Uniform limits preserve continuity

Try

A sequence of continuous functions can have a discontinuous pointwise limit: the powers x^n on [0,1] already demonstrate this. Uniform convergence prevents that particular failure. The key is not that each graph looks smooth but that one selected approximating function is close to the limit at both points being compared. Its continuity then bridges the remaining gap.

Predict

Predict the order of choices in a continuity proof for the limit f at a. First fix the output tolerance. Then choose one index N giving uniform approximation. Only after that use continuity of f_N to select an input radius. If N were allowed to change with the nearby input, the continuity radius for the chosen function could change too, leaving no single neighborhood guarantee.

Experiment

Break the target error into three pieces.

Error term	Controlled by
|f(x)−f_N(x)|	Uniform approximation
|f_N(x)−f_N(a)|	Continuity of the fixed f_N
|f_N(a)−f(a)|	Uniform approximation

Choose each term below ε/3 and add them. This table is the proof’s mechanism: two approximation bridges connect the unknown limit to one known continuous function. Draw those bridges as a three-step path in output space, keeping all four values labeled.

Understand

Here is the complete argument. Suppose f_n are continuous on a domain D and converge uniformly to f. Fix a∈D and ε>0. Choose N with |f_N(y)−f(y)|<ε/3 for every y∈D. Continuity of f_N at a supplies δ>0 so x∈D and |x−a|<δ imply |f_N(x)−f_N(a)|<ε/3. The triangle inequality using the three displayed errors gives |f(x)−f(a)|<ε. Thus f is continuous at every point of D.

The domain need not be compact or closed. Relative continuity is enough because the argument compares only inputs belonging to D. If each f_n is uniformly continuous, the same proof uses one δ controlling every pair x,y, so the uniform limit is uniformly continuous. Ordinary continuity of approximants cannot be silently substituted for that stronger hypothesis when proving this stronger conclusion.

Apply

Repair the converse “a continuous pointwise limit implies uniform convergence.” On R, let f_n(x)=max(0,1−n²|x−1/n|). Each function is a triangular peak of height one centered at 1/n, supported within distance 1/n² of that center. At every fixed nonzero x, the support eventually misses x; at zero, the value is eventually zero as well. The pointwise limit is the continuous zero function, yet the supremum error remains one at the moving centers.

This example distinguishes a condition sufficient to preserve continuity from a condition necessary for continuity of a particular limit. It also shows why following a few fixed input traces can miss behavior that moves through the domain.

Continue

Optional depth views continuous bounded functions as a closed subset of the complete bounded-function space. “Closed” here means that a supremum-norm limit stays in the subset. Combining this preservation theorem with completeness will later prove that C(K) is a complete metric space for compact K.

Save a proof that identifies which estimate is uniform and which is only local at a. Then change the conclusion to uniform continuity and mark the extra hypothesis needed. Next, the same worst-case error will control differences of integrals, but only after accounting for the size of the integration domain and integrability of the limit.