4
$\begingroup$

I just had a general question about differential forms.

Background. Let $f = f_k$ and define $f^{k-1}$ on $I^{k-1}$ by $f_{k-1}(x_1, \dots, x_{k-1})= \int_{a_k}^{b_k} f_{k}(x_1, \dots, x_{k-1}, x_k) \ dx_{k}$

Note that $f$ is real continuous on $I^{k}$.

Question. What is the intuitive meaning of this? Rudin says that after $k$ steps we arrive at a number $f_0$ which is the integral of $f$ over $I_k$. Is this similar to the notion of the exterior derivative? Basically if I want to integrate something in $\mathbb{R}^{100}$ I just have to work backwards?

Source. Principles of Mathematical Analysis by Rudin

  • 2
    ... the Stone-Weierstrass theorem. More specifically, let ${\cal A}$ be the algebra of functions on $I^{k}$ generated by all functions on $I^{k}$ of the form $h(x_1,\dots,x_k)=\prod_{i=1}^{k} h_i(x_i)$ where the $h_i$ are continuous complex-valued functions on $I$. The uniform closure of ${\cal A}$ is the algebra of all continuous complex-valued functions on $I^k$ by the Stone-Weierstrass theorem. (If you are unfamiliar with the Stone-Weierstrass theorem, then I suggest you look at chapter 7 of Rudin's *Principles of Mathematical Analysis* where he discusses this result in some detail.)2011-08-27

0 Answers 0