We divide the proof in three steps.

Integrand estimate. Let \( x', y' \in \mathbb{R}^{d-1} \). First, we estimate

\[ \frac{|u(x',0) - u(y',0)|^{2}}{|x' - y'|^{d}}. \]

Set \( \xi' = x' - y' \), and \( z = \left( \tfrac{1}{2}(x' + y'), |s'| \right) \).

Then

\begin{align*} |u(x',0) - u(y',0)| &\le |u(z) - u(x',0)| + |u(z) - u(y',0)|\\ &\le \sqrt{2}\, |\xi'| \left( \int_{0}^{1} |\nabla u(x' - t \xi', t |\xi'|)| \, dt + \int_{0}^{1} |\nabla u(y' + t \xi', t |\xi'|)| \, dt \right). \end{align*}

Thus, there exists a numerical constant \( c > 0 \) such that

\begin{equation}\label{eq:integrand_est} \frac{|u(x',0) - u(y',0)|^{2}}{|x' - y'|^{d}} \le c \left[ \left( \int_{0}^{1} \frac{|\nabla u(x' - t \xi', t |\xi'|)|}{|x' - y'|^{(d-2)/2}} \, dt \right)^{2} +\left( \int_{0}^{1} \frac{|\nabla u(y' + t \xi', t |\xi'|)|}{|x' - y'|^{(d-2)/2}} \, dt \right)^{2} \right]. \end{equation}

Employing Minkowski’s inequality. Integrating \eqref{eq:integrand_est} over \( x' \) and \( y' \), identifying the second integral on the right-hand side with the first one, and employing Minkowski’s integral inequality , yields

\begin{equation}\label{eq:h12_est_1} [u(\cdot, 0)]_{H^{1/2}(\mathbb{R}^{d-1})} \le c \int_{0}^{1} \left( \int_{\mathbb{R}^{d-1}} \int_{\mathbb{R}^{d-1}} \frac{|\nabla u(x' + t \xi', t |\xi'|)|^{2}}{|x' - y'|^{d-2}} \, dx' dy' \right)^{1/2} dt, \end{equation}

where \(c>0\) denotes a numerical constant.

Estimate of \eqref{eq:h12_est_1}. We set \( y' = x' + r \omega \), with \( \omega \in \mathbb{S}^{d-2} \) and \( r = |x' - y'| = 2|s'| \). Then

\begin{align*} \int_{\mathbb{R}^{d-1}} \int_{\mathbb{R}^{d-1}} \frac{|\nabla u(x' - t \xi', t |\xi'|)|^{2}}{|x' - y'|^{d-2}} \, dx' dy' &= \int_{\mathbb{S}^{d-2}} \int_{0}^{\infty} \int_{\mathbb{R}^{d-1}} |\nabla u(x' + t r \omega, t r / 2)|^{2} \, dx' dr\, d\omega \\ &= \int_{\mathbb{S}^{d-2}} \int_{0}^{\infty} \int_{\mathbb{R}^{d-1}} |\nabla u(x', t r / 2)|^{2} \, dx' dr \,d\omega\\ &= \frac{C}{t} \int_{\mathbb{R}^{d}_{+}} |\nabla u|^{2} \, dx, \end{align*}

for a constant \( C > 0 \) depending on \( d \). This, together with \eqref{eq:h12_est_1}, implies \eqref{eq:ineq}.