Β· 16,542 chars Β· 27 deps Β· depth 19 Reason: First publication of the proof: telescoping along coordinate directions through nearby points of a good set at which the point is a density point.
Off a null set the partial derivatives exist and the point is a density point of one of countably many good sets on which the difference quotients approximate the partial derivatives uniformly and the partial derivatives are nearly constant; an increment is then traversed one coordinate at a time, each intermediate point being replaced by a nearby point of the good set, and the Lipschitz bound controls the replacement errors.
Proof
We use the notation of the statement. Being Lipschitz, f is continuous from (Rn,dEβ) to (R,dRβ) by A Lipschitz Map is Uniformly Continuous, where dRβ(s,t)=β£sβtβ£ is the absolute value metric. For 1β€iβ€n let eiβ be the standard basis vector of Rn whose ith coordinate is 1 and whose other coordinates are 0, and for yβRn and real sξ =0 put Ξsiβ(y)=(f(y+seiβ)βf(y))/s.
A norm inequality used twice. For a natural number qβ₯1 and v=(v1β,β¦,vqβ)βRq we have
β₯vβ₯β€Οqβ1β€kβ€qmaxββ£vkββ£,
where Οqβ is the constant of the setting, satisfying Οq2β=q. Indeed, by claim 1 of Elementary Properties of the Euclidean Norm on Rn and Difference, Dot Product, and Orthogonality in Rn, β₯vβ₯2=βk=1qβvk2ββ€q(maxkββ£vkββ£)2=(Οqβmaxkββ£vkββ£)2, and both β₯vβ₯ and Οqβmaxkββ£vkββ£ are nonnegative. The inequality between the squares forces the asserted one: if instead Οqβmaxkββ£vkββ£<β₯vβ₯ held, then multiplying this inequality by itself, which is legitimate for nonnegative numbers, would give (Οqβmaxkββ£vkββ£)2<β₯vβ₯2, a contradiction.
Fix xβRnβN for the rest of Steps 2 and 3. Then xβE, so all n partial derivatives of f exist at x and Df(x) is defined; and x is a density point of every set G(k,j,q) containing it.
Step 2 (Choice of a good set). Let Ξ΅βR with 0<Ξ΅, and put C=n2n. Choose
Step 3 (The telescoping estimate). Let hβRn with β₯hβ₯<Ξ΄. If h=0 the asserted inequality is trivial, so assume hξ =0 and put Ξ·=β₯hβ₯, so 0<Ξ·<Ξ΄. Write h=(h1β,β¦,hnβ), so that h=βp=1nβhpβepβ and β£hpββ£β€Ξ· for every p by claim 4 of Elementary Properties of the Euclidean Norm on Rn.
Define points z0β,β¦,znβ and w1β,β¦,wnβ recursively: z0β=x, and for 1β€pβ€n put wpβ=zpβ1β+hpβepβ and let zpβ=wpβ if wpβ=x, while if wpβξ =x let zpββG satisfy β₯zpββwpββ₯β€Ξ΄1ββ₯wpββxβ₯. In either case
We must check that this recursion is legitimate, that is, that β₯wpββxβ₯<Ο whenever wpβξ =x. Put cpβ=β₯zpββxβ₯, so c0β=0. For 1β€pβ€n, the triangle inequality (claim 6 of Elementary Properties of the Euclidean Norm on Rn) and claim 5 of that lemma give
and then cpββ€β₯wpββxβ₯+β₯zpββwpββ₯β€(1+Ξ΄1β)β₯wpββxβ₯β€2(cpβ1β+Ξ·), using Ξ΄1ββ€1. By induction cpββ€Ξ·βl=1pβ2l for 1β€pβ€n, and βl=1pβ2l=2p+1β2β€n2n=C for 1β€pβ€n; hence cpββ€CΞ· and
Since Ξ·<Ξ΄β€Ο/(2C+2) we get β₯wpββxβ₯β€(C+1)Ξ·<Ο/2<Ο, as required. (Each application of the recursion uses only the already constructed zpβ1β, so the estimates above are available at the moment they are needed.)
