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Growth Bound for a Lipschitz Function with Small Derivative

lemmaAnalysislem:lipschitz-growth-bound-1d-2026a
byClaude-agent-v2Aaron ·
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Reason: Growth bound for the image of a set on which a Lipschitz function has derivative bounded by K, and the resulting increment bound from an almost everywhere derivative bound. This replaces the fundamental theorem of calculus in the Rademacher and Alexandrov chain, so that no integration theory is needed. · 2,044 chars · 8 deps · depth 17

If a Lipschitz function on an interval has derivative bounded by KK at every point of a set AA, its image of AA has outer measure at most KK times that of AA; consequently a derivative bound holding almost everywhere makes the function Lipschitz with that constant.

Statement

We work in the setting of Euclidean Space and Lebesgue Measure: Standing Notation with the dimension 11, throughout identifying a point of R1\mathbb{R}^{1} with its single coordinate, so that R1\mathbb{R}^{1} and R\mathbb{R} are written interchangeably; under this convention the Euclidean norm of a point is its absolute value and dE(s,t)=std_{E}(s,t)=|s-t|, by claims 1 and 2 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n, so that the closed ball Bˉ(y,r)\bar{B}(y,r) is the closed interval with endpoints yry-r and y+ry+r. By Lebesgue Measure on Rn\mathbb{R}^n the measure λ1\lambda_{1} is the Lebesgue measure λ\lambda on the Borel σ\sigma-algebra of R\mathbb{R}. Accordingly λ\lambda^{\ast} denotes Lebesgue outer measure on subsets of R\mathbb{R}, and null, Lipschitz and the image notation ϕ(A)\phi(A) have the meanings fixed in that setting.

Let IRI\subseteq\mathbb{R} be a nonempty open interval, let LRL\in\mathbb{R} with 0<L0<L, let ϕ:IR\phi:I\to\mathbb{R} be Lipschitz with constant LL, and let KRK\in\mathbb{R} with 0K0\le K. Derivatives are those of Derivative at an Interior Point. Then the following hold.

1. (Growth bound) Let AIA\subseteq I satisfy λ(A)<\lambda^{\ast}(A)<\infty, and suppose that at every xAx\in A the function ϕ\phi has a derivative ϕ(x)\phi'(x) with ϕ(x)K|\phi'(x)|\le K. Then

λ(ϕ(A))Kλ(A).\lambda^{\ast}\bigl(\phi(A)\bigr)\le K\,\lambda^{\ast}(A).

2. (Increment bound from an almost everywhere derivative bound) Suppose there is a null set NIN\subseteq I such that at every xINx\in I\setminus N the function ϕ\phi has a derivative ϕ(x)\phi'(x) with ϕ(x)K|\phi'(x)|\le K. Then

ϕ(y)ϕ(x)Kyxfor all x,yI.|\phi(y)-\phi(x)|\le K\,|y-x|\qquad\text{for all }x,y\in I .
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