uniform Cauchy criterion
For ordinary number sequences, the Cauchy criterion lets you prove convergence without knowing the limit in advance: just check the terms eventually huddle together. The uniform Cauchy criterion is the same idea lifted to functions — it lets you establish uniform convergence even when you have no formula for the limit function, by checking that the f_n bunch up uniformly.
Statement: a sequence f_n on E is uniformly Cauchy if for every epsilon > 0 there exists N such that |f_m(x) - f_n(x)| < epsilon for all m, n >= N and all x in E — that is, sup over x of |f_m(x) - f_n(x)| < epsilon. The theorem (valid for functions into the real or complex numbers, or any complete metric space) is: f_n converges uniformly on E if and only if it is uniformly Cauchy. Both directions matter.
Why it is true: pointwise, the Cauchy condition forces each f_n(x) to converge to some value, defining the limit f(x). Then holding m fixed and letting n -> infinity in |f_m(x) - f_n(x)| < epsilon gives |f_m(x) - f(x)| <= epsilon for all x at once, which is exactly uniform convergence. The completeness of the target space is what makes the pointwise limits exist — the criterion can fail for functions valued in an incomplete space like the rationals.
Consider the partial sums s_n(x) = sum_{k=1}^n cos(kx)/k^2 on the real line. For m > n, |s_m(x) - s_n(x)| <= sum_{k=n+1}^m 1/k^2, independent of x, and this tail goes to 0 since sum 1/k^2 converges. So (s_n) is uniformly Cauchy, hence converges uniformly — no need to identify the limit explicitly.
Proving uniform convergence without naming the limit.
In sup-norm language this is just the statement that the space of bounded functions is complete: a sequence converges uniformly iff it is Cauchy in the sup norm.