absolute convergence implies uniform convergence

Theorem 1.

Let $T$ be a topological space, $f$ be a continuous function from $T$ to $[0,\\infty)$ , andlet $\\{f_{k}\\}_{k=0}^{\\infty}$ be a sequence of continuous functions from $T$ to $[0,\\infty)$ such that, for all $x\\in T$ , the sum $\\sum_{k=0}^{\\infty}f_{k}(x)$ converges to $f(x)$ . Thenthe convergence of this sum is uniform on compact subsets of $T$ .

Proof.

Let $X$ be a compact subset of $T$ and let $\\epsilon$ be a positive real number. We willconstruct an open cover of $X$ . Because the series is assumed to converge pointwise, forevery $x\\in X$ , there exists an integer $n_{x}$ such that $\\sum_{k=n_{x}}^{\\infty}f_{k}(x)<\\epsilon/3$ . By continuity, there exists an open neighborhood $N_{1}$ of $x$ such that $|f(x)-f(y)|<\\epsilon/3$ when $y\\in N_{1}$ and an open neighborhood $N_{2}$ of $x$ such that $\\left|\\sum_{k=0}^{n_{x}}f_{k}(x)-\\sum_{k=0}^{n}f_{k}(y)\\right|<\\epsilon/3$ when $y\\in N_{2}$ .Let $N_{x}$ be the intersection of $N_{1}$ and $N_{2}$ . Then, for every $y\\in N$ , we have

f(y)-\\sum_{k=0}^{n_{x}}f_{k}(y)<|f(y)-f(x)|+\\left|f(x)-\\sum_{k=0}^{n_{x}}f_{k}%(x)\\right|+\\left|\\sum_{k=0}^{n_{x}}f_{k}(x)-\\sum_{k=0}^{n_{x}}f_{k}(y)\\right|<\\epsilon.

In this way, we associate to every point $x$ an neighborhood $N_{x}$ and an integer $n_{x}$ .Since $X$ is compact, there will exist a finite number of points $x_{1},\\ldots x_{m}$ suchthat $X\\subseteq N_{x_{1}}\\cup\\cdots\\cup N_{x_{m}}$ . Let $n$ be the greatest of $n_{x_{1}},\\ldots,n_{x_{m}}$ . Then we have $f(y)-\\sum_{k=0}^{n}f_{k}(y)<\\epsilon$ for all $y\\in X$ ,so, the functions $f_{k}$ being positive, $f(y)-\\sum_{k=0}^{h}f_{k}(y)<\\epsilon$ for all $h\\geq n$ , 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.