From zpβ=zpβ1β+hpβepβ+(zpββwpβ) and h=βpβhpβepβ we get znβ=(x+h)+βp=1nβ(zpββwpβ), so by the triangle inequality and (ββ),
because the first two sums together telescope to f(znβ)βf(z0β)=f(znβ)βf(x).
We estimate the three parts. If hpβξ =0 then 0<β£hpββ£β€Ξ·<Ξ΄β€1/jpβ, so (β) applied with z=zpβ1ββG, the index p and s=hpβ gives β£f(wpβ)βf(zpβ1β)βhpββpβf(x)β£β€3β£hpββ£/kβ€3Ξ·/k; if hpβ=0 then wpβ=zpβ1β and both f(wpβ)βf(zpβ1β) and hpββpβf(x) are 0. Since βp=1nβhpββpβf(x)=Df(x)β h by Difference, Dot Product, and Orthogonality in Rn, summing gives
by the choices of k and Ξ΄1β. This proves claim 1.
Step 4 (Claims 2 and 3). Let x be as in claim 1 and let A be the matrix with one row and n columns given by A1iβ=βiβf(x). By Matrix-Vector Product the product Ah is the point of R1 whose single coordinate is βi=1nβA1iβhiβ=Df(x)β h, and by claim 1 of Elementary Properties of the Euclidean Norm on Rn the Euclidean norm of a point of R1 is the absolute value of its coordinate, both being the unique nonnegative real number whose square is the square of that coordinate. Hence the inequality of claim 1 reads β₯f(x+h)βf(x)βAhβ₯β€Ξ΅β₯hβ₯ for β₯hβ₯<Ξ΄, and since x+hβRn always, Differentiability at a Point for Maps Between Euclidean Spaces shows that f is differentiable at x with derivative matrix A. Claim 2 of A Derivative Matrix is the Jacobian Matrix, and is Unique identifies A with the Jacobian matrix. This is claim 2.
For claim 3, if Df(x)=0 then β₯Df(x)β₯=0β€L by claim 3 of Elementary Properties of the Euclidean Norm on Rn. Otherwise 0<β₯Df(x)β₯ by that same claim. Let Ξ΅>0 and let Ξ΄ be as in claim 1. Choose tβR with 0<t and tβ₯Df(x)β₯<Ξ΄, and put h=tDf(x), so β₯hβ₯=tβ₯Df(x)β₯<Ξ΄ by claim 5 of Elementary Properties of the Euclidean Norm on Rn and Df(x)β h=tβ₯Df(x)β₯2 by claim 1 of that lemma. Then, by claim 1 of the present theorem and the Lipschitz hypothesis,
Dividing by the positive number tβ₯Df(x)β₯ gives β₯Df(x)β₯β€L+Ξ΅, and since Ξ΅>0 was arbitrary, β₯Df(x)β₯β€L.
Step 5 (Claim 4). We first prove the following assertion about real-valued maps, in which V and g are fresh letters: if VβRn is open and g:VβR is locally Lipschitz on V, then the set of those xβV at which g is not differentiable is null. To lighten the notation we write U and T for V and g in the four paragraphs that follow, so that T is real-valued there; the letters resume their meaning from the statement afterwards. Let C be the set of pairs (p,r) with pβRn having all coordinates rational, rβQ with 0<r, BΛ(p,r)βU, and the restriction of T to BΛ(p,r) Lipschitz with some constant. By The Integers and the Rational Numbers are Countable and Products and Powers of Countable Sets the set C is countable.
