Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 105 / 2 / a / i

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- Solution i

Solution

In polar coordinates, set

u (x, y) = (lo g \frac{1}{x ^{2} + y ^{2}})^{1/4}

away from the origin, assigning any value at the origin. This is unbounded as

r = x^{2} + y^{2} ↓ 0

. It belongs to

L^{2} (U)

because

\int_{0}^{1/2} r (lo g \frac{1}{r})^{1/2}, d r < \infty.

Moreover

∣\nabla u ∣^{2} = \frac{1}{16 r ^{2}} (lo g \frac{1}{r})^{- 3/2},

and hence

\int_{U} ∣\nabla u ∣^{2} = \frac{π}{8} \int_{0}^{1/2} \frac{d r}{r ( l o g ( 1/ r ) ) ^{3/2}} < \infty.

Thus

u \in H^{1} (U)

but

u \in / L^{\infty} (U)

, exhibiting the failure of first-order Sobolev embedding into Linfinity in two dimensions.

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