LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-21

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A bounded vector field makes the Picard operator preserve a sufficiently short closed curve ball

Statement

Let $C = [t_{0} - h, t_{0} + h] \times \overline{B} (x_{0}, r)$ be contained in the domain of a continuous vector field $F$ , with $h \geq 0$ and $r > 0$ . If $∥ F (t, x) ∥_{2} \leq M$ on $C$ and $h M \leq r$ , then the Picard operator maps the closed curve ball $B_{r}$ into itself.

Facts & Assumptions

Given: The cylinder, bound, and Picard operator in the Statement.

[L1]

For $u \leq v$ , the norm of a vector integral is at most the integral of the Euclidean norm; for reversed limits the oriented convention gives the same estimate with absolute value on the scalar integral (For $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable).

[L2]

Composites of continuous maps are continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[L3]

The integral function of a continuous real function on a nondegenerate interval is differentiable and hence continuous; a vector-valued map is continuous exactly when its coordinate maps are continuous (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ , A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions).

Proof

technique · direct

1.1givenL1L2L3

If $h = 0$ , the domain is the singleton ${t_{0}}$ , so $T x$ is automatically continuous and its displacement is zero. Assume $h > 0$ . For $x \in B_{r}$ , [L2] makes $t \mapsto F (t, x (t))$ continuous, and [L3] applied componentwise makes $T x$ continuous. The given bound and [L1], applied on the interval between $t_{0}$ and $t$ , then give $∥ (T x) (t) - x_{0} ∥_{2} \leq M ∣ t - t_{0} ∣$ for every $t$ , also when $M = 0$ .

2.1step 1.1algebra∎

Since $∣ t - t_{0} ∣ \leq h$ and $M h \leq r$ , step 1.1 gives $\sup_{t} ∥ (T x) (t) - x_{0} ∥_{2} \leq r$ , so $T x \in B_{r}$ .

Depends on

Used by

Dependency tree · two levels

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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)