Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 105 / 1 / a

Solution

Put

Q = Q_{r} (x)

and

v_{Q} = ∣ Q ∣^{- 1} \int_{Q} v

∣ v (y) - v_{Q} ∣ \leq \frac{1}{r ^{n}} \int_{Q} \int_{0}^{1} ∣ D v (y + t (z - y)) ∣ ∣ z - y ∣ d t d z .

For fixed

t

w = y + t (z - y)

t^{n}

. The transformed domain

Q_{t} = y + t (Q - y)

lies in

Q

, because a cube is a convex set. Moreover,

w \in Q_{t}

implies

∣ w - y ∣ \leq n r t

∣ v (y) - v_{Q} ∣ \leq \frac{1}{r ^{n}} \int_{Q} ∣ D v (w) ∣ ∣ w - y ∣ \int_{∣ w - y ∣/ (n r)}^{1} t^{- n - 1} d t d w \leq C_{n} \int_{Q} \frac{∣ D v ( w ) ∣}{∣ w - y ∣ ^{n - 1}} d w .

w = y

p^{'} = p / (p - 1)

now applies:

p > n

implies

(n - 1) p^{'} < n

, and integration in spherical coordinates bounds the kernel by

(\int_{Q} ∣ w - y ∣^{- (n - 1) p^{'}} d w)^{1/ p^{'}} \leq C_{n, p} r^{n / p^{'} - (n - 1)} = C_{n, p} r^{1 - n / p} .

This proves the Morrey inequality on a cube, uniformly even when

y

approaches its boundary:

∣ v_{Q} - v (y) ∣ \leq C_{1} r^{1 - n / p} ∥ D v ∥_{L^{p} (Q)} .

Apply this at both

y

and the center

x

and use the triangle inequality to obtain

∣ v (y) - v (x) ∣ \leq C_{2} r^{1 - n / p} ∥ D v ∥_{L^{p} (Q)}, C_{2} = 2 C_{1} .

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