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LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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A rectangular second difference equals a mixed partial times the side lengths

Statement

Let RR be a closed axis-parallel rectangle and suppose that fxf_x and fxyf_{xy} exist on an open neighbourhood of RR. If (x0,y0),(x1,y1)(x_0,y_0),(x_1,y_1) are opposite corners of a nondegenerate subrectangle of RR, then some ξ\xi strictly between x0,x1x_0,x_1 and some η\eta strictly between y0,y1y_0,y_1 satisfy

f(x1,y1)f(x1,y0)f(x0,y1)+f(x0,y0)=(x1x0)(y1y0)fxy(ξ,η).f(x_1,y_1)-f(x_1,y_0)-f(x_0,y_1)+f(x_0,y_0)=(x_1-x_0)(y_1-y_0)f_{xy}(\xi,\eta).

Facts & Assumptions

Given: The stated open-neighbourhood hypotheses and a nondegenerate subrectangle of RR.

[L1]

After ordering its two endpoints, the one-variable mean-value theorem gives g(v)g(u)=(vu)g(c)g(v)-g(u)=(v-u)g'(c) for a function continuous on the closed interval and differentiable on its interior, with cc strictly between the endpoints (The mean value theorem, as the case g(x)=xg(x) = x of Cauchy's: for ff continuous on [a,b][a,b] with a<ba < b and differentiable on (a,b)(a,b) there is c(a,b)c \in (a,b) with f(b)f(a)=f(c)(ba)f(b) - f(a) = f'(c)(b-a)).

Proof

technique · direct
1.1

Apply [L1] in the xx variable to xf(x,y1)f(x,y0)x\mapsto f(x,y_1)-f(x,y_0). The stated existence of fxf_x on an open neighbourhood gives the required one-variable regularity, and the rectangle difference is (x1x0)(fx(ξ,y1)fx(ξ,y0))(x_1-x_0)(f_x(\xi,y_1)-f_x(\xi,y_0)).

L1givenchoose
2.1

Apply [L1] in the yy variable to yfx(ξ,y)y\mapsto f_x(\xi,y). Since fxyf_{xy} exists on an open neighbourhood, this one-variable map is continuous on the closed interval and differentiable on its interior. This yields (y1y0)fxy(ξ,η)(y_1-y_0)f_{xy}(\xi,\eta) and proves the formula.

step 1.1L1givenchoose

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