proof of Weierstrass approximation theorem
To simplify the notation, assume that the function is defined on the interval . This involves no loss of generality because if is defined on some other interval, one can make a linear change of variable which maps the domain of to .
Let us start by demonstrating a few special cases of the theorem, starting with the case . In this case, we can use the ancient Babylonian method of computing square roots to construct polynomial approximations. Define the polynomials recursively as
It is an obvious consequence of this definition that, if then for all . It is equally obvious that each is a monotonically increasing function on the interval . By subtracting the recursion from itself, cancelling, and factoring, we obtain the relation
From this relation, we conclude that for all and all in . This implies that exists for all in . Taking the limit of both sides of the recursion that defines and simplifying, one sees that . The relation also implies that is also a monotonically increasing function of in the interval for all . Therefore,
Summing over and cancelling, one sees that
whenever . Taking the limit as approaches infinity, one concludes that
Next consider the special case , where . A little algebra shows that
By the case of the approximation theorem already proven, there exists a polynomial such that
The case of piecewise linear functions
A corollary of the result just proven is the Weierstrass appriximation theorem for piecewise linear functions. Any piecewise linear function can be expressed as
for suitable constants . By the result just proven, there exist polynomials such that
By the triangle inequality,
The general proof
Having succeeded in proving all these special cases, we now have the courage to attack the general theorem. In light of the case just proven, it suffices to show that if is continuous on [0,1] then for all there exists a piecewise-linear function such that for all in , . For, if such a function exists, then there also exists a polynomial such that , but then .
Define by the following two conditions: If is an integer between and , . On any interval , is linear.
For every point in the interval , there exists an integer such that lies in the subinterval . Since a linear function is bounded by its values at the endpoints, lies between and . Since , it follows that . Because , it is also true that . Hence, by the triangle inequality, .
|Title||proof of Weierstrass approximation theorem|
|Date of creation||2013-03-22 14:34:58|
|Last modified on||2013-03-22 14:34:58|
|Last modified by||rspuzio (6075)|