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 Incoming links: Fourier proof of Hölder regularity from a Sobolev norm

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 23F / c / Solution Created 2026-09-24 Updated 2026-09-25

We prove the estimate first for

u

in the Schwartz space. By the Fourier inversion theorem and the Cauchy-Schwarz inequality,

∣ u (x) ∣ \leq C \int_{R^{n}} ∣ u (ξ) ∣ d ξ \leq C (\int_{R^{n}} (1 + ∣ ξ ∣^{2})^{- n} d ξ)^{1/2} ∥ u ∥_{H^{n}} .

The integral is finite because

n > n /2

.

For

0 < α \leq 1

, the elementary bound

∣ e^{i t} - 1∣ \leq C_{α} ∣ t ∣^{α}

gives

∣ u (x) - u (y) ∣ \leq C ∣ x - y ∣^{α} \int_{R^{n}} ∣ ξ ∣^{α} ∣ u (ξ) ∣ d ξ \leq C ∣ x - y ∣^{α} (\int_{R^{n}} ∣ ξ ∣^{2 α} (1 + ∣ ξ ∣^{2})^{- n} d ξ)^{1/2} ∥ u ∥_{H^{n}} .

The last integral is finite near zero for every

α > 0

and at infinity exactly when

2 α < n

. Choose, for example,

α = 1/4

, which works for every

n \geq 1

. We obtain

∥ u ∥_{C^{0, 1/4}} \leq C_{n} ∥ u ∥_{H^{n}} .

Density of the Schwartz space in

H^{n}

extends the estimate and supplies a unique Hölder-continuous representative. Therefore

H^{n} (R^{n}) ↪ C^{0, 1/4} (R^{n})

continuously. This is the Fourier proof of Hölder regularity from a Sobolev norm.

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