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LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)
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Uniform limits respect sums and scalar multiples

Statement

Let XX be a set. Suppose fkff_k\to f and gkgg_k\to g uniformly on XX, where all functions are real valued. Then, for all α,βR\alpha,\beta\in\mathbb{R},

αfk+βgkαf+βg\alpha f_k+\beta g_k\longrightarrow \alpha f+\beta g

uniformly on XX. In particular, uniform convergence is preserved by sums, differences, and scalar multiples.

Facts & Assumptions

Given: A set XX, uniformly convergent sequences fkff_k\to f and gkgg_k\to g, and reals α,β\alpha,\beta.

[A1]

Uniform convergence gives, for every real η>0\eta>0, one index after which fk(x)f(x)<η|f_k(x)-f(x)|<\eta at every xx, and likewise for gkgg_k\to g (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[L1]

For reals u,vu,v, u+vu+v|u+v|\le |u|+|v|, while cu=cu|cu|=|c||u| and absolute values are nonnegative (The triangle inequality, Basic properties of the absolute value).

Proof

technique · direct
1.1

Let ε>0\varepsilon>0 be real and put η:=ε/(2(α+β+1))>0\eta:=\varepsilon/\bigl(2(|\alpha|+|\beta|+1)\bigr)>0.

constructL1
1.2

By uniform convergence choose Nf,NgN_f,N_g such that fk(x)f(x)<η|f_k(x)-f(x)|<\eta for kNfk\ge N_f and all xx, and gk(x)g(x)<η|g_k(x)-g(x)|<\eta for kNgk\ge N_g and all xx.

A1choose
2.1

Choose an index NN at least as large as NfN_f and NgN_g. For every kNk\ge N and xXx\in X, one has α(fk(x)f(x))+β(gk(x)g(x))αη+βη<ε|\alpha(f_k(x)-f(x))+\beta(g_k(x)-g(x))|\le |\alpha|\eta+|\beta|\eta<\varepsilon.

step 1.1step 1.2L1choosealgebra
3.1

The expression in step 2.1 is (αfk+βgk)(x)(αf+βg)(x)|(\alpha f_k+\beta g_k)(x)-(\alpha f+\beta g)(x)|, and the index NN serves every xx, so the asserted convergence is uniform.

step 2.1A1

Depends on

Used by

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Sources