Suppose that fn is a sequence of continuous functions on an

Suppose that {f_n} is a sequence of continuous functions on an interval [a,b] converging uniformly to a function f on the open interval (a,b). If f is also continuous on [a,b], show that the convergence is uniform on [a,b].

Solution

let us Assume that fn 0 pointwise on [a, b]. Together with the hypothesis about nonincreasing, this implies that for each x and n, we must have fn(x) 0. We’ll proceed by contradiction. (A direct proof using ideas from section 13 is sketched later). Suppose that fn does not converge to 0 uniformly on [a, b]. Then there is a fixed > 0 such that for each integer k, there exist nk > k and xk [a, b] with | fnk (xk)| . Let z = lim inf xk . Then z [a, b]

Since hfn(z)i 0, there is an N0 R so that (n > N0)(| fn(z)| < 2 ). Let m be the least integer > N0. Since fm is continuous at z, there is z > 0 so that (y [a, b] : |y z| < z)(| fm(y) fm(z)| < 2 ). Notice that by the triangle inequality, for every y we have | fm(y)| | fm(y) fm(z)| + | fm(z)|. So (y [a, b] : |y z| < z)(| fm(y)| < ). Since z = lim inf xk , there is an integer M m such that |xM z| < z . Consider fnM (xM). We’ve already assumed that | fnM (xM)| . However, we also have | fnM (xM)| = fnM (xM) (since each fn(x) 0) fm(xM) (since nM > M m > N0, and hfn(xM)i is decreasing) < (since |xM x| < z) This gives a contradiction. We conclude that if f is pointwise convergent to 0, it is also uniformly convergent to 0, given the hypotheses for this problem.

We can also give a direct proof of this problem using the Heine-Borel theorem of section 13. The idea is that if fn 0 pointwise, each fn is continuous, and > 0, then for each z [a, b], there is an integer nz with 0 fnz (z) < 2 ; and there is nz ,z > 0 so that |x z| < nz ,z implies | fnz (x) fnz (z)| < 2 . Since (by the Heine-Borel theorem) the closed interval [a, b] is compact, and the open sets of the form (z nz ,z ,z + nz ,z) cover [a, b], there is a finite sub-collection S of these open sets such that the sets in S cover [a, b]. Now choose N to be the max of all the nz’s mentioned in S . If n > N and x [a, b], then x is in one of the open intervals in S of the form (z nz ,z ,z + nz ,z). So 0 fn(x) fnz (x) | fnz (x) fnz (z)| + | fnz (z)| < 2 + 2 = . This shows directly that the convergence is uniform.