Well-Definedness of Derivatives for Smooth Extensions

Independence of Derivatives from Smooth Extension.

Let $U$ be an open subset of the upper half-space ${\mathbb{H}}^n=\{(x^1,\dots ,x^n)\in {\mathbb{R}}^n\mid x^n\ge 0\}$. Let $f:U\to \mathbb{R}$ be a smooth function. By definition, for any point $a\in U$, there exists an open neighborhood $\tilde{U}\subseteq {\mathbb{R}}^n$ of $a$ and a smooth function $\tilde{f}:\tilde{U}\to \mathbb{R}$ such that $\tilde{f}{|}_{U\cap \tilde{U}}=f$. Such a function $\tilde{f}$ is called a smooth extension of $f$. We prove that the partial derivatives of any smooth extension at a boundary point $a\in U\cap \partial {\mathbb{H}}^n$ are uniquely determined by $f$.

Theorem (Uniqueness of Derivatives at the Boundary).

Let $f:U\to \mathbb{R}$ be a smooth function on an open subset $U\subseteq {\mathbb{H}}^n$, and let $a\in U\cap \partial {\mathbb{H}}^n$. If ${\tilde{f}}_1$ and ${\tilde{f}}_2$ are two smooth extensions of $f$ defined on a neighborhood of $a$ in ${\mathbb{R}}^n$, then their partial derivatives at $a$ are identical. That is, for each $i\in \{1,\dots ,n\}$,

$$\frac{\partial {\tilde{f}}_1}{\partial x^i}(a)=\frac{\partial {\tilde{f}}_2}{\partial x^i}(a).$$

Proof. Let ${\tilde{f}}_1$ and ${\tilde{f}}_2$ be two smooth extensions of $f$ on an open neighborhood $\tilde{U}\subseteq {\mathbb{R}}^n$ of $a$. Define the difference function $h:\tilde{U}\to \mathbb{R}$ by $h={\tilde{f}}_1-{\tilde{f}}_2$. Since ${\tilde{f}}_1$ and ${\tilde{f}}_2$ are smooth, $h$ is also smooth on $\tilde{U}$. Furthermore, for any point $x\in \tilde{U}\cap {\mathbb{H}}^n$, we have $h(x)={\tilde{f}}_1(x)-{\tilde{f}}_2(x)=f(x)-f(x)=0$. Thus, $h$ vanishes identically on $\tilde{U}\cap {\mathbb{H}}^n$.

The proof reduces to showing that all partial derivatives of $h$ at $a$ are zero. This can be shown in two ways.

Argument from Continuity. Since $h$ is smooth, its partial derivatives $\frac{\partial h}{\partial x^i}$ are continuous functions on $\tilde{U}$. On the open set $\tilde{U}\cap \mathrm{Int}({\mathbb{H}}^n)=\{x\in \tilde{U}\mid x^n>0\}$, the function $h$ is identically zero. Consequently, all its partial derivatives vanish on this open set: $\frac{\partial h}{\partial x^i}(x)=0$ for all $x\in \tilde{U}\cap \mathrm{Int}({\mathbb{H}}^n)$. By the continuity of the partial derivatives, their values at the boundary point $a$ must be the limit of their values from the interior.

$$\frac{\partial h}{\partial x^i}(a)=\underset{\begin{array}{c}x\to a\\ x\in \mathrm{Int}({\mathbb{H}}^n)\end{array}}{\lim }\frac{\partial h}{\partial x^i}(x)=\underset{\begin{array}{c}x\to a\\ x^n>0\end{array}}{\lim }0=0.$$

Argument from Difference Quotients. Alternatively, we compute the derivatives at $a$ directly from their definition. Let $a=(a^1,\dots ,a^{n-1},0)$.

1. Tangential Derivatives ($1\le i<n$): The derivative is taken along a direction parallel to the boundary. For sufficiently small $t$, the point $a+t{e}_i$ lies in $\partial {\mathbb{H}}^n$, so $h(a+t{e}_i)=0$ and $h(a)=0$.

   $$\frac{\partial h}{\partial x^i}(a)=\underset{t\to 0}{\lim }\frac{h(a+t{e}_i)-h(a)}{t}=\underset{t\to 0}{\lim }\frac{0-0}{t}=0.$$

2. Normal Derivative ($i=n$): The derivative is taken in the direction normal to the boundary. Since we must remain in the domain of $h$, we can only take the limit from within ${\mathbb{H}}^n$.

   $$\frac{\partial h}{\partial x^n}(a)=\underset{t\to 0^{+}}{\lim }\frac{h(a+t{e}_n)-h(a)}{t}=\underset{t\to 0^{+}}{\lim }\frac{0-0}{t}=0.$$

Both arguments show that $\frac{\partial h}{\partial x^i}(a)=0$ for all $i=1,\dots ,n$. This implies $\frac{\partial {\tilde{f}}_1}{\partial x^i}(a)-\frac{\partial {\tilde{f}}_2}{\partial x^i}(a)=0$, so the derivatives are equal. As a consequence, for any tangent vector $w=\sum w^i\frac{\partial }{\partial x^i}{|}_a\in T_a{\mathbb{R}}^n$, the action $w(\tilde{f})=\sum w^i\frac{\partial \tilde{f}}{\partial x^i}(a)$ is independent of the chosen extension $\tilde{f}$. $\square$