Partial Derivatives: Why Closed Domains Don't Work

[lmedin02]

Homework Statement

A mapping f from an open subset S of Rⁿ into R^m is called smooth if it has continuous partial derivatives of all orders. However, when the domain S is not open one cannot usually speak of partial derivatives. Why?

Homework Equations

The Attempt at a Solution

In the 1 dimensional case there are not partial derivatives and we can consider the derivative of a function on a closed set by just using the derivative from the left if we are at the left boundary point of the interval. In 2 dimensions I tried creating a counter example, but no luck yet. In the definition of the partial derivative we already assume the domain to be open.

 

[rochfor1]

Can you come up with a natural well-defined analog of a one-sided limit in n-dimensions?

 

[lmedin02]

In n-dimensions the analog of the derivative is the total derivative (i.e., gradient). When we consider the partial derivative or directional derivative in the direction of a unit coordinate vector we use a similar definition to that of the 1 dimensional derivative. In the 1 sided limit in 1 dimension we can approach the a point from 1 direction only. What I have in my mind is that we can approach a point in n-dimensions, for example, along the right hand side of the line that passes through the point where the partial derivative is taken in the direction of the coordinate unit vector.

 

[rochfor1]

Surely you've seen an example of a function of the plane such that the limits along the x and y-axis exist and are different. Which one would you choose?

 

[lmedin02]

yes, f(x,y)=(x+y)/(x-y).