Smooth maps between Euclidean spaces

Question

There is a question that has bothered me for quite a long time. When we define smoothness of a function(or a map from one space to another), we define it as "has continuous partial derivatives of all orders". However this is not obvious to me. I can't see where the motivation comes from because I would think that the existence of first order derivative would be enough(because then the "graph" of the map would look smooth.). What pathological behaviour would the functions have if they don't have partial derivatives of all orders?

Thanks for everyone's time!

Cheers, Evariste

Answer 1

The meaning of "smooth" depends on context. I have seen it used for:

1. infinitely differentiable,
2. once continuously differentiable,
3. having "as many continuous derivatives as is needed in the proofs".

One of reasons to want second order derivatives is to define curvature. And if you are also interested in how the curvature changes from one point to another, you need 3rd order derivatives.

Answer 2

An "algebraic" motivation could be that the class of smooth function is closed with respect to taking derivatives: if you derive $C^\infty\!$-functions still obtain $C^\infty\!$-functions. So, if being discontinuous or nondifferentiable is pathological for functions, then a pathology of non-$C^\infty\!$-functions is that you derive them a finite number of times and get pathological functions.

Another motivation (somehow related to the previous) is the following.

In differential geometry it is standard (and useful) to define tangent spaces to a smooth (or $C^\infty$) manifold by means of derivations (see http://en.wikipedia.org/wiki/Tangent_space). Among other reasons, this definition turns to be nice since the space of derivations has the same dimension of the manifold (e.g. you get that the tangent space to a surface has dimension $2$; and you're glad). However, if we deal with $C^k\!$-manifolds ($k < \infty$), the spaces of derivations are always infinite-dimensional.