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Proof of Rademacher's Theorem in Rn\mathbb{R}^n

theoremthm:rademacher-rn-2026a
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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, ff is continuous from (Rn,dE)(\mathbb{R}^{n},d_{E}) to (R,dR)(\mathbb{R},d_{\mathbb{R}}) by A Lipschitz Map is Uniformly Continuous, where dR(s,t)=∣sβˆ’t∣d_{\mathbb{R}}(s,t)=|s-t| is the absolute value metric. For 1≀i≀n1\le i\le n let eie_{i} be the standard basis vector of Rn\mathbb{R}^{n} whose iith coordinate is 11 and whose other coordinates are 00, and for y∈Rny\in\mathbb{R}^{n} and real sβ‰ 0s\neq0 put Ξ”si(y)=(f(y+sei)βˆ’f(y))/s\Delta^{i}_{s}(y)=(f(y+se_{i})-f(y))/s.

A norm inequality used twice. For a natural number qβ‰₯1q\ge1 and v=(v1,…,vq)∈Rqv=(v_{1},\dots,v_{q})\in\mathbb{R}^{q} we have

βˆ₯vβˆ₯≀σqmax⁑1≀k≀q∣vk∣,\lVert v\rVert\le\sigma_{q}\max_{1\le k\le q}|v_{k}| ,

where Οƒq\sigma_{q} is the constant of the setting, satisfying Οƒq2=q\sigma_{q}^{2}=q. Indeed, by claim 1 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n and Difference, Dot Product, and Orthogonality in Rn\mathbb{R}^n, βˆ₯vβˆ₯2=βˆ‘k=1qvk2≀q(max⁑k∣vk∣)2=(Οƒqmax⁑k∣vk∣)2\lVert v\rVert^{2}=\sum_{k=1}^{q}v_{k}^{2}\le q\bigl(\max_{k}|v_{k}|\bigr)^{2}=\bigl(\sigma_{q}\max_{k}|v_{k}|\bigr)^{2}, and both βˆ₯vβˆ₯\lVert v\rVert and Οƒqmax⁑k∣vk∣\sigma_{q}\max_{k}|v_{k}| are nonnegative. The inequality between the squares forces the asserted one: if instead Οƒqmax⁑k∣vk∣<βˆ₯vβˆ₯\sigma_{q}\max_{k}|v_{k}|<\lVert v\rVert held, then multiplying this inequality by itself, which is legitimate for nonnegative numbers, would give (Οƒqmax⁑k∣vk∣)2<βˆ₯vβˆ₯2(\sigma_{q}\max_{k}|v_{k}|)^{2}<\lVert v\rVert^{2}, a contradiction.

Step 1 (The exceptional set). For 1≀i≀n1\le i\le n let EiE_{i} be the set of points at which βˆ‚if\partial_{i}f exists. By The Partial Derivatives of a Lipschitz Function on Rn\mathbb{R}^n Exist Almost Everywhere Β§ae each Rnβˆ–Ei\mathbb{R}^{n}\setminus E_{i} is a Borel null set, so by Null Set of a Measure and claim 2 of Basic Properties of a Measure it satisfies Ξ»n(Rnβˆ–Ei)=0\lambda_{n}(\mathbb{R}^{n}\setminus E_{i})=0. Put E=β‹‚i=1nEiE=\bigcap_{i=1}^{n}E_{i}; then Rnβˆ–E=⋃i=1n(Rnβˆ–Ei)\mathbb{R}^{n}\setminus E=\bigcup_{i=1}^{n}(\mathbb{R}^{n}\setminus E_{i}) is Borel, since a Οƒ\sigma-algebra contains the union of finitely many of its members, and it is null by claim 5 of Elementary Properties of Lebesgue Outer Measure on Rn\mathbb{R}^n.

Since ff is continuous, Borel Structure of the Set Where a Partial Derivative Exists applies for each index ii; write Ej,k(i)E^{(i)}_{j,k} for the set denoted Ej,kE_{j,k} there when it is applied with that index.

