absolute convergence implies uniform convergence
Theorem 1.
Let T be a topological space, f be a continuous function
from T to [0,∞), and
let {fk}∞k=0 be a sequence of continuous functions from T to [0,∞)
such that, for all x∈T, the sum ∑∞k=0fk(x) converges
to f(x). Then
the convergence of this sum is uniform on compact subsets of T.
Proof.
Let X be a compact subset of T and let ϵ be a positive real number. We will
construct an open cover of X. Because the series is assumed to converge pointwise, for
every x∈X, there exists an integer nx such that ∑∞k=nxfk(x)<ϵ/3. By continuity, there exists an open neighborhood N1 of x such that |f(x)-f(y)|<ϵ/3 when y∈N1 and an open neighborhood N2 of x such that |∑nxk=0fk(x)-∑nk=0fk(y)|<ϵ/3 when y∈N2.
Let Nx be the intersection
of N1 and N2. Then, for every y∈N, we have
f(y)-nx∑k=0fk(y)<|f(y)-f(x)|+|f(x)-nx∑k=0fk(x)|+|nx∑k=0fk(x)-nx∑k=0fk(y)|<ϵ. |
In this way, we associate to every point x an neighborhood Nx and an integer nx. Since X is compact, there will exist a finite number of points x1,…xm such that X⊆Nx1∪⋯∪Nxm. Let n be the greatest of nx1,…,nxm. Then we have f(y)-∑nk=0fk(y)<ϵ for all y∈X, so, the functions fk being positive, f(y)-∑hk=0fk(y)<ϵ for all h≥n, which means that the sum converges uniformly. ∎
Note: This result can also be deduced from Dini’s theorem, since the partial sums of positive functions are monotonically increasing.
Title | absolute convergence implies uniform convergence |
---|---|
Canonical name | AbsoluteConvergenceImpliesUniformConvergence |
Date of creation | 2013-03-22 18:07:27 |
Last modified on | 2013-03-22 18:07:27 |
Owner | rspuzio (6075) |
Last modified by | rspuzio (6075) |
Numerical id | 9 |
Author | rspuzio (6075) |
Entry type | Theorem |
Classification | msc 40A30 |