Reading a paper I came across the following estimate: given $u \in C^{1,1}(\Omega)$, where $\Omega \subset \mathbb{R}^n$ is an open, bounded Lipschitz domain, then
$$ |u(x)-u(x_0)- \nabla u(x_0) (x-x_0)| \le \| u \|_{1,+\infty} |x-x_0|^2 \qquad \forall x \in \Omega \ \ \forall x_0 \in \partial \Omega $$
Looking at the LHS, it would be natural to try with a Taylor expansion at $x_0$, truncated at the first order with Lagrange remainder; the problem is that $u$ does not admit classical second derivatives (at least not everywhere). We may instead argue that $\nabla u $ is almost everywhere differentiable and use (in some way) the definition, but the estimates seems to hold "everywhere". Is someone able to give an insight on this one? Thanks in advance.
These things tend to become more clear if we subtract off the linear part. Let $v(x) = u(x)-u(x_0)-\nabla u(x_0)(x-x_0)$. Then $\nabla v$ has the same Lipschitz constant $L$ as $\nabla u$. Since $\nabla v(x_0)=0$, it follows that $|\nabla v|\le Lr$ in the ball $B(x_0, r)$. By the Mean Value Inequality, $$ |v(x)-v(x_0)|\le r \sup_{B(x_0, r)}|\nabla v| \le Lr^2 $$ for $v\in B(x_0, r)$. This is the desired estimate.