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What discuss below is a reference form the text "Analysis on Manifolds" by James Munkres.

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First of all we observe the the interior of $U$ is not empty when $U$ is too: for details see here. So by the first definition to we can suppose that the derivative of $\alpha$ exist only in $\text{int}(U)$ but strangerly Munkres state that $D\beta(x)=D\alpha(x)$ for any $x\in U$. So how explain this statement? Could it be a typo? So could someone help me, please?

Antonio Maria Di Mauro
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I'm not sure I properly understand what your problem is, but Munkres nowhere explicitly writes down the equation $\ D\alpha(x)=D\beta(x)\ $, or implies that $\ D\alpha(x)\ $ exists at a boundary point of $\ U\ $ in any sense other than what he is here showing can be taken to be its definition there. The point of his Lemma $23.2$ is that if $\ \beta_1\ $ and $\ \beta_2\ $ are any $\ C^r\ $ extensions of $\ \alpha\ $ to an open subset of $\ \mathbf{R}^k\ $, then $\ D\beta_1(x)= D\beta_2(x)\ $ for all $\ x\in U\ $. Thus, even though the normal definition of $\ D\alpha(x)\ $ cannot be used to define it on the intersection of $\ U\ $ with the boundary of $\ \mathbf{H}^k\ $ , you can nevertheless take its definition to be $\ D\beta(x)\ $ there, where $\ \beta\ $ is any extension of $\ \alpha\ $ to an open subset of $\ \mathbf{R}^k\ $.

lonza leggiera
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  • Munkres states that if $\beta_1$ and $\beta_2$ are $C^r$ extensions of $\alpha$ to two open subset $U_1$ and $U_2$ in $\Bbb R^k$ then substantially $D\beta_1(x)=D\beta_2(x)$ for $x\in U$ and not in $H^k$ where $U$ is the domain of $\alpha$. – Antonio Maria Di Mauro Sep 19 '20 at 06:18
  • Thank you for picking up the error.. The fact that $\ \alpha\ $ was not defined on the whole of $\ \mathbf{H}^k\ $ had slipped my mind. I have now corrected my answer. – lonza leggiera Sep 19 '20 at 08:27
  • So actually using Munkres definition it has not sense to calculate the derivative of $\alpha$ at the boundary points of $U$, right?. Anyway in this case it seems possible to calculate the limit $$\lim_{h\rightarrow 0^+}\frac{\alpha(x_0+he_j)-\alpha(x_0)}{h}$$ for any $j=1,...,n$ and for any $x_0\in\text{Bd}(U)$ so in this particular case we can think to extend the definition of derivative of $\alpha$. What do you think about? – Antonio Maria Di Mauro Sep 19 '20 at 09:32
  • I don't understand what you're trying to say in your first sentence, but it's not necessarily true that you can calculate that limit at any point of $\ \text{Bd}{\mathbf{R}^k}(U)\ $. You can, however, certainly do it at those points of $\ \text{Bd}{\mathbf{R}^k}(U)\ $ which do not also lie in $\ \text{Bd}_{\mathbf{H}^k}(U)\ $—that is, at those points of $\ U\ $ lying in the boundary of $\ \mathbf{H}^k $. – lonza leggiera Sep 20 '20 at 02:45
  • I wanted say what to follow. So $\text{Bd}{\pmb{R}^k}(U)={x\in U: x_k=0}$ and fortunately if $x_0\in\text{Bd}{\pmb{R}^k}(U)$ then $0$ is a limit point of the set $H_j:={h\in\pmb R:(x_0+he_j)\in U}$ for any $j=1,...,n$ so that we can calculate the quantity $$\lim_{h\rightarrow 0}\frac{\alpha(x_0+he_j)-\alpha(x_0)}h$$ for any $j=1,..,k$ but in the case $j=k$ we have necessarly $h\rightarrow 0^+$ and when the latter quantity exists we can call it the derivative of $\alpha$ respect $x_j$ extendig in this particular case the definition of derivative that Munkres gave above, that's all. – Antonio Maria Di Mauro Sep 20 '20 at 07:41
  • I only paraphrased what you said in your answer adding some topological observations. Do you see any mistake in what I said? – Antonio Maria Di Mauro Sep 20 '20 at 07:46
  • My answer had an error which I noticed and then corrected after I had read your paraphrase. But, yes, I think I now understand what you meant, and, if so, I don't any mistakes other than my own. – lonza leggiera Sep 20 '20 at 09:21
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This, "First of all we observe the the interior of U is not empty when U is too". That is not true, the singleton set U= {x} is non-empty but its interior is empty.

user247327
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  • In our case $U$ is open and instead the singletons are not open. – Antonio Maria Di Mauro Sep 18 '20 at 21:41
  • Oh, I missed that! But in that case, your statement is trivial! The interior of an open set is that open set. – user247327 Sep 18 '20 at 22:55
  • Sorry, but my statement is not trivial. Munkres state that $U$ is open in $H^k$ but is not open in $\Bbb R^k$. So if $U$ is open in $H^k$ then there exist an open set $V$ in $\Bbb R^k$ such that $U=V\cap H^k$ so that the Munkres statement is equivalent to state that $\text{int}(U)=\text{int}(V)\cap\text{int}(H^k)=V\cap\text{int}(H^k)$ is not empty that is $U\subseteq\text{cl}\big(\text{int}(H^k)\big)$ and this it is clearly not trivia I think. – Antonio Maria Di Mauro Sep 19 '20 at 06:25
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    I think the ambiguity of the term "interior of $\ U\ $" has caused some communication problems here. The interior of $\ U\ $ in $\ \mathbf{H}^k\ $ is $\ U\ $ just as user247327 says. However, I presume by "the interior of $\ U\ $" in your statement you mean the interior of $\ U\ $ in $\ \mathbf{R}^k\ $, which, of course is not the same thing (as you have explained). – lonza leggiera Sep 20 '20 at 02:19