Write Q\mathbb{Q} for the rational numbers. For k∈Nk\in\mathbb{N}, j=(j1,…,jn)\mathbf{j}=(j_{1},\dots,j_{n}) with every ji∈Nj_{i}\in\mathbb{N}, and q=(q1,…,qn)\mathbf{q}=(q_{1},\dots,q_{n}) with every qi∈Qq_{i}\in\mathbb{Q}, put

G(k,j,q)={ y∈Eβ€…β€Š:β€…β€Šy∈Eji,k(i)Β andΒ βˆ£βˆ‚if(y)βˆ’qiβˆ£β‰€1kΒ Β forΒ everyΒ iΒ withΒ 1≀i≀n }.G(k,\mathbf{j},\mathbf{q})=\Bigl\{\,y\in E\;:\;y\in E^{(i)}_{j_{i},k}\ \text{and}\ |\partial_{i}f(y)-q_{i}|\le\tfrac{1}{k}\ \text{ for every }i\text{ with }1\le i\le n\,\Bigr\}.

By The Integers and the Rational Numbers are Countable and Products and Powers of Countable Sets the set of triples (k,j,q)(k,\mathbf{j},\mathbf{q}) is countable, so these sets form a countable family. For each triple, The Lebesgue Density Theorem in Rn\mathbb{R}^n Β§ae shows that the set of those y∈G(k,j,q)y\in G(k,\mathbf{j},\mathbf{q}) which are not density points of G(k,j,q)G(k,\mathbf{j},\mathbf{q}) is null; let ZZ be the union of these sets over all triples, a null set by claim 5 of Elementary Properties of Lebesgue Outer Measure on Rn\mathbb{R}^n. The set (Rnβˆ–E)βˆͺZ(\mathbb{R}^{n}\setminus E)\cup Z is then null, so by Null Set of a Measure there is N∈B(Rn)N\in\mathcal{B}(\mathbb{R}^{n}) with (Rnβˆ–E)βˆͺZβŠ†N(\mathbb{R}^{n}\setminus E)\cup Z\subseteq N and Ξ»n(N)=0\lambda_{n}(N)=0.

Fix x∈Rnβˆ–Nx\in\mathbb{R}^{n}\setminus N for the rest of Steps 2 and 3. Then x∈Ex\in E, so all nn partial derivatives of ff exist at xx and Df(x)Df(x) is defined; and xx is a density point of every set G(k,j,q)G(k,\mathbf{j},\mathbf{q}) containing it.

Step 2 (Choice of a good set). Let Ρ∈R\varepsilon\in\mathbb{R} with 0<Ξ΅0<\varepsilon, and put C=n 2nC=n\,2^{n}. Choose

δ1=min⁑{1, Ρ4Ln(C+1)+4},\delta_{1}=\min\Bigl\{1,\ \frac{\varepsilon}{4Ln(C+1)+4}\Bigr\},

a positive real number with Ξ΄1≀1\delta_{1}\le1 and 2Ln(C+1)Ξ΄1≀Ρ/22Ln(C+1)\delta_{1}\le\varepsilon/2, and use The Archimedean Property of the Real Numbers to choose k∈Nk\in\mathbb{N} with 1/k<Ξ΅/(6n)1/k<\varepsilon/(6n), so that 3n/k≀Ρ/23n/k\le\varepsilon/2.

For each ii, since x∈Eix\in E_{i}, Borel Structure of the Set Where a Partial Derivative Exists Β§exhaustion provides ji∈Nj_{i}\in\mathbb{N} with x∈Eji,k(i)x\in E^{(i)}_{j_{i},k}, and The Rational Numbers are Dense in the Real Numbers provides qi∈Qq_{i}\in\mathbb{Q} with βˆ£βˆ‚if(x)βˆ’qiβˆ£β‰€1/k|\partial_{i}f(x)-q_{i}|\le1/k. Write G=G(k,j,q)G=G(k,\mathbf{j},\mathbf{q}); then x∈Gx\in G, so xx is a density point of GG.

