Helly's Selection Theorem
By Bolzano-Weierstrass’ theorem, we know that any bounded sequence of real values has a convergent subsequence. This result can be extended to finite dimensional space, i.e, any bounded sequence in \(\mathbb{R}^n\) has a convergent subsequence. However, it is not true in infinite dimensional space, say, \(\mathbb{R}^{[0,1]}\). Nevertheless, there are two well-known theorem to establish the convergence of a sequence of functions, the Arzelà–Ascoli theorem and Helly's selection theorem. The main difference between these two results is the notion of convergence of a function sequence. Arzelà–Ascoli theorem deals with the uniformly convergence and Helly's selection theorem deals with the pointwise convergence.
Statement
Lemma (Helly). Suppose that \(\{f_n\}_{n\in\mathbb{N}}\) is a uniformly bounded sequence of increasing functions on an interval. Then there is a subsequence converging pointwise to an increasing function.
In probability theory, this theorem is often stated in the following manner.
Theorem (Helly). For every sequence of \(\{F_n\}\) of distribution functions from \(\mathbb{R}\) to \([0,1]\), there exists a subsequence \(\{F_{n_k}\}\) and a nondecreasing, right-continuous function \(F\) such that \(\lim_k F_{n_k}(x) = F(x)\) at continuity points \(x\) of \(F\).
Proof. The proof is based on the diagonal argument. See here for the complete proof given in Billingsley's book.
Remark. The limiting funciton \(F\) may not be a distribution function. For example, let \(F_n(x) = \mathbb{1}(x \geq n)\) be the distribution function corresponding to a unit mass at \(n\). Then the limiting function \(F(x)\equiv0\) is clearly not a distribution function.
Remark. The proof of this version can be adapted to prove the original version. First, it is necessary that \(0 \leq F_n \leq 1\) but \(F_n\) need not to be a distribution function. Second, if we do not require \(F\) to be right-continuous, then we can redefine the values of \(F\) at discontinuity points, and select a finer subsequence such that \(\lim_k F_{n_k}(x)=F(x)\) at every \(x\). This is because the set of discontinuity points of a monotone function is countable.
Problems
- Is there a subsequence of \((\sin (x/k))_{k=1}^\infty\) converges pointwise?
- Is there a subsequence of \((\sin (x+k))_{k=1}^\infty\) converges pointwise?
- Is there s subsequence of \((\sin (kx))_{k=1}^\infty\) converges pointwise?
Answers to the first two questions are positive, which can be shown easily by Arzelà–Ascoli theorem. The answer to the last question, however, is negative; see this discussion. Hence, the last one is a perfect example that shows the monotone condition is essential in Helly's selection theorem.
Further discussion
Pointwise convergence is weaker than uniformly convergence. In \(\mathbb{R}^{\mathbb{N}}\), let \(e^n=(e^n_k)_{k=1}^\infty\) be the sequence with all zero entries except the \(n\)-th entry be 1. Then the sequence \((e^n)_{n=1}^\infty\) is bounded under the sup norm. Moreover, it converges pointwise to \(e^*\equiv0\). However, The sequence \((e^n)\) is clearly divergent under the sup norm. Indeed, by the diagonal argument, any uniformly bounded sequence in \(\mathbb{R}^{\mathbb{N}}\) has a subsequence which converges pointwise.