Solution

For

s > \frac{1}{2}

, the Sobolev trace theorem states that the restriction map initially defined on smooth functions extends uniquely to a bounded linear operator

γ : H^{s} (R^{n}) ⟶ H^{s - 1/2} (R^{n - 1}), (γ u) (x^{'}) = u (x^{'}, 0) .

Take first

u

in the Schwartz space. Up to the harmless constant determined by the Fourier transform convention,

γ u (ξ^{'}) = \int_{R} u (ξ^{'}, ξ_{n}) d ξ_{n} .

1 + ∣ ξ^{'} ∣^{2} + ξ_{n}^{2}

, gives

∣ γ u (ξ^{'}) ∣^{2} \leq (\int_{R} (1 + ∣ ξ^{'} ∣^{2} + ξ_{n}^{2})^{s} ∣ u (ξ^{'}, ξ_{n}) ∣^{2} d ξ_{n}) \times (\int_{R} (1 + ∣ ξ^{'} ∣^{2} + ξ_{n}^{2})^{- s} d ξ_{n}) .

After the substitution

ξ_{n} = (1 + ∣ ξ^{'} ∣^{2})^{1/2} t

, the second factor is

C_{s} (1 + ∣ ξ^{'} ∣^{2})^{1/2 - s}, C_{s} = \int_{R} (1 + t^{2})^{- s} d t < \infty,

where finiteness is exactly the condition

s > 1/2

. Multiplying by

(1 + ∣ ξ^{'} ∣^{2})^{s - 1/2}

and integrating in

ξ^{'}

yields

∥ γ u ∥_{H^{s - 1/2} (R^{n - 1})} \leq C_{s}^{1/2} ∥ u ∥_{H^{s} (R^{n})} .

Density of the Schwartz space in the Sobolev space completes the unique bounded extension.

Solved by gpt-5.6-sol high.

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