We record the key property of GG: for every z∈Gz\in G, every ii with 1≀i≀n1\le i\le n and every s∈Rs\in\mathbb{R} with 0<∣sβˆ£β‰€1/ji0<|s|\le1/j_{i},

∣f(z+sei)βˆ’f(z)βˆ’sβ€‰βˆ‚if(x)βˆ£β‰€3∣s∣k.(βˆ—)\bigl|f(z+se_{i})-f(z)-s\,\partial_{i}f(x)\bigr|\le\frac{3|s|}{k}. \tag{$\ast$}

Indeed z∈Ei∩Eji,k(i)z\in E_{i}\cap E^{(i)}_{j_{i},k}, so Borel Structure of the Set Where a Partial Derivative Exists Β§uniform gives βˆ£Ξ”si(z)βˆ’βˆ‚if(z)βˆ£β‰€1/k|\Delta^{i}_{s}(z)-\partial_{i}f(z)|\le1/k, while βˆ£βˆ‚if(z)βˆ’qiβˆ£β‰€1/k|\partial_{i}f(z)-q_{i}|\le1/k and βˆ£βˆ‚if(x)βˆ’qiβˆ£β‰€1/k|\partial_{i}f(x)-q_{i}|\le1/k give βˆ£βˆ‚if(z)βˆ’βˆ‚if(x)βˆ£β‰€2/k|\partial_{i}f(z)-\partial_{i}f(x)|\le2/k by claim 5 of Properties of the Absolute Value in an Ordered Field; adding and multiplying by ∣s∣|s| gives (βˆ—)(\ast).

By Near a Density Point Every Direction Meets the Set Closely Β§nearby, applied to GG, to the density point xx and to Ξ΄1\delta_{1}, there is ρ∈R\rho\in\mathbb{R} with 0<ρ0<\rho such that for every w∈Rnw\in\mathbb{R}^{n} with 0<βˆ₯wβˆ’xβˆ₯<ρ0<\lVert w-x\rVert<\rho there is g∈Gg\in G with βˆ₯gβˆ’wβˆ₯≀δ1βˆ₯wβˆ’xβˆ₯\lVert g-w\rVert\le\delta_{1}\lVert w-x\rVert. Put

Ξ΄=min⁑{1j1,…,1jn, ρ2C+2},\delta=\min\Bigl\{\tfrac{1}{j_{1}},\dots,\tfrac{1}{j_{n}},\ \frac{\rho}{2C+2}\Bigr\},

a positive real number.

Step 3 (The telescoping estimate). Let h∈Rnh\in\mathbb{R}^{n} with βˆ₯hβˆ₯<Ξ΄\lVert h\rVert<\delta. If h=0h=0 the asserted inequality is trivial, so assume hβ‰ 0h\neq0 and put Ξ·=βˆ₯hβˆ₯\eta=\lVert h\rVert, so 0<Ξ·<Ξ΄0<\eta<\delta. Write h=(h1,…,hn)h=(h_{1},\dots,h_{n}), so that h=βˆ‘p=1nhpeph=\sum_{p=1}^{n}h_{p}e_{p} and ∣hpβˆ£β‰€Ξ·|h_{p}|\le\eta for every pp by claim 4 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n.

Define points z0,…,znz_{0},\dots,z_{n} and w1,…,wnw_{1},\dots,w_{n} recursively: z0=xz_{0}=x, and for 1≀p≀n1\le p\le n put wp=zpβˆ’1+hpepw_{p}=z_{p-1}+h_{p}e_{p} and let zp=wpz_{p}=w_{p} if wp=xw_{p}=x, while if wpβ‰ xw_{p}\neq x let zp∈Gz_{p}\in G satisfy βˆ₯zpβˆ’wpβˆ₯≀δ1βˆ₯wpβˆ’xβˆ₯\lVert z_{p}-w_{p}\rVert\le\delta_{1}\lVert w_{p}-x\rVert. In either case

zp∈Gandβˆ₯zpβˆ’wpβˆ₯≀δ1βˆ₯wpβˆ’xβˆ₯.(βˆ—βˆ—)z_{p}\in G\qquad\text{and}\qquad\lVert z_{p}-w_{p}\rVert\le\delta_{1}\lVert w_{p}-x\rVert . \tag{$\ast\ast$}

