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Uniform Cauchy sequence of functions

A sequence (f_n) such that sup_x d(f_m(x),f_n(x))→0 as m,n→∞.
Uniform Cauchy sequence of functions

Let XX be a set and let (Y,dY)(Y,d_Y) be a metric space. A sequence of functions fn:X→Yf_n:X\to Y is uniformly Cauchy if

∀ε>0, ∃N∈N such that ∀m,n≥N, ∀x∈X, dY(fm(x),fn(x))<ε.\forall \varepsilon>0,\ \exists N\in\mathbb{N}\ \text{such that}\ \forall m,n\ge N,\ \forall x\in X,\ d_Y\bigl(f_m(x),f_n(x)\bigr)<\varepsilon.

Equivalently,

sup⁡x∈XdY(fm(x),fn(x))→m,n→∞0.\sup_{x\in X} d_Y\bigl(f_m(x),f_n(x)\bigr)\xrightarrow[m,n\to\infty]{}0.

Uniform Cauchy-ness is the Cauchy criterion for uniform convergence: in complete codomains (e.g., Y=RY=\mathbb{R}), uniformly Cauchy sequences converge uniformly.

Examples:

  • If fn→ff_n\to f uniformly, then (fn)(f_n) is uniformly Cauchy.
  • On X=[0,1]X=[0,1], define fn(x)=∑k=1nxk2kf_n(x)=\sum_{k=1}^n \frac{x^k}{2^k}. Then (fn)(f_n) is uniformly Cauchy since the tail is bounded by a geometric series.
  • Pointwise Cauchy need not be uniformly Cauchy: fn(x)=xnf_n(x)=x^n is pointwise Cauchy on [0,1][0,1] but not uniformly Cauchy.