \qquad f \in C(\mathbb{T}),$$ is bounded. The norm of $$\varphi_{N,x},$$ in the dual of $$C(\mathbb{T}),$$ is the norm of the signed measure $$(2(2 \pi)^{-1} D_N(x - t) d t,$$ namely $$\left\|\varphi_{N,x}\right\| = \frac{1}{2 \pi} \int_0^{2 \pi} \left|D_N(x-t)\right| \, dt = \frac{1}{2 \pi} \int_0^{2 \pi} \left|D_N(s)\right| \, ds = \left\|D_N\right\|_{L^1(\mathbb{T})}.$$

It can be verified that $$\frac{1}{2 \pi} \int_0 ^{2 \pi} |D_N(t)| \, dt \geq \frac{1}{2\pi}\int_0^{2\pi} \frac{\left|\sin\left( (N + \tfrac{1}{2})t \right)\right|}{t/2} \, dt \to \infty.$$

So the collection $$\left(\varphi_{N, x}\right)$$ is unbounded in $$C(\mathbb{T})^{\ast},$$ the dual of $$C(\mathbb{T}).$$ Therefore, by the uniform boundedness principle, for any $$x \in \mathbb{T},$$ the set of continuous functions whose Fourier series diverges at $$x$$ is dense in $$C(\mathbb{T}).$$

More can be concluded by applying the principle of condensation of singularities. Let $$\left(x_m\right)$$ be a dense sequence in $$\mathbb{T}.$$ Define $$\varphi_{N, x_m}$$ in the similar way as above. The principle of condensation of singularities then says that the set of continuous functions whose Fourier series diverges at each $$x_m$$ is dense in $$C(\mathbb{T})$$ (however, the Fourier series of a continuous function $$f$$ converges to $$f(x)$$ for almost every $$x \in \mathbb{T},$$ by Carleson's theorem).

Generalizations
In a topological vector space (TVS) $$X,$$ "bounded subset" refers specifically to the notion of a von Neumann bounded subset. If $$X$$ happens to also be a normed or seminormed space, say with (semi)norm $$\|\cdot\|,$$ then a subset $$B$$ is (von Neumann) bounded if and only if it is, which by definition means $\sup_{b \in B} \|b\| < \infty.$

Barrelled spaces
Attempts to find classes of locally convex topological vector spaces on which the uniform boundedness principle holds eventually led to barrelled spaces. That is, the least restrictive setting for the uniform boundedness principle is a barrelled space, where the following generalized version of the theorem holds :

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Uniform boundedness in topological vector spaces
A family $$\mathcal{B}$$ of subsets of a topological vector space $$Y$$ is said to be in $$Y,$$ if there exists some bounded subset $$D$$ of $$Y$$ such that $$B \subseteq D \quad \text{ for every } B \in \mathcal{B},$$ which happens if and only if $$\bigcup_{B \in \mathcal{B}} B$$ is a bounded subset of $$Y$$; if $$Y$$ is a normed space then this happens if and only if there exists some real $$M \geq 0$$ such that $\sup_{\stackrel{b \in B}{B \in \mathcal{B}}} \|b\| \leq M.$ In particular, if $$H$$ is a family of maps from $$X$$ to $$Y$$ and if $$C \subseteq X$$ then the family $$\{h(C) : h \in H\}$$ is uniformly bounded in $$Y$$ if and only if there exists some bounded subset $$D$$ of $$Y$$ such that $$h(C) \subseteq D \text{ for all } h \in H,$$ which happens if and only if $H(C) := \bigcup_{h \in H} h(C)$ is a bounded subset of $$Y.$$

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Generalizations involving nonmeager subsets
Although the notion of a nonmeager set is used in the following version of the uniform bounded principle, the domain $$X$$ is assumed to be a Baire space.

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Every proper vector subspace of a TVS $$X$$ has an empty interior in $$X.$$ So in particular, every proper vector subspace that is closed is nowhere dense in $$X$$ and thus of the first category (meager) in $$X$$ (and the same is thus also true of all its subsets). Consequently, any vector subspace of a TVS $$X$$ that is of the second category (nonmeager) in $$X$$ must be a dense subset of $$X$$ (since otherwise its closure in $$X$$ would a closed proper vector subspace of $$X$$ and thus of the first category).

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Sequences of continuous linear maps
The following theorem establishes conditions for the pointwise limit of a sequence of continuous linear maps to be itself continuous.

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If in addition the domain is a Banach space and the codomain is a normed space then $$\|h\| \leq \liminf_{n \to \infty} \left\|h_n\right\| < \infty.$$

Complete metrizable domain
proves a weaker form of this theorem with Fréchet spaces rather than the usual Banach spaces.

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