TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31

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Uniform Abel test: a uniformly convergent function series times a uniformly bounded pointwise monotone family gives a uniformly convergent product series

Statement

Let $X$ be a set and let $u_{k}, v_{k} : X \to R$ . Suppose $\sum u_{k}$ converges uniformly on $X$ , there is $M \geq 0$ with $∣ v_{k} (x) ∣ \leq M$ for every $k, x$ , and for each fixed $x$ the real sequence $(v_{k} (x))$ is monotone. Its direction may depend on $x$ . Then $\sum u_{k} v_{k}$ converges uniformly on $X$ .

Facts & Assumptions

Given: Functions $u_{k}, v_{k} : X \to R$ satisfying the hypotheses in the Statement.

[L1]

Uniform convergence of $\sum u_{k}$ is equivalent to uniformly small tails (A series of real-valued functions converges uniformly if and only if its tails are uniformly small).

[L2]

For real sequences $(a_{j}), (b_{j})$ and $A_{n} : = \sum_{j < n} a_{j}$ , Abel summation by parts says that, for every $n \geq 1$ , $\sum_{j < n} a_{j} b_{j} = A_{n} b_{n - 1} - \sum_{j < n - 1} A_{j + 1} (b_{j + 1} - b_{j})$ (Abel summation by parts: with $A_{n} = \sum_{k < n} a_{k}$ one has $\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k})$ for every $n \geq 1$ ).

[L3]

Finite sums split and telescope, and repeated triangle inequalities bound the absolute value of a finite sum by the sum of the absolute values (Finite sums and finite products, by recursion, Laws of finite sums and finite products, The triangle inequality, Basic properties of the absolute value).

Proof

technique · direct

1.1

Let $ε > 0$ and put $η : = ε / (3 M + 1) > 0$ . By [L1] choose $N$ such that $∣ \sum_{k = m + 1}^{n} u_{k} (x) ∣ < η$ for every $n > m \geq N$ and every $x \in X$ .

L1choose

1.2

For every $x \in X$ and naturals $p \leq q$ , monotonicity makes all successive differences $v_{k} (x) - v_{k + 1} (x)$ have one sign, so $\sum_{k = p}^{q - 1} ∣ v_{k} (x) - v_{k + 1} (x) ∣ = ∣ v_{p} (x) - v_{q} (x) ∣ \leq 2 M$ .

givenL3algebra

2.1

Fix $n > m \geq N$ and $x \in X$ , put $p : = m + 1$ , $q : = n$ , and define $B_{r} (x) : = \sum_{k = p}^{r} u_{k} (x)$ for $p \leq r \leq q$ . Then $∣ B_{r} (x) ∣ < η$ for every such $r$ .

step 1.1construct

3.1

For $0 \leq j \leq q - p$ put $a_{j} : = u_{p + j} (x)$ and $b_{j} : = v_{p + j} (x)$ . Their partial sums satisfy $A_{j + 1} = B_{p + j} (x)$ , so [L2] with $n = q - p + 1$ gives $\sum_{k = p}^{q} u_{k} (x) v_{k} (x) = B_{q} (x) v_{q} (x) + \sum_{k = p}^{q - 1} B_{k} (x) (v_{k} (x) - v_{k + 1} (x))$ .

step 2.1L2L3

4.1

By steps 2.1, 3.1, and 1.2, $∣ \sum_{k = p}^{q} u_{k} (x) v_{k} (x) ∣ \leq η ∣ v_{q} (x) ∣ + η \sum_{k = p}^{q - 1} ∣ v_{k} (x) - v_{k + 1} (x) ∣ \leq 3 M η < ε$ .

step 2.1step 3.1step 1.2L3algebra

5.1

The estimate is uniform in $x$ and holds for every $n > m \geq N$ , so [L1] proves uniform convergence of $\sum u_{k} v_{k}$ .

step 4.1L1∎

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Sources

University of Alberta Math 317, Infinite Series of Functions (standard reference, not scraped)