We must check that this recursion is legitimate, that is, that βˆ₯wpβˆ’xβˆ₯<ρ\lVert w_{p}-x\rVert<\rho whenever wpβ‰ xw_{p}\neq x. Put cp=βˆ₯zpβˆ’xβˆ₯c_{p}=\lVert z_{p}-x\rVert, so c0=0c_{0}=0. For 1≀p≀n1\le p\le n, the triangle inequality (claim 6 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n) and claim 5 of that lemma give

βˆ₯wpβˆ’xβˆ₯≀cpβˆ’1+∣hpβˆ£β€‰βˆ₯epβˆ₯=cpβˆ’1+∣hpβˆ£β‰€cpβˆ’1+Ξ·,\lVert w_{p}-x\rVert\le c_{p-1}+|h_{p}|\,\lVert e_{p}\rVert=c_{p-1}+|h_{p}|\le c_{p-1}+\eta ,

and then cp≀βˆ₯wpβˆ’xβˆ₯+βˆ₯zpβˆ’wpβˆ₯≀(1+Ξ΄1)βˆ₯wpβˆ’xβˆ₯≀2(cpβˆ’1+Ξ·)c_{p}\le\lVert w_{p}-x\rVert+\lVert z_{p}-w_{p}\rVert\le(1+\delta_{1})\lVert w_{p}-x\rVert\le 2(c_{p-1}+\eta), using Ξ΄1≀1\delta_{1}\le1. By induction cpβ‰€Ξ·βˆ‘l=1p2lc_{p}\le\eta\sum_{l=1}^{p}2^{l} for 1≀p≀n1\le p\le n, and βˆ‘l=1p2l=2p+1βˆ’2≀n 2n=C\sum_{l=1}^{p}2^{l}=2^{p+1}-2\le n\,2^{n}=C for 1≀p≀n1\le p\le n; hence cp≀CΞ·c_{p}\le C\eta and

βˆ₯wpβˆ’xβˆ₯≀cpβˆ’1+η≀(C+1)Ξ·(1≀p≀n).\lVert w_{p}-x\rVert\le c_{p-1}+\eta\le(C+1)\eta\qquad(1\le p\le n).

Since Ξ·<δ≀ρ/(2C+2)\eta<\delta\le\rho/(2C+2) we get βˆ₯wpβˆ’xβˆ₯≀(C+1)Ξ·<ρ/2<ρ\lVert w_{p}-x\rVert\le(C+1)\eta<\rho/2<\rho, as required. (Each application of the recursion uses only the already constructed zpβˆ’1z_{p-1}, so the estimates above are available at the moment they are needed.)

From zp=zpβˆ’1+hpep+(zpβˆ’wp)z_{p}=z_{p-1}+h_{p}e_{p}+(z_{p}-w_{p}) and h=βˆ‘phpeph=\sum_{p}h_{p}e_{p} we get zn=(x+h)+βˆ‘p=1n(zpβˆ’wp)z_{n}=(x+h)+\sum_{p=1}^{n}(z_{p}-w_{p}), so by the triangle inequality and (βˆ—βˆ—)(\ast\ast),

βˆ₯x+hβˆ’znβˆ₯β‰€βˆ‘p=1nβˆ₯zpβˆ’wpβˆ₯≀δ1βˆ‘p=1nβˆ₯wpβˆ’xβˆ₯≀δ1 n (C+1) η.\lVert x+h-z_{n}\rVert\le\sum_{p=1}^{n}\lVert z_{p}-w_{p}\rVert\le\delta_{1}\sum_{p=1}^{n}\lVert w_{p}-x\rVert\le\delta_{1}\,n\,(C+1)\,\eta .

Moreover

f(x+h)βˆ’f(x)=βˆ‘p=1n(f(wp)βˆ’f(zpβˆ’1))+βˆ‘p=1n(f(zp)βˆ’f(wp))+(f(x+h)βˆ’f(zn)),f(x+h)-f(x)=\sum_{p=1}^{n}\bigl(f(w_{p})-f(z_{p-1})\bigr)+\sum_{p=1}^{n}\bigl(f(z_{p})-f(w_{p})\bigr)+\bigl(f(x+h)-f(z_{n})\bigr),

because the first two sums together telescope to f(zn)βˆ’f(z0)=f(zn)βˆ’f(x)f(z_{n})-f(z_{0})=f(z_{n})-f(x).

