Uniform convergence

In pointwise convergence each \(x\) has its own \(N\), which may grow without bound as \(x\) varies. In uniform convergence, a single \(N\) serves all of \(A\). A useful way to check it is the maximal distance \(\|f_n-f\|_A=\sup_{x\in A}|f_n(x)-f(x)|\): convergence is uniform if and only if this quantity tends to \(0\).

The classical example is \(f_n(x)=x^n\) on \([0,1]\). Its pointwise limit is \(0\) for \(x<1\) and \(1\) for \(x=1\), and \(\sup_{[0,1)}x^n=1\) for every \(n\), so the convergence is not uniform. On \([0,a]\) with \(a<1\), by contrast, the supremum is \(a^n\to0\) and the convergence is uniform. Another instructive case is \(nxe^{-nx}\), which tends to \(0\) pointwise but keeps a peak of height \(1/e\) sliding toward the origin.

The importance of the concept lies in what it preserves. If each \(f_n\) is continuous and \(f_n\to f\) uniformly, then \(f\) is continuous; the proof uses a three-\(\varepsilon\) argument: \(|f(x)-f(y)|\le|f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|\). This also explains why \(x^n\) cannot converge uniformly on \([0,1]\): its limit is not continuous.

Moreover, the Cauchy criterion has a uniform version: \((f_n)\) converges uniformly on \(A\) if and only if \(\sup_A|f_m-f_n|\to0\) as \(m,n\to\infty\). With it one proves convergence without knowing the limit, and it is the tool that allows swapping limits with integrals and with series of functions.