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# absolute convergence implies uniform convergence

###### Theorem 1.

Let $T$ be a topological space, $f$ be a continuous function from $T$ to $[0,\infty)$, and let $\{f_{k}\}_{{k=0}}^{\infty}$ 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 $f(x)$. Then the 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 will construct an open cover of $X$. Because the series is assumed to converge pointwise, for every $x\in X$, 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 $x_{1},\ldots x_{m}$ such that $X\subseteq N_{{x_{1}}}\cup\cdots\cup N_{{x_{m}}}$. 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.

## Mathematics Subject Classification

40A30*no label found*

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