Some stuffs to note

I got really sick yesterday. I'm glad that the midterm wasn't so bad.

Anyways, somethings to note.
Question: If r_{1} = R + a r, what is the relationship between \nabla_{r_{1}} and
\nabla_{R}, \nabla_{r}?

It is important to know that \nabla_{r_{1}} is a differentiation along r_{1} direction, i.e. how much a function f(x_{1}, x_{2}, \cdots, x_{n}) changes as we move the coordinates (x_{1}, cdots , x_{n}) in r_{1} direction. So, let's define a function which moves in r_{1} direction.
g(t) = x + tr_{1}, where x = (x_{1}, cdots , x_{n}) or a point at which we differentiate the function. Then the original function becomes f(g_{t}).
Since changing t corresponds to changing the coodinates in r_{1} direction, it must be that \nabla_{r_{1}} = \frac{d f(g_{t})}{dt}. Now we are in a good shape. Using the chain rule,

\frac{d f(g_{t})}{dt} = \frac{\partial f(x)}{\partial x} \frac{\partial g_{t}}{\partial t}
The derivitive has to be taken at x = g_{t} or rather, g_{0}.
Since we know \frac{\partial g_{t}}{\partial t} = r_{1}, the result is
\nabla _{r_{1}} f = \nabra f r_{1}.
Using this relation, it is easy to answer the question above.

Question: A linear map is continuous if vector spaces are finite-dimensional. However, a linear map does not have to be continuous if vector spaces are infinite dimensional.

Take continuous smooth function as our vector spaces, C^{\infty}[(0,1)]. This is not finite-dimensional. (the basis can be \cos(nt), \sin(nt)) Take dirivitive as our linear map. If we define the matric by inner product \int^{1}_{0} f(t) \bar{g(t)}, then we can make a function getting closer to 0, whereas the derivitive need not get close to zero. But since a linear map takes 0 at f=0, the derivitive does not have to be continuous function.