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TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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 XX be a set and let uk,vk:X→Ru_k,v_k:X\to\mathbb{R}. Suppose ∑uk\sum u_k converges uniformly on XX, there is M≥0M\ge0 with ∣vk(x)∣≤M|v_k(x)|\le M for every k,xk,x, and for each fixed xx the real sequence (vk(x))(v_k(x)) is monotone. Its direction may depend on xx. Then ∑ukvk\sum u_kv_k converges uniformly on XX.

Facts & Assumptions

Given: Functions uk,vk:X→Ru_k,v_k:X\to\mathbb{R} satisfying the hypotheses in the Statement.

[L1]

Uniform convergence of ∑uk\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 (aj),(bj)(a_j),(b_j) and An:=∑j<najA_n:=\sum_{j<n}a_j, Abel summation by parts says that, for every n≥1n\ge1, ∑j<najbj=Anbn−1−∑j<n−1Aj+1(bj+1−bj)\sum_{j<n}a_jb_j=A_nb_{n-1}-\sum_{j<n-1}A_{j+1}(b_{j+1}-b_j) (Abel summation by parts: with An=∑k<nakA_n = \sum_{k<n} a_k one has ∑k<nakbk=Anbn−1−∑k<n−1Ak+1 (bk+1−bk)\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≥1n \ge 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\varepsilon>0 and put η:=ε/(3M+1)>0\eta:=\varepsilon/(3M+1)>0. By [L1] choose NN such that ∣∑k=m+1nuk(x)∣<η\left|\sum_{k=m+1}^{n}u_k(x)\right|<\eta for every n>m≥Nn>m\ge N and every x∈Xx\in X.

L1choose
1.2

For every x∈Xx\in X and naturals p≤qp\le q, monotonicity makes all successive differences vk(x)−vk+1(x)v_k(x)-v_{k+1}(x) have one sign, so ∑k=pq−1∣vk(x)−vk+1(x)∣=∣vp(x)−vq(x)∣≤2M\sum_{k=p}^{q-1}|v_k(x)-v_{k+1}(x)|=|v_p(x)-v_q(x)|\le2M.

givenL3algebra
2.1

Fix n>m≥Nn>m\ge N and x∈Xx\in X, put p:=m+1p:=m+1, q:=nq:=n, and define Br(x):=∑k=pruk(x)B_r(x):=\sum_{k=p}^{r}u_k(x) for p≤r≤qp\le r\le q. Then ∣Br(x)∣<η|B_r(x)|<\eta for every such rr.

step 1.1construct
3.1

For 0≤j≤q−p0\le j\le q-p put aj:=up+j(x)a_j:=u_{p+j}(x) and bj:=vp+j(x)b_j:=v_{p+j}(x). Their partial sums satisfy Aj+1=Bp+j(x)A_{j+1}=B_{p+j}(x), so [L2] with n=q−p+1n=q-p+1 gives ∑k=pquk(x)vk(x)=Bq(x)vq(x)+∑k=pq−1Bk(x)(vk(x)−vk+1(x))\sum_{k=p}^{q}u_k(x)v_k(x)=B_q(x)v_q(x)+\sum_{k=p}^{q-1}B_k(x)\bigl(v_k(x)-v_{k+1}(x)\bigr).

step 2.1L2L3
4.1

By steps 2.1, 3.1, and 1.2, ∣∑k=pquk(x)vk(x)∣≤η∣vq(x)∣+η∑k=pq−1∣vk(x)−vk+1(x)∣≤3Mη<ε\left|\sum_{k=p}^{q}u_k(x)v_k(x)\right|\le\eta|v_q(x)|+\eta\sum_{k=p}^{q-1}|v_k(x)-v_{k+1}(x)|\le3M\eta<\varepsilon.

step 2.1step 3.1step 1.2L3algebra
5.1

The estimate is uniform in xx and holds for every n>m≥Nn>m\ge N, so [L1] proves uniform convergence of ∑ukvk\sum u_kv_k.

step 4.1L1∎

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