|
|
Consider a particular case, namely
f:[0,1]--->R and x(j)=j/n , j in {0,1,...,n}.
In the following
[a,b,c;f] or [a,b,c;f(t)]_t denotes the
divided difference of f,with respect to variable t,
at the (distinct) knots a,b,c from [0,1].More
precisely
[a,b,c;f]=[a,b,c;f(t)]_t =
=f(a)/((a-b)(a-c)) + f(b)/((b-a)(b-c)) +
+f(c)/((c-a)(c-b)).
It's well known that for g in C^2[0,1] there exists
at least a point z in [0,1] such that
(1) [a,b,c;g]= g"(z)/2! .
Therefore , in case when g is in C^2[0,1],
||g"||:= max_{x in [0,1]}|g"(x)| ,
we have
(2) | [a,b,c;g] | =< ||g"||/2.
Further let
h_x(t):=|x-t|,
{.}:=fractional part,
[.]:=integer part,
e_j(t):=t^j ,
D_n(x):= {nx}(1-{nx})/n^2 .
If c_n:=1/n , x in [0,1], define
===========================
(S_nf)(x)=
=c_n*SUM_{k=0 to k=n}[(k-1)/n,k/n,(k+1)/n; h_x(t)]_t f(k/n).
==========================
Observe following facts:
1) S_n: f-->S_nf is a linear positive operator.
2) For j in {0,1} we have S_ne_j = e_j .
3) (S_ne_2)(x)= e_2(x) + D_n(x)
4) 0=< D_n(x) =< 1/(4n^2) for x in [0,1].
5) The restrictions of S_nf at intervals [(j-1)/n,j/n],
jin {1,2,...,n}, are linear functions.
6) (S_nf)(k/n)=f(k/n) , k in {0,1,...,n}.
7) For x in [0,1] , a_n(x):=[nx]/n , we have
(S_nf)(x) -f(x) =
(#)
= D_n(x)*[a_n(x),x,a_n(x)+ 1/n ;f] .
Now assume that f is in C^2[0,1] and x in [0,1].
Using (#) together with 4), (1)-(2), one finds
| (S_nf)(x)-f(x) | =< ||f"||/(8*n^2) , n in {1,2,...}.
Of course, a generalisation (to arbitrary knots x(j))
is possible. Perhaps help/Alex
======
Note: If you assume that f is only in C[0,1] you can use the
modulus of continuity (of the first order) that is
W(f;d):=max_{|t-x|=< d}|f(t)-f(x)| .
One finds (x in [0,1])
|f(x)-(S_nf)(x)|=< 2*W(f;sqrt({nx}(1-{nx}))/n) .
|
|