We estimate the three parts. If hpβ‰ 0h_{p}\neq0 then 0<∣hpβˆ£β‰€Ξ·<δ≀1/jp0<|h_{p}|\le\eta<\delta\le1/j_{p}, so (βˆ—)(\ast) applied with z=zpβˆ’1∈Gz=z_{p-1}\in G, the index pp and s=hps=h_{p} gives ∣f(wp)βˆ’f(zpβˆ’1)βˆ’hpβˆ‚pf(x)βˆ£β‰€3∣hp∣/k≀3Ξ·/k|f(w_{p})-f(z_{p-1})-h_{p}\partial_{p}f(x)|\le3|h_{p}|/k\le3\eta/k; if hp=0h_{p}=0 then wp=zpβˆ’1w_{p}=z_{p-1} and both f(wp)βˆ’f(zpβˆ’1)f(w_{p})-f(z_{p-1}) and hpβˆ‚pf(x)h_{p}\partial_{p}f(x) are 00. Since βˆ‘p=1nhpβˆ‚pf(x)=Df(x)β‹…h\sum_{p=1}^{n}h_{p}\partial_{p}f(x)=Df(x)\cdot h by Difference, Dot Product, and Orthogonality in Rn\mathbb{R}^n, summing gives

βˆ£βˆ‘p=1n(f(wp)βˆ’f(zpβˆ’1))βˆ’Df(x)β‹…hβˆ£β‰€3nΞ·k.\Bigl|\sum_{p=1}^{n}\bigl(f(w_{p})-f(z_{p-1})\bigr)-Df(x)\cdot h\Bigr|\le\frac{3n\eta}{k}.

The Lipschitz hypothesis and (βˆ—βˆ—)(\ast\ast) give

βˆ‘p=1n∣f(zp)βˆ’f(wp)βˆ£β‰€Lβˆ‘p=1nβˆ₯zpβˆ’wpβˆ₯≀L δ1 n (C+1) η,\sum_{p=1}^{n}\bigl|f(z_{p})-f(w_{p})\bigr|\le L\sum_{p=1}^{n}\lVert z_{p}-w_{p}\rVert\le L\,\delta_{1}\,n\,(C+1)\,\eta ,

and likewise ∣f(x+h)βˆ’f(zn)βˆ£β‰€Lβˆ₯x+hβˆ’znβˆ₯≀L δ1 n (C+1) η|f(x+h)-f(z_{n})|\le L\lVert x+h-z_{n}\rVert\le L\,\delta_{1}\,n\,(C+1)\,\eta. Adding the three estimates,

∣f(x+h)βˆ’f(x)βˆ’Df(x)β‹…hβˆ£β‰€3nk η+2L δ1 n (C+1) η≀Ρ2Ξ·+Ξ΅2Ξ·=Ρ βˆ₯hβˆ₯,\bigl|f(x+h)-f(x)-Df(x)\cdot h\bigr|\le\frac{3n}{k}\,\eta+2L\,\delta_{1}\,n\,(C+1)\,\eta\le\frac{\varepsilon}{2}\eta+\frac{\varepsilon}{2}\eta=\varepsilon\,\lVert h\rVert ,

by the choices of kk and Ξ΄1\delta_{1}. This proves claim 1.

