Reason: Proof of lem:segment-derivative-c1-2026a: interval neighborhood by infimum/supremum of the excluded parameter sets; first derivative by coordinate telescoping with the one-dimensional mean value theorem on each slice; continuity from Euclidean continuity of f and its partials; second derivative by applying claim 2 to each partial and summing.
Second, βk=1nβhk2ββ€sh2β, since expanding sh2β=βkββlββ£hkββ£β£hlββ£ yields the terms β£hkββ£2=hk2β together with nonnegative cross terms. In particular, if a real Ξ»>0 satisfies β£ΟβΟβ£(1+shβ)<Ξ», then Q(x+Οh,x+Οh)β€(ΟβΟ)2sh2β<Ξ»2, the last step by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field.
Claim 1. If h=0, then x+Οh=xβW for every Ο, and r=1 works. So assume hξ =0; then shβ>0.
Let I={ΟβR:x+ΟhβW}; the hypothesis says that every Ο with 0β€Οβ€1 lies in I. Every Ο0ββI has a surrounding interval inside I: since W is open there is Ο>0 with {y:Q(y,x+Ο0βh)<Ο2}βW, and then by the preliminary estimate every Ο with β£ΟβΟ0ββ£<Ο/(1+shβ) lies in I; write Ξ·(Ο0β)=Ο/(1+shβ)>0 for one such choice.
Let A={ΟβR:Ο>1,Β Οβ/I} and B={ΟβR:Ο<0,Β Οβ/I}.
If Aξ =β : A is bounded below by 1, so Ξ²=infA exists by Existence of the Infimum of a Nonempty Subset of R Bounded Below. Every element of A is β₯1+Ξ·(1), since ΟβA with Ο<1+Ξ·(1) would satisfy β£Οβ1β£<Ξ·(1) and hence ΟβI. Thus 1+Ξ·(1) is a lower bound of A, so Ξ²β₯1+Ξ·(1)>1. Moreover every Ο with 1<Ο<Ξ² lies in I: otherwise ΟβA, so Ξ²β€Ο, a contradiction. If A=β , set Ξ²=2; then again every Ο with 1<Ο<Ξ² lies in I.
Symmetrically, if Bξ =β , then B is nonempty and bounded above by 0, so it has a least upper bound Ξ³ by the Dedekind completeness of R (The Real Numbers and Standard Notation). Every element of B is β€βΞ·(0) (as above, using the interval around 0βI), so Ξ³β€βΞ·(0)<0, and every Ο with Ξ³<Ο<0 lies in I. If B=β , set Ξ³=β1, with the same conclusion.
Let r be the least of 1, Ξ²β1, and βΞ³; it is positive as the least of three positive reals (claim 9 of Elementary Order Arithmetic in an Ordered Field, applied twice). Let βr<Ο<1+r. If 0β€Οβ€1, then ΟβI by hypothesis; if 1<Ο<1+r, then Ο<Ξ² since rβ€Ξ²β1, so ΟβI; if βr<Ο<0, then Ξ³β€βr<Ο, so ΟβI. This proves claim 1.
Claim 2. Let Ο0β be an interior point of J, and set z=x+Ο0βh and L=βi=1nββiβf(z)hiβ.
Since W is open and zβW, fix Ο>0 such that Bzβ={yβRn:Q(y,z)<Ο2}βW. Let Ξ΅>0 be real. For each kβ{1,β¦,n}, continuity of βkβf at z (preamble), applied with the positive real Ξ΅/(n(1+β£hkββ£)), gives Ξ΄kβ>0 such that every yβW with Q(y,z)<Ξ΄k2β has
Let tβR with 0<β£tβ£<Ξ΄ and Ο0β+tβJ. For kβ{0,1,β¦,n} let wkββRn be the point whose lth coordinate is zlβ+thlβ for lβ€k and zlβ for l>k; thus w0β=z and wnβ=x+(Ο0β+t)h. More generally, consider any point y whose lth coordinate is zlβ+thlβ for l<k, is zkβ+u with β£uβ£β€β£tβ£β£hkββ£ for l=k, and is zlβ for l>k; then
Fix k with hkβξ =0; if hkβ=0 then wkβ=wkβ1β and the kth term vanishes. Let p be the lesser and q the greater of 0 and thkβ, so p<q; let K={uβR:pβ€uβ€q}, an interval; and define gkβ:KβR by gkβ(u)=f(ykβ(u)), where ykβ(u) is the point whose lth coordinate is zlβ+thlβ for l<k, is zkβ+u for l=k, and is zlβ for l>k. By the preceding paragraph ykβ(u)βBzβ for every uβK, so gkβ is well defined, with gkβ(0)=f(wkβ1β) and gkβ(thkβ)=f(wkβ).
