Kenneth Kuttler

26.6. RADEMACHER’S THEOREM 961

If x ∈Ω, then for all w ∈Ω,

h(w)+K |x−w| ≥ h(x)

by 26.6.20. This shows h(x) ≤ h(x). But also you could take w = x in 26.6.21 whichyields h(x)≤ h(x). Therefore h(x) = h(x) if x ∈Ω.

Now suppose x,y ∈ Rn and consider∣∣h(x)−h(y)

∣∣. Without loss of generality assumeh(x)≥ h(y) . (If not, repeat the following argument with x and y interchanged.) Pick w∈Ω

such thath(w)+K |y−w|− ε < h(y).

Then ∣∣h(x)−h(y)∣∣= h(x)−h(y)≤ h(w)+K |x−w|−

[h(w)+K |y−w|− ε]≤ K |x−y|+ ε.

Since ε is arbitrary, ∣∣h(x)−h(y)∣∣≤ K |x−y|

and this proves the theorem.With this theorem, here is the main result called Rademacher’s theorem.

Theorem 26.6.7 Let h : Ω→Rm be Lipschitz on Ω where Ω is some nonempty measurableset in Rn. Then Dh(x) exists for a.e. x ∈Ω. If Ω = Rn, then for each ei,

limh→0

h(·+hei)−h(·)h

= h,i weak ∗ in L∞ (Rn)

and whenever φ ε is a mollifier,

(h∗φ ε),i→ h,i in Lp (Rn;Rm) .

Proof: The last two claims follow from the above argument applied to the componentsof h. By Theorem 26.6.6 the function can be extended to a Lipschitz function defined on allof Rn, still denoted as h. Let Ωr ≡Ω∩B(0,r) . Now let ψ ∈C∞

c (B(0,2r)) such that ψ = 1on B

(0, 3

2 r). Then ψh is Lipschitz on Rn and vanishes off a bounded set. It follows from

Lemma 26.6.5 applied to the components of h that this function has a derivative off a setof measure zero Nr. If x ∈Ωr \Nr it follows since ψ = 1 near x that Dh(x) exists. LettingN = ∪∞

r=1Nr, it follows that if x ∈Ω\N, then Dh(x) exists. This proves the theorem.For u Lipschitz as described above, the limit of the difference quotient u,i is called the

weak partial derivative of u. For p > n and an assertion that the difference quotients arebounded in Lp everything done above would work out the same way and one can thereforegeneralize parts of the above theorem. The extension is problematic but one can give thefollowing results with essentially the same proof as the above.

Lemma 26.6.8 Let u ∈ Lp (Rn). There exists u,i ∈ Lp (Rn) such that

limh→0

u(·+hei)−u(·)h

= u,i weakly in Lp (Rn)

if and only if the difference quotients u(·+hei)−u(·)h are bounded in Lp (Rn) for all nonzero h.

26.6. RADEMACHER’S THEOREM 961If x € Q, then for all w € Q,h(w)+K|x—w| > A(x)by 26.6.20. This shows h(x) < h(x). But also you could take w =x in 26.6.21 whichyields h(x) < h(x). Therefore h(x) = h(x) ifx € Q.Now suppose x, y € IR” and consider lh (x) —h(y) |. Without loss of generality assumeh(x) >h(y). (If not, repeat the following argument with x and y interchanged.) Pick w € Qsuch that 7h(w)+Kl|y—w|—e€ <h(y).Then _ _ _ _|i (x) —A(y)| = h(x) -h(y) Sh(w) +K|x—w|—[h(w) + Kly—w|—€] <K|x—y|+e.Since € is arbitrary,(x) —h(y)| <K|x—y|and this proves the theorem.With this theorem, here is the main result called Rademacher’s theorem.Theorem 26.6.7 Leth: Q— R” be Lipschitz on Q where Q is some nonempty measurableset in R". Then Dh (x) exists for a.e. x € Q. If Q = R", then for each e;,lim h(-+he;) —h(-)lim h =h, weak * in L® (R")and whenever 6, is a mollifier,(hx be); —h, in L? (R";R”).Proof: The last two claims follow from the above argument applied to the componentsof h. By Theorem 26.6.6 the function can be extended to a Lipschitz function defined on allof R”, still denoted as h. Let Q, = QNB(0,r) . Now let y € Ce (B(0,2r)) such that y = |on B (0,37) . Then wh is Lipschitz on R” and vanishes off a bounded set. It follows fromLemma 26.6.5 applied to the components of h that this function has a derivative off a setof measure zero N,. If x € Q;\ N, it follows since y = | near x that Dh (x) exists. LettingN =U?_,N,, it follows that if x € Q\ N, then Dh (x) exists. This proves the theorem.For u Lipschitz as described above, the limit of the difference quotient wu; is called theweak partial derivative of u. For p > n and an assertion that the difference quotients arebounded in L’? everything done above would work out the same way and one can thereforegeneralize parts of the above theorem. The extension is problematic but one can give thefollowing results with essentially the same proof as the above.Lemma 26.6.8 Let u € L? (IR"). There exists u; € L? (IR") such thattim Wot hes) = uC)= uj in LP (R"lim A uj; weakly in L? (R")if and only if the difference quotients whe) uC) are bounded in L? (IR") for all nonzero h.