Let $a \in \partial \mathbb{H}^n$ and let $w \in T_a\mathbb{R}^n$. Show that $v \in T_a\mathbb{H}^n$ given by $v(f) = w(\tilde{f})$ is well-defined. 
Let $\mathbb{H}^n$ denote closed Euclidean half-space. Let $a \in \partial \mathbb{H}^n$ and let $w \in T_a\mathbb{R}^n$. Show that $v \in T_a\mathbb{H}^n$ given by $v(f) = w(\tilde{f})$ where $\tilde{f}$ is any smooth extension of $f \in C^{\infty}(\mathbb{H}^n)$ to all of $\mathbb{R}^n$ is well-defined.

Here are my thoughts. Let $\tilde{f}$ and $\tilde{g}$ be any two smooth extensions of $f : \mathbb{H}^n \to \mathbb{R}$ to all of $\mathbb{R}^n$. If I can show that in some open set $U$ in $\mathbb{R}^n$ which contains $a$ we have that $\tilde{f}|_U = \tilde{g}|_U$, then it would follow that $w(\tilde{f}) = w(\tilde{g})$ and hence that $v$ would be well-defined. 
But the problem is this, the way smooth extensions work, we can only guarantee that $\tilde{f}|_{\mathbb{H}^n} = \tilde{g}|_{\mathbb{H}^n} = f$ and any neighbourhood of $a$ in $\mathbb{H}^n$ is always going to be of the form $W \cap \mathbb{H}^n$ where $W$ is an open set in $\mathbb{R}^n$, so there's no way I can find an open set $U$ in $\mathbb{R}^n$ such that $U \subseteq W \cap \mathbb{H}^n$. Hence it doesn't seem possible to guarantee that there exists some open set $U$ in $\mathbb{R}^n$ which contains $a$ for which $\tilde{f}|_U = \tilde{g}|_U$.
With that being said how can I show that $v$ would be well-defined? For what it's worth this definition of $v$ pops up in Lemma 3.11 in John Lee's Introduction to Smooth Manifolds and I'm just trying to verify that it is well-defined here.
 A: By using local coordinates, 
$$
vf:=w^i \frac{\partial \tilde{f}}{\partial x^i}(a).
$$
So we need to check that for every extension $\tilde{f}_1$ and $\tilde{f}_2$ of $f$ we have
$$
w^i \frac{\partial \tilde{f}_1}{\partial x^i}(a) = w^i \frac{\partial \tilde{f}_2}{\partial x^i}(a).
$$
In fact we only need to check that
$$
\frac{\partial \tilde{f}_1}{\partial x^i}(a) = \frac{\partial \tilde{f}_2}{\partial x^i}(a)
$$
for all $i=1,\dots,n$. For $i\neq n$, by definition we have
$$
\lim_{h\to 0} \frac{\tilde{f}_1(a+he_i) - \tilde{f}_1(a)}{h} = \lim_{h\to 0} \frac{f(a+he_i) - f(a)}{h} = \lim_{h\to 0} \frac{\tilde{f}_2(a+he_i) - \tilde{f}_2(a)}{h},
$$
since $a+he_i \in \mathbb{H}^n$. For the case $i=n$, by definition, the derivate $\partial_n\tilde{f}_j(a)$ of any extension $\tilde{f}_j$ of $f$ is 
$$
\lim_{h\to 0} \frac{\tilde{f}_j(a+he_n) - \tilde{f}_j(a)}{h} = \lim_{|h|\to 0} \frac{\tilde{f}_j(a+|h|e_n) - \tilde{f}_j(a)}{|h|} = \lim_{|h|\to 0} \frac{f(a+|h|e_n) - f(a)}{|h|}.
$$
Where the first equality is due to the continuity of partial derivative. The last expression is the definition of $\partial_n f(a)$. So we are done.
