 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/23f/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

(x, y)

∣ u (x, y) ∣^{2} = \int_{- \infty}^{y} \partial_{t} ∣ u (x, t) ∣^{2} d t \leq 2 \int_{R} ∣ u (x, t) ∣ ∣ \partial_{2} u (x, t) ∣ d t \leq 2 A (x),

where

A (x) = \int_{R} ∣ u (x, t) ∣ ∣\nabla u (x, t) ∣ d t .

Applying the same argument in the other coordinate gives

∣ u (x, y) ∣^{2} \leq 2 B (y), B (y) = \int_{R} ∣ u (t, y) ∣ ∣\nabla u (t, y) ∣ d t .

\int_{R^{2}} ∣ u (x, y) ∣^{4} d x d y \leq 4 \int_{R^{2}} A (x) B (y) d x d y = 4 (\int_{R^{2}} ∣ u ∣ ∣\nabla u ∣)^{2} .

(\int_{R^{2}} ∣ u ∣ ∣\nabla u ∣)^{2} \leq (\int_{R^{2}} ∣ u ∣^{2}) (\int_{R^{2}} ∣\nabla u ∣^{2}) .

Thus the required estimate holds with $C = 4$ ; it is the Ladyzhenskaya inequality in two dimensions.

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