Supremum norm and the uniform Cauchy criterion

Try

Think of a bounded function as one point in a space of functions. Its distance from another function is the greatest discrepancy over the domain: d(f,g)=sup |f(x)−g(x)|. On [0,1], the functions x and x² differ by x−x², whose maximum is 1/4 at x=1/2. This distance measures worst-case error rather than an average or the error at a chosen sample.

Predict

Predict what happens if the domain is all of R. The difference between x and x² is unbounded, so the proposed distance is not finite. A metric must have finite values. The supremum norm belongs naturally to bounded functions on a nonempty domain, or to an appropriate space of bounded differences. Naming a formula a norm without checking finiteness obscures which mathematical space is actually being used.

Experiment

Compare three function errors on [0,1].

Difference	Supremum norm	Attained?
x−x²	1/4	At x=1/2
Constant 1/n	1/n	Everywhere
x on [0,1), a different domain	1	No

The last row emphasizes that a supremum need not be a maximum. Draw the first difference and verify its maximum by completing the square, x−x²=1/4−(x−1/2)². Then explain why the norm definition still makes sense when no maximizing point exists.

Understand

For bounded real functions on a nonempty set D, define ||f||∞=sup over D of |f|. It is nonnegative, vanishes only for the zero function, scales by absolute scalar magnitude and satisfies the triangle inequality. The latter follows pointwise from |f+g|≤|f|+|g|≤||f||∞+||g||∞, then taking a supremum. Uniform convergence is equivalent to ||f_n−f||∞→0 whenever these norms are finite.

A sequence is uniformly Cauchy when every ε>0 admits N such that all m,n≥N and every x∈D satisfy |f_m(x)−f_n(x)|<ε. If it converges uniformly, the triangle inequality through its limit proves this condition. Conversely, scalar completeness gives a pointwise limit f(x). Choose a common Cauchy bound ε/2, hold n in that tail and let m→∞. The result is |f_n(x)−f(x)|≤ε/2 for every x, proving uniform convergence with the desired strict ε bound.

Apply

If all f_n are bounded, their uniform limit is bounded too. Choose one N with |f−f_N|<1 everywhere. A bound M for that single f_N gives |f|≤M+1. There is no need to assume one common bound for every term before the proof. This establishes completeness of the bounded-function space B(D) under its supremum norm.

Repair “pointwise Cauchy means uniformly Cauchy.” At each fixed point, a cutoff may be chosen differently. The powers x^n on [0,1] are pointwise convergent and therefore pointwise Cauchy, but not uniformly convergent. The proved equivalence rules out uniform Cauchy behavior for them.

Continue

Optional depth observes how completeness enters twice in this discussion: the real numbers supply each pointwise limit, and the uniform estimates then prove that those limits form a member of the function space. If outputs lie in an incomplete metric space, the first step may fail even when pairwise tail errors shrink.

Write the converse proof with the order “choose N, fix x and n, then let m tend to infinity” visible. Using ε/2 prevents an unjustified claim that strict inequalities necessarily remain strict after a limit. Next, continuity will identify a closed subspace preserved by uniform limits.