 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/23f/c/solution

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

The relation

1/ q = 1/ p - α / n > 0

implies

α p < n

. Fix

r > 0

and split the Riesz potential into the regions

∣ x - y ∣ < r

and

∣ x - y ∣ \geq r

For the near part, decompose the ball into the dyadic annuli

2^{- k - 1} r \leq ∣ x - y ∣ < 2^{- k} r, k \geq 0.

\int_{∣ x - y ∣ < r} \frac{∣ f ( y ) ∣}{∣ x - y ∣ ^{n - α}} d y \leq C_{n, α} r^{α} M f (x) \sum_{k = 0}^{\infty} 2^{- k α} \leq C r^{α} M f (x) .

For the far part, Hölder's inequality gives

\int_{∣ x - y ∣ \geq r} \frac{∣ f ( y ) ∣}{∣ x - y ∣ ^{n - α}} d y \leq ∥ f ∥_{p} (\int_{∣ z ∣ \geq r} ∣ z ∣^{- (n - α) p^{'}} d z)^{1/ p^{'}} \leq C ∥ f ∥_{p} r^{α - n / p},

where convergence follows from

α p < n

M f (x) > 0

, choose

r^{n / p} = \frac{∥ f ∥ _{p}}{M f ( x )} .

The two bounds then have the same size. If

M f (x) = 0

, the result is immediate. Consequently the Hedberg inequality is

I_{α} ∣ f ∣ (x) \leq C_{n, p, α} ∥ f ∥_{p}^{α p / n} (M f (x))^{1 - α p / n} .

Here the dimensional constant is required for the unnormalized kernel and Lebesgue measure used in the question.

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