Continuity survives a uniform limit
Here is the payoff. The uniform limit theorem states: if each f_n is continuous at a point a, and f_n -> f uniformly, then f is continuous at a. The proof is the celebrated epsilon/3 argument — three errors, each squeezed below epsilon/3, glued by the triangle inequality. Uniformity is exactly what makes two of those three errors independent of the point.
Uniform limit theorem (continuity at a).
Given: f_n -> f uniformly on E, each f_n continuous at a in E.
Claim: |f(x) - f(a)| < epsilon for x near a.
Fix epsilon > 0.
(1) UNIFORMITY: choose N with ||f_N - f|| < epsilon/3, so for ALL x
|f(x) - f_N(x)| < epsilon/3.
(2) CONTINUITY of f_N at a: choose delta > 0 with
|f_N(x) - f_N(a)| < epsilon/3 whenever |x - a| < delta.
Now for |x - a| < delta, split with the triangle inequality:
|f(x) - f(a)|
<= |f(x) - f_N(x)| + |f_N(x) - f_N(a)| + |f_N(a) - f(a)|
< epsilon/3 + epsilon/3 + epsilon/3
= epsilon.
Hence f is continuous at a. QED
(Errors 1 and 3 used the SAME N for all x — that is uniformity.)Integrate term by term
On a closed bounded interval [a,b], if Riemann integrable functions f_n converge uniformly to f, then f is integrable and the integral commutes with the limit. This is term-by-term integration. The proof is a one-line bound: the gap between the integrals is at most the length times the sup-norm error.
Term-by-term integration on [a,b], f_n -> f uniformly. | integral_a^b f_n - integral_a^b f | = | integral_a^b (f_n - f) | <= integral_a^b |f_n - f| <= integral_a^b ||f_n - f|| (constant bound) = (b - a) * ||f_n - f|| -> 0. So lim_n integral_a^b f_n = integral_a^b lim_n f_n = integral_a^b f. WARNING (needs uniform!): the spike g_n(x) = n x (1-x)^n on [0,1] from Guide 2 converges pointwise to 0, yet integral_0^1 g_n -> a nonzero constant, NOT 0. Without uniformity the swap is false.
Differentiation is touchier
Uniform convergence of f_n does not give f_n' -> f'. Derivatives feel slopes, and a tiny-amplitude wiggle can have a huge slope. The correct term-by-term differentiation theorem flips the hypotheses: assume the derivatives f_n' converge uniformly, and that (f_n) converges at even one point. Then f_n -> f uniformly and, crucially, f' = lim f_n'.
Counterexample: amplitude small, slope not. f_n(x) = sin(n x) / sqrt(n) on R. ||f_n|| = 1/sqrt(n) -> 0, so f_n -> 0 UNIFORMLY. But f_n'(x) = sqrt(n) cos(n x), and f_n'(0) = sqrt(n) -> infinity. So f_n' does NOT converge to (0)' = 0; it diverges. Moral: control the DERIVATIVES uniformly, not the functions, to differentiate term by term. Even scarier (Guide 5 preview): a uniform limit (a Weierstrass series sum b^k cos(a^k pi x), 0<b<1, ab>1) can be continuous yet NOWHERE differentiable.