Convergence of Step Functions Generated by Uniformly Distributed Random Points

Question

I have encountered a surprisingly complicated problem to solve and I'm looking for some help. It could be difficult because I don't have a background in probability and so don't know the appropriate terms to search, or it is a genuinely difficult thing to prove.

Preamble: Let $f:[0,1] \to \mathbb{R}$ be a continuous function. Then, for $n\geq 1$ we consider evenly spaced points $0 \leq x_1 < x_2 < \dots < x_n \leq 1$ and a partition $\mathcal{P}_n$ of the interval $[0,1]$ into exactly $n$ disjoint intervals so that each $x_i$ belongs to exactly one of these partition intervals. Using these points and the partition we can construct a step function approximation of $f$, denoted $f_n$, by assigning $f_n(x) = f(x_k)$ for all $x$ in the $k$th partition interval. Then, a typical exercise one might encounter in real analysis is to prove that $\lim_{n\to \infty} \|f_n - f\|_\infty \to 0$, which follows form the uniform continuity of $f$.

My Problem: Now, suppose that for $n\geq 1$ we have $u_1,\dots,u_n$ drawn independently form the uniform distribution on $[0,1]$ and we set the $x_1,\dots,x_n$ to be the corresponding order statistics. My question is: if we construct the step functions $f_n$ as above, can we still get uniform convergence to the original function $f$ with high probability?

What I have so Far: A previous question on here has provided that if the $x_1,\dots,x_n$ are the order statistics of the i.i.d. points $u_1,\dots,u_n$ we have that $$\mathbb{P}\bigg(\max_{1 \leq k \leq n} |x_{k+1} - x_{k}| \leq \delta \bigg) = \sum_{k = 0}^{\lfloor 1/\delta\rfloor} (-1)^k{n+1\choose k}(1 - k\delta)^n $$ where $x_0 = 0$, $x_{n+1} = 1$, and for all $\delta \in [1/(n+1),1]$.

Of course, once one can argue that with high probability the points become close together, one can again use uniform continuity of $f$ on $[0,1]$ to get the desired result. However, I'm unable to get anything asymptotic out of the above formula. I've run simulations with $\delta = 1/\sqrt{n}$, for example, and the above probability quickly converges to 1 as $n \to \infty$, but again, I can't prove it.

Question: It seemed to me that my problem would have been answered at some point in the mathematical literature, but I can't find any reference. Therefore, could someone either point me to a reference or help me out with a proof here? Thank you!

Answer 1

For any $k\ge 1$, as $0\le 1 - k\delta<1$ we have [1]:

$$\lim_{n\to \infty} (-1)^k{n+1\choose k}(1 - k\delta)^n=\lim_{n\to \infty}\frac{(-1)^kn^k(1 - k\delta)^n}{k!}=0.$$

Hence,

$$\lim_{n\to \infty}\sum_{k = 0}^{\lfloor 1/\delta\rfloor} (-1)^k{n+1\choose k}(1 - k\delta)^n=1+\lim_{n\to \infty}\sum_{k = 1}^{\lfloor 1/\delta\rfloor} (-1)^k{n+1\choose k}(1 - k\delta)^n=1.$$

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Therefore, using the above result, by the uniform continuity of $f$ (which is continuous on the compact interval $[0,1]$), as $n\to \infty$ we have

$$\|f_n - f\|_\infty \xrightarrow{\color{blue}{p}} 0,$$

which also implies

$$\|f_n - f\|_\infty \xrightarrow{\color{blue}{L_1}} 0,$$

using the dominated convergence theorem as $\|f_n - f\|$ can be bounded by some constant over $[0,1]$.

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For any $\delta>0$, we have

$$\sum_{n=\lfloor 1/\delta\rfloor-1}^\infty \mathbb{P}\bigg(\|\Delta_n\|_\infty=\max_{1 \leq k \leq n} |x_{k+1} - x_{k}| > \delta \bigg) = \sum_{n=\lfloor 1/\delta\rfloor-1}^\infty \left (1-\sum_{k = 0}^{\lfloor 1/\delta\rfloor} (-1)^k{n+1\choose k}(1 - k\delta)^n \right)=\sum_{k = 0}^{\lfloor 1/\delta\rfloor} (-1)^k \sum_{n=\lfloor 1/\delta\rfloor-1}^\infty {n+1\choose k}(1 - k\delta)^n<\infty,$$

because each of the inner summations is a converging series, for example, for $\delta=\frac 1 2$, we have

$$\sum_{n=1}^\infty \mathbb{P}\bigg(\|\Delta_n\|_\infty=\max_{1 \leq k \leq n} |x_{k+1} - x_{k}| > \delta \bigg) = \sum_{n=1}^\infty \left (1-\sum_{k = 0}^{\lfloor 1/\delta\rfloor} (-1)^k{n+1\choose k}(1 - k\delta)^n \right)= \sum_{n=1}^\infty (n+1) (0.5)^n=3 <\infty.$$

Thus, using Borel-Cantelli lemma [2] $\|\Delta_n\|_\infty \xrightarrow{a.s.} 0$. As $f$ is also a Lipschitz continuous function on $[0,1]$, for some $C>0$ we have

$$0\le\|f_n - f\|_\infty \le C \|\Delta_n\|_\infty$$

Finally, by applying the squeeze theorem for almost sure convergence [3], as $n\to \infty$ we have

$$ \|f_n - f\|_\infty \xrightarrow{\color{blue}{\text{a.s.}}} 0.$$