proof of Weierstrass approximation theorem in R^n
Define as follows:
Consider the function defined as follows:
We shall now show that whenever lies in the cubical region. By way that was defined, only two of the terms in the sum defining will differ from zero for any particular value of , and hence
Next, we will use the Weierstrass approximation theorem in dimensions and in one dimesnsion to approximate by a polynomial. Since is continuous and the cubical region is compact, must be bounded on this region. Let be an upper bound for the absolute value of on the cubical region. Using the Weierstrass approximation theorem in one dimension, we conclude that there exists a polynoial such that for all in the region. Using the Weierstrass approximation theorem in dimensions, we conclude that there exist polynomials , such that . Then one has the following inequality:
As a finite sum of products of polynomials, this is a polynomial. From the above inequality, we conclude that , hence .
It is a simple matter of rescaling variables to conclude the Weirestrass approximation theorem for arbitrary parallelopipeds. Any compact subset of can be embedded in some paralleloped and any continuous function on the compact subset can be extended to a continuous function on the parallelopiped. By approximating this extended function, we conclude the Weierstrass approximation theorem for arbitrary compact subsets of .
|Title||proof of Weierstrass approximation theorem in R^n|
|Date of creation||2013-03-22 15:40:03|
|Last modified on||2013-03-22 15:40:03|
|Last modified by||rspuzio (6075)|