Step 4 (Claims 2 and 3). Let xx be as in claim 1 and let AA be the matrix with one row and nn columns given by A1i=βˆ‚if(x)A_{1i}=\partial_{i}f(x). By Matrix-Vector Product the product AhAh is the point of R1\mathbb{R}^{1} whose single coordinate is βˆ‘i=1nA1ihi=Df(x)β‹…h\sum_{i=1}^{n}A_{1i}h_{i}=Df(x)\cdot h, and by claim 1 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n the Euclidean norm of a point of R1\mathbb{R}^{1} 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 βˆ₯f(x+h)βˆ’f(x)βˆ’Ahβˆ₯≀Ρβˆ₯hβˆ₯\lVert f(x+h)-f(x)-Ah\rVert\le\varepsilon\lVert h\rVert for βˆ₯hβˆ₯<Ξ΄\lVert h\rVert<\delta, and since x+h∈Rnx+h\in\mathbb{R}^{n} always, Differentiability at a Point for Maps Between Euclidean Spaces shows that ff is differentiable at xx with derivative matrix AA. Claim 2 of A Derivative Matrix is the Jacobian Matrix, and is Unique identifies AA with the Jacobian matrix. This is claim 2.

For claim 3, if Df(x)=0Df(x)=0 then βˆ₯Df(x)βˆ₯=0≀L\lVert Df(x)\rVert=0\le L by claim 3 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n. Otherwise 0<βˆ₯Df(x)βˆ₯0<\lVert Df(x)\rVert by that same claim. Let Ξ΅>0\varepsilon>0 and let Ξ΄\delta be as in claim 1. Choose t∈Rt\in\mathbb{R} with 0<t0<t and tβˆ₯Df(x)βˆ₯<Ξ΄t\lVert Df(x)\rVert<\delta, and put h=t Df(x)h=t\,Df(x), so βˆ₯hβˆ₯=tβˆ₯Df(x)βˆ₯<Ξ΄\lVert h\rVert=t\lVert Df(x)\rVert<\delta by claim 5 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n and Df(x)β‹…h=tβˆ₯Df(x)βˆ₯2Df(x)\cdot h=t\lVert Df(x)\rVert^{2} by claim 1 of that lemma. Then, by claim 1 of the present theorem and the Lipschitz hypothesis,

tβˆ₯Df(x)βˆ₯2β‰€βˆ£f(x+h)βˆ’f(x)∣+Ξ΅βˆ₯hβˆ₯≀Lβˆ₯hβˆ₯+Ξ΅βˆ₯hβˆ₯=(L+Ξ΅) t βˆ₯Df(x)βˆ₯.t\lVert Df(x)\rVert^{2}\le\bigl|f(x+h)-f(x)\bigr|+\varepsilon\lVert h\rVert\le L\lVert h\rVert+\varepsilon\lVert h\rVert=(L+\varepsilon)\,t\,\lVert Df(x)\rVert .

Dividing by the positive number tβˆ₯Df(x)βˆ₯t\lVert Df(x)\rVert gives βˆ₯Df(x)βˆ₯≀L+Ξ΅\lVert Df(x)\rVert\le L+\varepsilon, and since Ξ΅>0\varepsilon>0 was arbitrary, βˆ₯Df(x)βˆ₯≀L\lVert Df(x)\rVert\le L.

Step 5 (Claim 4). We first prove the following assertion about real-valued maps, in which VV and gg are fresh letters: if VβŠ†RnV\subseteq\mathbb{R}^{n} is open and g:Vβ†’Rg:V\to\mathbb{R} is locally Lipschitz on VV, then the set of those x∈Vx\in V at which gg is not differentiable is null. To lighten the notation we write UU and TT for VV and gg in the four paragraphs that follow, so that TT is real-valued there; the letters resume their meaning from the statement afterwards. Let C\mathcal{C} be the set of pairs (p,r)(p,r) with p∈Rnp\in\mathbb{R}^{n} having all coordinates rational, r∈Qr\in\mathbb{Q} with 0<r0<r, BΛ‰(p,r)βŠ†U\bar{B}(p,r)\subseteq U, and the restriction of TT to BΛ‰(p,r)\bar{B}(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\mathcal{C} is countable.

