This sequence of functions is uniformly convergent throughout a region R if, for every ε there is n such that fj(x) is within ε (g(x), for every x in R and for every j ≥ n. The functions all approach g(R) together, one n fits all. This is similar to uniform continuity, where one δ fits all.
If the range space is complex, or a real vector space, the sequence of functions is uniformly convergent iff all the component functions are uniformly convergent. Given an ε, find fn that is close to g, and the components of fn must be close to the components of g, for all x. Conversely, if the components are within ε then the n dimensional function is within nε, for all x.
Without uniform convergence, g is rather unpredictable. Let the domain be the closed interval [0,1], and let fn = xn. Note that the sequence f approaches a function g that is identically 0, except for g(1) = 1. Each function in the sequence is uniformly continuous, yet the limit function isn't even continuous.
In contrast, let a sequence of continuous functions converge uniformly to g. Given p and ε, find n so that functions beyond fn are all within ε/3 of g. Then find δ so that fn(x) is within ε/3 of fn(p) when x is within δ of p. Now step from g(x) to fn(x) to fn(p) to g(p), a distance that is less than ε by the triangular inequality. Therefore g is continuous at p.
If each fn is uniformly continuous then δ will not depend on p. As a result, g is uniformly continuous. In summary, uniform convergence carries continuity or uniform continuity out to the limit function g.
What about the converse? If a sequence of uniform functions has a uniform limit, do we have uniform convergence? Consider the following example. Let fn start at the origin, increase with slope 1/n up to a height of 1, then decrease with slope -1/n back to a height of 0. The function is 0 thereafter. All functions are uniform, and the limit function of 0 is uniform, yet convergence is not uniform. Points far out on the x axis require ever larger values of n, if fj maps x below ½ for all j beyond n.
This example can be modified so that the domain is the unit interval, a compact metric space. Each fn is flat, except for a bump that rises to 1 and falls back down to 0. Move the bump farther to the right of the unit interval, and make it thinner, as n approaches infinity. The limit is still the 0 function, yet convergence is not uniform.
If each function in the series is continuous, each partial sum is continuous. Similarly, uniform functions lead to uniform partial sums. If the partial sums exhibit uniform convergence, the limit function is continuous or uniform, as shown above.