LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Weissinger's fixed-point criterion for summably contracting iterates

Statement

Let $(X, d)$ be a nonempty complete metric space and let $T : X \to X$ . Suppose nonnegative reals $(a_{m})_{m \geq 1}$ satisfy $\sum_{m \geq 1} a_{m} < \infty$ and

$d (T^{m} x, T^{m} y) \leq a_{m} d (x, y) (x, y \in X, m \geq 1) .$

Summably contracting iterates have a unique fixed point and the iteration tail bounds its error. Explicitly, for $x_{k + 1} = T x_{k}$ and the fixed point $x^{*}$ ,

$d (x_{n}, x^{*}) \leq d (x_{1}, x_{0}) \sum_{m \geq n} a_{m} (n \geq 1) .$

Facts & Assumptions

Given: The complete metric space, map, constants, and starting point in the Statement.

[L1]

Absolute convergence implies convergence for a real series (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges).

[L2]

In a complete metric space every Cauchy sequence converges to a point of the space (Complete metric space: every Cauchy sequence converges in the space).

Proof

technique · direct

1.1givenL1algebra

For $p > n$ , telescoping and the iterate estimate give $d (x_{p}, x_{n}) \leq d (x_{1}, x_{0}) \sum_{m = n}^{p - 1} a_{m}$ ; by [L1] the tails tend to zero, so $(x_{n})$ is Cauchy.

2.1step 1.1L2algebra∎

By [L2], $x_{n} \to x^{*} \in X$ ; letting $p \to \infty$ in step 1.1 gives the stated tail bound. The $m = 1$ estimate makes $T$ continuous, hence $T x^{*} = x^{*}$ , and since summability gives some $a_{m} < 1$ , two fixed points $u, v$ satisfy $d (u, v) = d (T^{m} u, T^{m} v) \leq a_{m} d (u, v)$ and are equal.

Depends on

Used by

Picard iteration converges with geometric short-time and factorial cylinder error bounds Proposition

Dependency tree · two levels

32 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald Teschl, Ordinary Differential Equations and Dynamical Systems, Ch. 2 (standard reference, not scraped)
Jiri Lebl, Basic Analysis I, Section 6.3 (standard reference, not scraped)