Definition of smooth functions on arbitrary subsets of $\mathbb{R}^n$ and partial derivatives

Question

Let $A$ be an arbitrary subset of $\mathbb{R}^n$, and let $f:A\to \mathbb{R}$ be a function.

We say that $f$ is smooth if for each point $p$ in $A$ there exists an open subset $U$ of $\mathbb{R}^n$ containing $p$ and a smooth map $g:U\to \mathbb{R}$ such that $f$ and $g$ agrees on $A \cap U$.

Now let $U$ be an open subset of $\mathbb{H}^n$, with $\mathbb{H}^n=\{x\in \mathbb{R}^n:x^n\ge0 \}$

Let $f:U\to \mathbb{R}$ be a smooth function in the above sense. Let $p\in U \cap \partial\mathbb{H}^n$. Thus there are $\tilde U$ open subset of $\mathbb{R}^n$ and a smooth map $g:\tilde U\to \mathbb{R}$ such that $g$ and $f$ agrees on $U \cap \tilde U$.

I want to show that the partial derivatives of $f$ at $p$ are determined by their values in Int$\mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension i.e. of the choice of $g$.

Here is my argument.

$$\lim_{t\to0^+}\frac{f(p+te_i)-f(p)}{t}=\lim_{t\to0^+}\frac{g(p+te_i)-g(p)}{t}\qquad [1]$$ beacuse $f(p)=g(p)$ and for sufficient positive small $t$ we have $p+te_i \in U \cap \tilde U$ and so $f(p+te_i)=g(p+te_i)$.

Now, since $g$ is smooth at $p$ by hypothesis, we have that the right hand side of the above equation exists, and we have

$$\lim_{t\to0^+}\frac{g(p+te_i)-g(p)}{t}=\lim_{t\to0^-}\frac{g(p+te_i)-g(p)}{t}=:\partial_i|_p g$$

So I can define $\partial_i|_p f:=\partial_i|_p g$ and this value is indipendent of the choice of $g$ by $[1]$.

So I have shown the claim in the above yellow box.

Is my argument right? In John M. Lee Textbook Introduction to smooth manifolds, on page 27, he says that By continuity, all partial derivatives of $f$ at points of $U\cap \partial\mathbb{H}^n$ are determined by their values in Int$\mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension.

But it seems I dont'use the continuity of any function, so I'm not convinced of having understood the matter.

Answer 1

Yes, your argument works. The argument I had in mind was this: Given $p\in \partial \mathbb H^n$, there's a sequence of points $q_j\in \text{Int}\ \mathbb H^n$ such that $q_j\to p$. Because $\partial g/\partial x^i$ is continuous, $$ \frac{\partial g}{\partial x^i}(p) = \lim_{j\to \infty} \frac{\partial g}{\partial x^i}(q_j) = \lim_{j\to \infty} \frac{\partial f}{\partial x^i}(q_j), $$ and this last expression manifestly depends only on the original function $f$.

This argument might be a little simpler, and it has the advantage that it applies to any subset of $\mathbb R^n$ that is the closure of its interior, not just to $\mathbb H^n$. But there's nothing wrong with your argument.