First, gkβ is differentiable at every interior pointu of K with gkβ²β(u)=βkβf(ykβ(u)): the partial derivative of f with respect to the kth variable exists at ykβ(u)βW with value βkβf(ykβ(u)) (the C1 hypothesis), and for wβR the point obtained from ykβ(u) by adding w to its kth coordinate is exactly ykβ(u+w); hence the difference quotients of gkβ at u coincide with those of Partial Derivative on a Euclidean Open Set, and the requirement of the derivative definition follows by restricting its quantifier to those w with u+wβK.
Second, gkβ is continuous at every point of K relative to K: given real Ξ΅β²>0, continuity of f at ykβ(u) (preamble) gives Ξ΄β²>0 with β£f(y)βf(ykβ(u))β£<Ξ΅β² whenever yβW and Q(y,ykβ(u))<Ξ΄β²2; for vβK with dRβ(v,u)=β£vβuβ£<Ξ΄β² one has Q(ykβ(v),ykβ(u))=(vβu)2<Ξ΄β²2, whence dRβ(gkβ(v),gkβ(u))<Ξ΅β².
By the mean value theorem applied to gkβ on K, there is ΞΎkβ with p<ΞΎkβ<q and gkβ(q)βgkβ(p)=gkβ²β(ΞΎkβ)(qβp). Whether thkβ is q (for thkβ>0) or p (for thkβ<0), this rearranges to
the terms with hkβ=0 of both sums vanishing. Since β£ΞΎkββ£β€β£tβ£β£hkββ£, the point ykβ(ΞΎkβ) satisfies Q(ykβ(ΞΎkβ),z)<ΞΌ2β€Ξ΄k2β, so
using β£hkββ£<1+β£hkββ£. As Ξ΅>0 was arbitrary and the estimate covers every t with 0<β£tβ£<Ξ΄ and Ο0β+tβJ, the function F is differentiable at Ο0β with Fβ²(Ο0β)=L, which is claim 2.
Claim 3. Let ΟβJ, write z=x+Οh, and let g be either f or βiβf for some iβ{1,β¦,n}, with Ξ¦:JβR given by Ξ¦(Ο)=g(x+Οh). Given real Ξ΅>0, continuity of g at z (preamble) gives Ξ΄β²>0 such that every yβW with Q(y,z)<Ξ΄β²2 has β£g(y)βg(z)β£<Ξ΅. Set Ξ·=Ξ΄β²/(1+shβ)>0. For ΟβJ with dRβ(Ο,Ο)<Ξ·, the preliminary estimate gives Q(x+Οh,z)<Ξ΄β²2, so dRβ(Ξ¦(Ο),Ξ¦(Ο))=β£g(x+Οh)βg(z)β£<Ξ΅. Hence Ξ¦ is continuous at Ο relative to J in the sense of Continuous Map Between Metric Spaces, which gives claim 3 for F and for each Οβ¦βiβf(x+Οh).
Claim 4. Suppose f is of class C2 on W. By clause 2 of C^k Maps on a Euclidean Open Set (with k=1), each βiβf:WβR is of class C1 on W, and by clause 4 there the partial derivative of βiβf with respect to the jth variable at any point of W is βjββiβf evaluated there. Fix an interior point Ο0β of J. Applying claim 2 with βiβf in place of f (the hypothesis that x+ΟhβW for every ΟβJ is unchanged), the function uiβ:JβR, uiβ(Ο)=βiβf(x+Οh), is differentiable at Ο0β with
Since G(Ο)=βi=1nβhiβuiβ(Ο) for ΟβJ, claim 2 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives shows that each constant multiple hiβuiβ is differentiable at Ο0β with derivative hiβuiβ²β(Ο0β), and then, by induction on the number of summands β the same claim applied to the sum of the first i summands and the summand hi+1βui+1β β the function G is differentiable at Ο0β with