Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In the book by Guillemin & Pollack they define a function $f$ from an open subset $U\subset R^n$ to $R^m$ to be smooth if it has continuous partial derivatives of all orders. Then they say "However, when the domain of $f$ is not open, one usually cannot speak of partial derivatives. (Why?)" Can someone explain this a little bit more?

share|improve this question
    
maybe because for the point on the border of $U$ the partial derivatives may not be defined? I don't know though, I'd love a nice answer to this –  Ant Jul 7 at 21:35
3  
Taking a derivative involves taking a limit, and limits should (usually) only be taken over open sets so that you can "approach" the point from any which way possible. –  Santiago Canez Jul 7 at 21:35
    
    
Your title says "if the domain is closed" but the book you quote says "if the domain is not open". That's not the same thing. The whole space $\mathbb R^n$ is closed, but it's also open, so "closed" does not imply "not open". And lots of sets are not open but also not closed. –  Michael Hardy Jul 7 at 22:20
add comment

3 Answers 3

up vote 2 down vote accepted

The definition of the $i$th partial derivative is really as follows: at a point $x = (x_1,\dots,x_n)$, we say $$ \left.\frac{\partial f}{\partial x_i}\right|_{x} = \lim_{t \to 0} \frac{f(x_1,\dots,x_{i-1},x_i+t,x_{i+1},\dots,x_n) - f(x_1,\dots,x_{i-1},x_i,x_{i+1},\dots,x_n)}{t} $$

However, in order for this limit to make sense, $f$ needs to be defined over $x + t(0,\dots,0,1,0,\dots,0)$ for $t$ such that $|t|$ is small enough. That is, $x + t(0,\dots,0,1,0,\dots,0)$ needs to be in $U$ for $t$ close enough to $0$.

One way to guarantee that this definition will make sense for every point $x \in U$ is to say that for each $x$, there is some $t$ so that the "cube" $$ (x_1-t,x_1+t) \times \cdots \times (x_n - t, x_n + t) $$ is contained in $U$. That is, as long as $U$ is open, we can talk about partial derivatives at every point.

share|improve this answer
add comment

The same happens with usual derivatives. If you work on a closed interval, say $[-1,1],$ then the function $f(x)=\sqrt{1-x^2}$ is $C^{\infty}$ in $(-1,1).$ However, it is not $\require{cancel} \cancel{\textrm{derivable}}$ differentiable at $x=-1,x=1.$

So, if you want to have derivatives (partial derivatives in higher dimensions) you have to consider the function defined in an larger open set, say $(-1-\epsilon,1+\epsilon)$ for some $\epsilon>0.$

From the example above you can consider $f(x_1,\cdots,x_n)\sqrt{1-x_1^2-\cdots -x_n^2}.$ This refers only to the function, but can happen to some of the (partial) derivatives at every (some) points of the boundary.

share|improve this answer
add comment

Actually, under some favorable conditions one do it for functions defined on subsets which are not open (and it is even useful).

Let $A$ be a subset of $R^n$ and $f: A\to R$ be a function such that there exists an open subset $B$ of $R^n$ (containing $A$) such that $f$ extends to a differentiable function $F: B\to R$. Then to define partial derivatives of $f$ on $A$ you just compute partial derivatives $D_iF$ and then restrict $D_iF$ back to $A$.

For instance, this appears when you integrate vector fields/differential forms over curves and surfaces (especially when using Green's and Stokes' theorems). Also, when talking about von Neumann boundary value problem, one talks about normal derivative of a function on the boundary of a domain in $R^n$.

One drawback of this definition is that at some points $a\in A$ the partials thus defined $D_if(a)$ might depend on the extension $F$. (For the von Neumann BVP this is not an issue since the boundary is typically assumed to be smooth and, hence, the normal derivative is independent of the extension.) For instance, if $A$ is the $x$-axis in the plane then $D_xf(a)$, $a\in A$, is not well-defined. Same for the subset $A$ which is the cuspidal curve $y^2=x^3$. This is related to a bounty question which was asked at MSE recently (maybe I will find it).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.