Show that $C^\infty(\overline{\Omega}) \subseteq C^{0,1} (\overline{\Omega})$

Question

Let $\Omega$ be a bounded, connected, open domain in $\mathbb{R}^d$ with smooth boundary. Denote by $C^{0,1} (\overline{\Omega})$ the space of continuous functions $u$ on $\overline{\Omega}$ such that

$$ \sup_{\substack{x,y \in \overline{\Omega}: \\ 0 < |x-y| \le 1 } }\frac{|u(x) - u(y)|}{|x-y|} < + \infty.$$

Let $C^\infty(\Omega)$ denote the space of smooth ($C^\infty$) functions $u$ on $\Omega$ such that all derivatives of $u$ have continuous extensions to $\overline{\Omega}$.

I would like to know whether $C^\infty(\overline{\Omega}) \subseteq C^{0,1} (\overline{\Omega})$. I would also like to know whether certain conditions can be relaxed, e.g., do we only need to assume $u \in C^1(\overline{\Omega})$, and must the boundary of $\Omega$ be smooth?

If $\overline{\Omega}$ is simply a closed ball $B$ in $\mathbb{R}^d$, then we immediately get (from the convexity of $B$ and the multidimensional mean value theorem) that $|u(x) - u(y)| \le \| \nabla u \|_{L^\infty(B)} |x - y|$ for all $x,y \in B$. However, it seems trickier to get such an estimate for general $\overline{\Omega}$, and this is where I'm stuck. I expect that $\overline{\Omega}$ will need to be covered by suitable small neighborhoods (this is where $\Omega$ being smooth may need to be used), and then finitely many local Lipschitz estimates will need to be pasted together.

Answer 1

This requires some care in how you define these spaces, but it would hold if we adopt for instance the following definition.

Definition. We define $C^1(\overline{\Omega})$ to be the set of $u \in C^1(\Omega)$ such that for each $x \in \overline\Omega,$ there is a neighborhood $U$ of $x$ and an extension $g \in C^1(U)$ such that $u=g$ on $U \cap \Omega.$

In this case we have $C^1(\overline\Omega) \subset C^{0,1}(\overline\Omega)$ by using the following result.

Proposition. If $u \in C^1(\overline{\Omega}),$ there is a global extension $v \in C^1_c(\Bbb R^n)$ such that $v = u$ on ${\Omega}.$

Proof: By definition and compactness of $\overline\Omega,$ we can find a finite covering $U_1,\dots,U_m$ of $\overline{\Omega}$ and extensions $v_i \in C^1(U_i)$ such that $v_i = u$ on $U_i$ for each $i.$ Then we can take a partition of unity of $\overline\Omega$ subordinate to the cover $\{U_i\};$ this gives $\rho_i \in C^{\infty}_c(U_i)$ such that $0 \leq \rho_i \leq 1$ and $$ \sum_{i=1}^m \rho_i \equiv 1 \text{ on } \overline{\Omega}. $$ Then we have $$ v = \sum_{i=1}^m \rho_i v_i $$ is well-defined, lies in $C^1_c(\Bbb R^n)$ and satisfies $v=u$ on $\Omega$ as required.

Now this is not quite how you defined $C^1(\overline\Omega).$ The two definitions are equivalent however, thanks to the following result.

Proposition 2. Suppose $u \in C^1(\Omega)$ such that $u, \partial_{x_i} \in C^0(\overline\Omega)$ for each $i.$ Then $u \in C^1(\overline\Omega).$

This is a consequence of the more general Whitney extension theorem. I'm not sure if there is a simpler way to prove this, but the above also allows you to consider higher derivatives and unbounded domains.