Singular Integrals. Weierstrass' Approximation Theorem
[19/100%]If the functions
are continuous on the interval and if they satisfy the conditions
then the terms of the sequence
are between the minimum and the maximum of the continuous function . (Cf. Problems 65, 79, 83 of Part I.)
Let be a fixed point of the interval considered in Problem 128. In order that
holds for all functions continuous on it is necessary and sufficient that
for all positive values of for which (if or the first or second integral resp. under the limit sign has to be omitted.) (Cf. Problems 66, 80, 84 of Part I.)
We have
provided that the integral on the left and the limit on the right hand side exist.
If the integral
converges for and for , , it converges for and it represents a continuous function of on that interval.
We assume that
are continuous functions of and , , and that for each
Let denote a continuous function. The functions
lie between the minimum and the maximum of on for any , ; i.e. . Furthermore
provided
uniformly for , fixed and positive. The convergence is uniform on any closed subinterval of .
Let denote a continuous function on . The convergence of
is uniform for , , fixed.
Let be a continuous periodic function with period . Then
the convergence is uniform for all .
Every function defined and continuous on the finite interval can be approximated uniformly on by polynomials to any degree of accuracy. (Weierstrass' approximation theorem.)
Every continuous function that is periodic with period can be uniformly approximated by trigonometric polynomials (see §2 of Chapter 1, Part VI) to any assigned degree of accuracy. (Weierstrass' approximation theorem.)
Let denote a function that is properly integrable over (). Two polynomials (trigonometric polynomials), and , can then be found for any positive so that for ()
and
The -th moment of a function is given by
If all the moments of a function that is defined and continuous on the finite interval vanish then the function vanishes identically.
If all the moments
of a function that is properly integrable over the interval vanish then the function vanishes at every point of continuity.
If the first moments vanish,
of a function defined and continuous on the finite or infinite interval then the function changes sign (see §2 of Chapter 1, Part V) at least times in the interval unless it is identically .
The -th and -th trigonometric moment (Fourier constants, cf. §4 of Chapter 1, Part VI) of a function with period are defined as
If the first trigonometric moments of a continuous function with period vanish then changes sign at least times in any interval of length (see §2 of Chapter 1, Part V) unless is identically .
Let the function be defined and continuous for . Suppose that the integral
converges for and that it vanishes for a sequence of 's increasing in arithmetic progression:
Then vanishes identically.
The -function
can be written as an integral (Problem 31). Use this fact to prove that does not have any zeroes.
(Problem 142).
Work out the polynomials , for
Let be any point on and
where refers to the subscripts for which and to those for which , . Then
Let be continuous on . The polynomials converge uniformly to on . (New proof of Weierstrass' theorem, Problem 135.)