From: israel@math.ubc.ca (Robert Israel)
Subject: Re: x^k (1-x)^{1-k}
Date: 10 Jun 1999 20:35:59 GMT
Newsgroups: sci.math.research
Keywords: Spanning a space of functions: Stone's Approximation Theorem, etc.
In article <7jor85$na2$1@nnrp1.deja.com>,
iawah@my-deja.com writes:
> > Does there exist a countable subset of the class of functions
> > of the form x^k (1-x)^{1-k} for k in [0,1] that span the
> > space of functions on [0,1]?
> -^-
> I see I omitted a key word: *convex* functions on [0,1].
> And I only need nonnegative and convex functions.
More to the point:
1) You want continuous functions, I assume (a convex function on a
closed interval need not be continuous at the endpoints of the interval).
2) I don't think you really mean "span" in the algebraic sense (or the
answer would clearly be no). You mean the closed span in some topology,
and I'll assume it's the usual topology on C[0,1] (the continuous functions
on [0,1]), given by the supremum norm.
3) Any C^2 function on [0,1] is the difference of two convex functions
(consider the positive and negative parts of the second derivative, and
integrate twice). C^2 functions are dense in C[0,1]. So the restriction
to nonnegative convex functions is no restriction, given (1) and (2).
4) Since your class of functions is separable, it suffices to show that
the linear span of your class of functions is dense; you can automatically
take "a countable subset of ...".
> Remark: The above class is obviously not an algebra, and thus
> Stone's Approximation Theorem does not apply.
Not directly, but it still may be useful. I don't know how to solve
your problem, but I can show that the span S of your functions is dense in
C[a,b] if 0 < a < b < 1. Note that
d/dk x^k (1-x)^(1-k) = ln(x/(1-x)) x^k (1-x)^(1-k) is a uniform limit
of members of S (for fixed k with 0 < k < 1). The same goes for the n'th
derivative ln(x/(1-x))^n x^k (1-x)^(1-k). By the Stone-Weierstrass Theorem,
polynomials in ln(x/(1-x)) are dense in C[a,b], and the same is true
after multiplying by x^k (1-x)^(1-k) (which is invertible in C[a,b]).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
==============================================================================
From: Allan Adler
Subject: Re: x^k (1-x)^{1-k}
Date: 10 Jun 1999 17:22:34 -0400
Newsgroups: sci.math.research
iawah@my-deja.com writes:
> > Does there exist a countable subset of the class of functions
> > of the form x^k (1-x)^{1-k} for k in [0,1] that span the
> > space of functions on [0,1]?
> -^-
> I see I omitted a key word: *convex* functions on [0,1].
> And I only need nonnegative and convex functions.
>
> Intuitively, one can get arbitrarily steep functions at 0 or 1
> but subtracting very small multiples of x^k (1-x)^{1-k} for
> k close enough to 0 or 1.
>
> Remark: The above class is obviously not an algebra, and thus
> Stone's Approximation Theorem does not apply.
I haven't thought about this at all and am not familiar with
any of the subtleties of the problem. But a fraction of an idea
occurs to me, namely that one might be able to imitate the techniques
used to prove the Muntz-Szasz theorem. In that theorem, one also has
a one parameter family of functions, namely x^k, k>0, and one wants to know
when a countable set of them spans a dense subspace of the space of
continuous functions on the unit interval. The answer is that they
span iff the sum of the reciprocals of the 1/k for that set diverges.
Moreover, if the sum converges, the clsure of the span doesn't contain x^k.
There is a proof of the Muntz-Szasz theorem in Rudin's Real and Complex
Analysis. It looks as though one can at least get started on a similar
proof, e.g. for any complex borel measure m on [0,1], consider the
function f(z) = integral from 0 to 1 of x^z (1-x)^(1-z)
(defined as exp(z long x + (1-z) log (1-x))).
Can this approach be carried through to completion?
Allan Adler
ara@altdorf.ai.mit.edu