I want to prove the following statement:
Let $u\in L^2(\mathbb{R}^n\setminus\bar{\Gamma})$, $\nabla u \in L^2(\mathbb{R}^n\setminus\bar{\Gamma})$.
Define $[u]=\lim\limits_{h\to 0^+}u(x+hn(x))-u(x-hn(x))$, where $n(x)$ is the unit vector normal to $\Gamma$ at $x$.
Then $[u]\in H^{1/2}(\Gamma)$.
I am using the definition $$H^{1/2}(\Gamma)=\left\{u\in L^2(\Gamma)\,:\,\int_\Gamma(1+|y|^2)^{1/2}|\hat{u}(y)|^2dy<\infty\right\}.$$
where $$\hat{u}(y)=\int_{\mathbb{R^n}}[u](x)e^{-2\pi ixy}dy$$ is the Fourier Transform of $u$.
Essentially I want to show that
$$\int_{\Gamma}(1+|y|^2)^{1/2}\int_\Gamma \left(\lim_{h\to 0^+}u(x+hn(x))-u(x-hn(x))\right)e^{-2\pi i x y}dxdy<\infty.$$
The reason that the inner integral is over $\Gamma$ rather than $\mathbb{R}^n$ because $[u]$ is zero wherever $\nabla u$ is defined and by assumption $\nabla u$ is defined almost everywhere on $\mathbb{R}^n\setminus \overline{\Gamma}$
I am stuck because I don't know how to rewrite $$\lim_{h\to 0^+}u(x+hn(x))-u(x-hn(x))$$ which uses the fact that $u\in L^2(\mathbb{R}^n\setminus\overline{\Gamma})$, $\nabla u \in L^2(\mathbb{R}^n\setminus\bar{\Gamma})$.
One way that I could introduce $\nabla u$ is with the mean value theorem: $$u(x+hn(x))-u(x-hn(x))=2h\nabla u(x+h\theta(h)n(x))\cdot n(x),$$ where $\theta(h)\in [-1,1]$.
But $u$ is not differentiable in $\Gamma$, so the gradient is not defined at $h=0$ and for the mean value theorem I need the gradient to exist for $\forall h\in [-1,1]$.
It would be especially nice if you could give me some general tips about proofs involving the function $[u]$ (for example some nice properties of $[u]$)
Thanks, Andrew Murdza