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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-105/1/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 105 / 1 / c / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Choose

x

at which

D u (x)

is finite and the

p

-mean version of the Lebesgue differentiation theorem holds:

ε_{x} (r) := (\frac{1}{∣ Q _{r} ( x ) ∣} \int_{Q_{r} (x)} ∣ D u (z) - D u (x) ∣^{p} d z)^{1/ p} ⟶ 0.

These conditions hold almost everywhere. On a cube about

x

, subtract the affine function and put

v (y) = u^{*} (y) - u^{*} (x) - D u (x) \cdot (y - x)

. Its weak derivative is

D v = D u - D u (x)

and

v (x) = 0

. Approximate

u

as in part (b) and subtract the same affine function from the approximants. Uniform convergence and convergence of their weak derivatives extend part (a)'s estimate to

v

∣ v (y) ∣ \leq C r^{1 - n / p} ∥ D u - D u (x) ∥_{L^{p} (Q_{r} (x))} = C r ε_{x} (r) (y \in Q_{r} (x)) .

For

h \neq = 0

, choose

r = 4∣ h ∣

and

y = x + h

. It follows that

\frac{∣ u ^{*} ( x + h ) - u ^{*} ( x ) - D u ( x ) \cdot h ∣}{∣ h ∣} \leq 4 C ε_{x} (4∣ h ∣) ⟶ 0.

This is precisely classical Fréchet derivative differentiability, not merely existence of coordinate derivatives. Therefore the differentiability almost everywhere of supercritical Sobolev functions gives

D_{classical} u^{*} (x) = D_{weak} u (x) for almost every x .

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