We claim U=⋃(p,r)∈CB(p,r)U=\bigcup_{(p,r)\in\mathcal{C}}B(p,r). Each B(p,r)βŠ†BΛ‰(p,r)βŠ†UB(p,r)\subseteq\bar{B}(p,r)\subseteq U. Conversely let x∈Ux\in U; by local Lipschitz continuity there are rx,Lx∈Rr_{x},L_{x}\in\mathbb{R} with 0<rx0<r_{x}, 0≀Lx0\le L_{x}, BΛ‰(x,rx)βŠ†U\bar{B}(x,r_{x})\subseteq U and TT Lipschitz with constant LxL_{x} on BΛ‰(x,rx)\bar{B}(x,r_{x}). By The Rational Numbers are Dense in the Real Numbers choose r∈Qr\in\mathbb{Q} with rx/4<r<rx/2r_{x}/4<r<r_{x}/2, and choose rational numbers p1,…,pnp_{1},\dots,p_{n} with ∣piβˆ’xi∣<rx/(4Οƒn)|p_{i}-x_{i}|<r_{x}/(4\sigma_{n}) for every ii; then p=(p1,…,pn)p=(p_{1},\dots,p_{n}) satisfies βˆ₯pβˆ’xβˆ₯<rx/4\lVert p-x\rVert<r_{x}/4 by the norm inequality recorded at the start. Hence βˆ₯pβˆ’xβˆ₯<r\lVert p-x\rVert<r, so x∈B(p,r)x\in B(p,r); and if y∈BΛ‰(p,r)y\in\bar{B}(p,r) then βˆ₯yβˆ’xβˆ₯≀βˆ₯yβˆ’pβˆ₯+βˆ₯pβˆ’xβˆ₯≀r+rx/4<rx\lVert y-x\rVert\le\lVert y-p\rVert+\lVert p-x\rVert\le r+r_{x}/4<r_{x}, so BΛ‰(p,r)βŠ†BΛ‰(x,rx)βŠ†U\bar{B}(p,r)\subseteq\bar{B}(x,r_{x})\subseteq U and TT is Lipschitz with constant LxL_{x} on BΛ‰(p,r)\bar{B}(p,r). Thus (p,r)∈C(p,r)\in\mathcal{C} and x∈B(p,r)x\in B(p,r).

Fix (p,r)∈C(p,r)\in\mathcal{C} and let Lβ€²L' be a Lipschitz constant for TT on the nonempty set BΛ‰(p,r)\bar{B}(p,r). By McShane Extension of a Real-Valued Lipschitz Function on a Metric Space Β§well-defined, McShane Extension of a Real-Valued Lipschitz Function on a Metric Space Β§extends and McShane Extension of a Real-Valued Lipschitz Function on a Metric Space Β§lipschitz, applied in the metric space (Rn,dE)(\mathbb{R}^{n},d_{E}), there is F:Rnβ†’RF:\mathbb{R}^{n}\to\mathbb{R} which is Lipschitz with constant Lβ€²L' and agrees with TT on BΛ‰(p,r)\bar{B}(p,r). Applying claims 1 and 2 of the present theorem to FF in place of ff and Lβ€²L' in place of LL, there is a Borel set Nβ€²N' with Ξ»n(Nβ€²)=0\lambda_{n}(N')=0 such that FF is differentiable at every point of Rnβˆ–Nβ€²\mathbb{R}^{n}\setminus N'.

Let x∈B(p,r)βˆ–Nβ€²x\in B(p,r)\setminus N' and let Aβ€²A' be the derivative matrix of FF at xx. Since B(p,r)B(p,r) is open there is Ξ΄2>0\delta_{2}>0 with B(x,Ξ΄2)βŠ†B(p,r)B(x,\delta_{2})\subseteq B(p,r). Given Ξ΅>0\varepsilon>0, take the smaller of Ξ΄2\delta_{2} and the radius supplied by the differentiability of FF at xx; for hh with 0<βˆ₯hβˆ₯0<\lVert h\rVert less than that number we have x+h∈B(p,r)βŠ†Ux+h\in B(p,r)\subseteq U, so T(x+h)=F(x+h)T(x+h)=F(x+h) and T(x)=F(x)T(x)=F(x), whence βˆ₯T(x+h)βˆ’T(x)βˆ’Aβ€²hβˆ₯≀Ρβˆ₯hβˆ₯\lVert T(x+h)-T(x)-A'h\rVert\le\varepsilon\lVert h\rVert. Thus TT is differentiable at xx. Consequently the set of those x∈B(p,r)x\in B(p,r) at which TT is not differentiable is contained in Nβ€²N' and hence null, and the set of those x∈Ux\in U at which TT is not differentiable, being contained in the union over the countable set C\mathcal{C} of these sets, is null by claim 5 of Elementary Properties of Lebesgue Outer Measure on Rn\mathbb{R}^n.