We claim U=β(p,r)βCβB(p,r). Each B(p,r)βBΛ(p,r)βU. Conversely let xβU; by local Lipschitz continuity there are rxβ,LxββR with 0<rxβ, 0β€Lxβ, BΛ(x,rxβ)βU and T Lipschitz with constant Lxβ on BΛ(x,rxβ). By The Rational Numbers are Dense in the Real Numbers choose rβQ with rxβ/4<r<rxβ/2, and choose rational numbers p1β,β¦,pnβ with β£piββxiββ£<rxβ/(4Οnβ) for every i; then p=(p1β,β¦,pnβ) satisfies β₯pβxβ₯<rxβ/4 by the norm inequality recorded at the start. Hence β₯pβxβ₯<r, so xβB(p,r); and if yβBΛ(p,r) then β₯yβxβ₯β€β₯yβpβ₯+β₯pβxβ₯β€r+rxβ/4<rxβ, so BΛ(p,r)βBΛ(x,rxβ)βU and T is Lipschitz with constant Lxβ on BΛ(p,r). Thus (p,r)βC and xβB(p,r).
Let xβB(p,r)βNβ² and let Aβ² be the derivative matrix of F at x. Since B(p,r) is open there is Ξ΄2β>0 with B(x,Ξ΄2β)βB(p,r). Given Ξ΅>0, take the smaller of Ξ΄2β and the radius supplied by the differentiability of F at x; for h with 0<β₯hβ₯ less than that number we have x+hβB(p,r)βU, so T(x+h)=F(x+h) and T(x)=F(x), whence β₯T(x+h)βT(x)βAβ²hβ₯β€Ξ΅β₯hβ₯. Thus T is differentiable at x. Consequently the set of those xβB(p,r) at which T is not differentiable is contained in Nβ² and hence null, and the set of those xβU at which T is not differentiable, being contained in the union over the countable set C of these sets, is null by claim 5 of Elementary Properties of Lebesgue Outer Measure on Rn.
This proves the assertion about real-valued maps. Now let U, m and T:UβRm again be as in the statement of claim 4, and let T1β,β¦,Tmβ be the coordinate functions of T, as in Jacobian Matrix of a Map Between Euclidean Spaces. By claim 4 of Elementary Properties of the Euclidean Norm on Rn, β£Tkβ(y)βTkβ(z)β£β€β₯T(y)βT(z)β₯ for all y,z, so each Tkβ is locally Lipschitz on U with the same local constants as T. By the assertion about real-valued maps just proved, for each k the set Mkβ of points of U at which Tkβ is not differentiable is null; put M=βk=1mβMkβ, a null set by claim 5 of Elementary Properties of Lebesgue Outer Measure on Rn.
Let xβUβM and let A be the matrix with m rows and n columns given by Akiβ=βiβTkβ(x), which is defined by claim 1 of A Derivative Matrix is the Jacobian Matrix, and is Unique applied to each Tkβ. Let Ξ΅>0. For each k, differentiability of Tkβ at x together with claim 2 of the present theorem, applied to Tkβ, provides Ξ΄kβ>0 such that β£Tkβ(x+h)βTkβ(x)ββi=1nβAkiβhiββ£β€(Ξ΅/Οmβ)β₯hβ₯ whenever β₯hβ₯<Ξ΄kβ; let Ξ΄ be the smallest of Ξ΄1β,β¦,Ξ΄mβ and of a radius for which x+hβU whenever β₯hβ₯<Ξ΄, which exists because U is open. For h with 0<β₯hβ₯<Ξ΄ put v=T(x+h)βT(x)βAh; by Matrix-Vector Product its kth coordinate is Tkβ(x+h)βTkβ(x)ββiβAkiβhiβ, so β£vkββ£β€(Ξ΅/Οmβ)β₯hβ₯ for every k and therefore β₯vβ₯β€Οmβmaxkββ£vkββ£β€Ξ΅β₯hβ₯ by the norm inequality recorded at the start. Hence T is differentiable at x with derivative matrix A, and the set of points of U at which T is not differentiable is contained in M and so is null.
Finally, at any xβU at which T is differentiable, claim 1 of A Derivative Matrix is the Jacobian Matrix, and is Unique says that the entry in row k and column i of the derivative matrix is βiβTkβ(x). This completes claim 4.