= Solution
Choose a standard <mollifier> $\rho\in C_0^\infty(\mathbb R^n)$ with $\rho\geq0$, $\int\rho=1$ and support in the unit ball, and set $\rho_\varepsilon(z)=\varepsilon^{-n}\rho(z/\varepsilon)$. Define the <convolution>
$$
\boxed{u_\varepsilon(x)=(\rho_\varepsilon*u)(x)=\int_{\mathbb R^n}\rho_\varepsilon(x-z)u(z)\,dz.}
$$
The smoothing properties of a <mollifier> give $u_\varepsilon\in C^\infty$ and $\partial_ju_\varepsilon=\rho_\varepsilon*D_ju$. Apply the <chain rule> to $t\mapsto u_\varepsilon(y+t(x-y))$ and integrate its derivative by the <fundamental theorem of calculus>:
$$
\boxed{u_\varepsilon(x)-u_\varepsilon(y)=\int_0^1\nabla u_\varepsilon(y+t(x-y))\cdot(x-y)\,dt.}
$$
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