This proves the assertion about real-valued maps. Now let UU, mm and T:Uβ†’RmT:U\to\mathbb{R}^{m} again be as in the statement of claim 4, and let T1,…,TmT_{1},\dots,T_{m} be the coordinate functions of TT, as in Jacobian Matrix of a Map Between Euclidean Spaces. By claim 4 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n, ∣Tk(y)βˆ’Tk(z)βˆ£β‰€βˆ₯T(y)βˆ’T(z)βˆ₯|T_{k}(y)-T_{k}(z)|\le\lVert T(y)-T(z)\rVert for all y,zy,z, so each TkT_{k} is locally Lipschitz on UU with the same local constants as TT. By the assertion about real-valued maps just proved, for each kk the set MkM_{k} of points of UU at which TkT_{k} is not differentiable is null; put M=⋃k=1mMkM=\bigcup_{k=1}^{m}M_{k}, a null set by claim 5 of Elementary Properties of Lebesgue Outer Measure on Rn\mathbb{R}^n.

Let x∈Uβˆ–Mx\in U\setminus M and let AA be the matrix with mm rows and nn columns given by Aki=βˆ‚iTk(x)A_{ki}=\partial_{i}T_{k}(x), which is defined by claim 1 of A Derivative Matrix is the Jacobian Matrix, and is Unique applied to each TkT_{k}. Let Ξ΅>0\varepsilon>0. For each kk, differentiability of TkT_{k} at xx together with claim 2 of the present theorem, applied to TkT_{k}, provides Ξ΄k>0\delta_{k}>0 such that ∣Tk(x+h)βˆ’Tk(x)βˆ’βˆ‘i=1nAkihiβˆ£β‰€(Ξ΅/Οƒm)βˆ₯hβˆ₯|T_{k}(x+h)-T_{k}(x)-\sum_{i=1}^{n}A_{ki}h_{i}|\le(\varepsilon/\sigma_{m})\lVert h\rVert whenever βˆ₯hβˆ₯<Ξ΄k\lVert h\rVert<\delta_{k}; let Ξ΄\delta be the smallest of Ξ΄1,…,Ξ΄m\delta_{1},\dots,\delta_{m} and of a radius for which x+h∈Ux+h\in U whenever βˆ₯hβˆ₯<Ξ΄\lVert h\rVert<\delta, which exists because UU is open. For hh with 0<βˆ₯hβˆ₯<Ξ΄0<\lVert h\rVert<\delta put v=T(x+h)βˆ’T(x)βˆ’Ahv=T(x+h)-T(x)-Ah; by Matrix-Vector Product its kkth coordinate is Tk(x+h)βˆ’Tk(x)βˆ’βˆ‘iAkihiT_{k}(x+h)-T_{k}(x)-\sum_{i}A_{ki}h_{i}, so ∣vkβˆ£β‰€(Ξ΅/Οƒm)βˆ₯hβˆ₯|v_{k}|\le(\varepsilon/\sigma_{m})\lVert h\rVert for every kk and therefore βˆ₯vβˆ₯≀σmmax⁑k∣vkβˆ£β‰€Ξ΅βˆ₯hβˆ₯\lVert v\rVert\le\sigma_{m}\max_{k}|v_{k}|\le\varepsilon\lVert h\rVert by the norm inequality recorded at the start. Hence TT is differentiable at xx with derivative matrix AA, and the set of points of UU at which TT is not differentiable is contained in MM and so is null.

Finally, at any x∈Ux\in U at which TT is differentiable, claim 1 of A Derivative Matrix is the Jacobian Matrix, and is Unique says that the entry in row kk and column ii of the derivative matrix is βˆ‚iTk(x)\partial_{i}T_{k}(x). This completes claim 4.

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