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LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-03
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Newton maps are uniform contractions near a point with invertible derivative

Statement

Let n1n\ge1, let URnU\subseteq\mathbb R^n be open, let f:URnf:U\to\mathbb R^n be C1C^1, and let aUa\in U. Suppose A:=Df(a)A:=Df(a) is invertible and put B:=A1B:=A^{-1}. Then there are R>0R>0, 0q<10\le q<1, and C>0C>0 such that B(a,R)U\overline B(a,R)\subseteq U, Bv2Cv2\|Bv\|_2\le C\|v\|_2, and, for every yRny\in\mathbb R^n, the Newton map

Ty(x):=x+B(yf(x))T_y(x):=x+B(y-f(x))

satisfies

Ty(x)Ty(z)2qxz2(x,zB(a,R)).\|T_y(x)-T_y(z)\|_2\le q\|x-z\|_2\qquad(x,z\in\overline B(a,R)).

Moreover Df(x)Df(x) is invertible for every xB(a,R)x\in\overline B(a,R) and

Df(x)1v2C1qv2.\|Df(x)^{-1}v\|_2\le\frac C{1-q}\|v\|_2.

Facts & Assumptions

Given: The dimensions, open set, C1C^1 map, point, and invertible derivative in the statement.

[L2]

The entries of DfDf are continuous by the definition of C1C^1 (Continuously differentiable maps, local inverses, and local diffeomorphisms).

[L7]

Invertibility means having a two-sided linear inverse (Invertible Euclidean linear maps).

Proof

technique · contraction
1.1

By [L1], choose C>0C>0 with Bv2Cv2\|Bv\|_2\le C\|v\|_2. Matrix-entry continuity [L2], [L6], and [L5] give R>0R>0 such that B(a,R)B(a,2R)U\overline B(a,R)\subset B(a,2R)\subseteq U and B(Df(w)A)v212v2\|B(Df(w)-A)v\|_2\le\tfrac12\|v\|_2 for wB(a,2R)w\in B(a,2R) and vRnv\in\mathbb R^n. Fix q:=1/2q:=1/2.

L1L2L5L6algebra
2.1

The chain rule gives DTy(w)=IBDf(w)=B(ADf(w))DT_y(w)=I-BDf(w)=B(A-Df(w)), independently of yy. The convex open ball B(a,2R)B(a,2R) contains the closed ball, so [L3] and step 1.1 yield Ty(x)Ty(z)2qxz2\|T_y(x)-T_y(z)\|_2\le q\|x-z\|_2 for x,zB(a,R)x,z\in\overline B(a,R).

step 1.1L3
3.1

Fix ww in the ball, put L:=Df(w)L:=Df(w), and fix vRnv\in\mathbb R^n. The map Sv(u):=u+B(vLu)S_v(u):=u+B(v-Lu) is a contraction of the complete space Rn\mathbb R^n with constant qq, by the same estimate as step 2.1. By [L4] it has a unique fixed point uu, and the fixed-point equation is equivalent to Lu=vLu=v. Thus LL is surjective. If Lu=0Lu=0, both uu and 00 are fixed by S0S_0, so uniqueness gives u=0u=0; hence LL is injective. The solution map is linear by uniqueness, so it is L1L^{-1}.

step 1.1L4L7
4.1

From Lu=vLu=v and step 1.1, u2uBLu2+Bv2qu2+Cv2\|u\|_2\le\|u-BLu\|_2+\|Bv\|_2 \le q\|u\|_2+C\|v\|_2. Therefore L1v2C(1q)1v2\|L^{-1}v\|_2\le C(1-q)^{-1}\|v\|_2.

step 1.1step 3.1algebra
5.1

Steps 1.1--4.1 give every asserted constant, contraction estimate, invertibility claim, and inverse bound.

step 1.1step 2.1step 3.1step 4.1

Depends on

Used by

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