Polynomials times a Gaussian are dense in L^2
John Baez
functions times the Gaussian exp(-x^2) are dense in L^2(R)?
I know a general-purpose "L^2 Stone-Weierstrass theorem"
that does the job, but it requires knowing the Fourier
transform is unitary, which takes a while to prove.
I also know the eigenfunctions of the harmonic oscillator
Hamiltonian are an orthonormal basis of L^2(R), and this
also does the job, but again it relies on some stuff that
takes work to prove.
For a class I plan to teach, I'd like a quick proof
that doesn't use much machinery, if one exists.
Nemo
For each compact interval I, polynomials times exp(-x^2) are dense in
the continuous functions on I with the sup norm; you can appeal to
Stone-Weierstrass, or its special case for polynomials. And continuous
functions with compact support are dense in L^2(R).
The above does not use Fourier transform being unitary.
Is this low-brow enough?
Nemo
David C. Ullrich
Possibly a little too low-brow, since it doesn't _quite_ give a
proof. Say f is in L^2. You can choose g continuous with
compact support so ||f-g||_2 < epsilon. And now you show
that one can approximate g _on_ the support of g by a
function of the required type, but you actually have to
approximate g in L^2(R).
>Nemo
************************
David C. Ullrich
Nemo
ooops
William F Hammond
> What's an easy low-brow way to prove that polynomial
> functions times the Gaussian exp(-x^2) are dense in L^2(R)?
See chapter 1 of Igusa's (Springer Ergebnisse, 70's) book
"Theta Functions" for a low-brow proof. I suspect that this
book will also have other things you may find useful for
your course.
-- Bill
ArtflDodgr
ba...@math.removethis.ucr.andthis.edu (John Baez) wrote:
> What's an easy low-brow way to prove that polynomial
> functions times the Gaussian exp(-x^2) are dense in L^2(R)?
Suppose that f is in L^2(R) and is orthogonal to all such products.
Expanding exp(zx) in a power series and using Fubini's theorem, check
that
int_R f(x) exp(zx) exp(-x^2) dx = 0,
first for all complex z of sufficiently small modulus, then for all
complex z by analytic continuation. In particular,
int_R f(x) exp(-itx) exp(-x^2) dx = 0,
for all real t. As f(x)exp(-x^2) is clearly an element of L^1(R),
this last display means that the Fourier transform of f vanishes.
Thus f = 0, almost everywhere.
--
A.
David C. Ullrich
wrote:
Yeah, I thought of suggesting more or less that (don't see why
we need to restrict to z of small modulus.) But it uses the
uniqueness theorem for the Fourier transform; since the OP
specified he didn't want to use the Plancherel theorem I
decided this was insufficiently low-brow.
There must be a proof that doesn't depend on anything
remotely non-trivial about the Fourier transform, along
the lines of Nemo's almost-proof. Not that I see what
it is...
************************
David C. Ullrich
David C. Ullrich
It's possible that there is no proof as elementary as
what you want. The following fact surprised me:
Fact: There exists a function G such that
(*) G > 0, G is continuous, and x^n G(x) -> 0
as x -> infinity, for n = 0, 1, ...,
such that the functions P(x) G(x) are _not_ dense in
L^2(R) (note that the fact that x^n G(x) -> 0 for all n
shows that P(x) G(x) is in L^2 for every P.)
My point is that thinking about a totally low-brow proof
of what you want I assumed it would follow just from
the fact that G(x) = exp(-x^2) satisfies (*). But it
doesn't follow from (*), so any proof must use
some more special property of exp(-x^2). (In particular
it's not clear that, for example, Nemo's argument
can be fixed - that was the sort of argument I was
looking for for a while...)
Proof of the Fact: Let phi be infinitely differentiable
on R, with support(phi) contained in [1,2] (and phi
not identically 0.) Let F^ = phi (where F^ is the
Fourier transform of F.)
Then F is a Schwarz function, so in particular
F(x) (1 + |x|^n) is in L^2 for n = 0, 1, ... .
So we can choose numbers c_n > 0 such that
f is in L^2(R), where
f(x) = F(x) sum(c_n (1 + |x|^n))
(and also such that sum(c_n (1 + |x|^n)) is
continuous, in case that doesn't follow.) Let
G(x) = 1 / sum(c_n (1 + |x|^n)).
It's clear that G satisfies (*). But the P(x) G(x)
are not dense in L^2, because f is orthogonal to
all of them:
<f, PG>
= <F, P>
= linear combination of derivatives of F^ at the origin
= 0.
************************
David C. Ullrich
Michael J. Doré
> What's an easy low-brow way to prove that polynomial
> functions times the Gaussian exp(-x^2) are dense in L^2(R)?
The way we were given in lectures was as follows. Since the
trigonometric polynomials are uniformly dense in C[a,b] it is easy to
see that the space of finite linear combinations of functions of the
form:
e^(iax)e^(-x^2/2) (a = 0,1,2,...)
is uniformly dense in C_0(R) (the space of continuous functions on R
tending to 0). From here it is possible to deduce that the space of
functions of the form p(x)e^(-x^2/2) (where p is a polynomial) is
uniformly dense in C_0(R).
Now given f in L^2(R) first approximate it in L^2 by continuous g with
compact support (as in previous messages) then uniformly approximate
g(x)e^(x^2/2) by functions of the form p(x)e^(-x^2/2) where p is a
polynomial. It is easy to see this gives an L^2 approximation of g by
functions of the form p(x)e^(-x^2) so we're done.
Michael
Herman Rubin
=?ISO-8859-1?Q?Michael_J._Dor=E9?= <md...@cam.ac.uk> wrote:
>ba...@math.removethis.ucr.andthis.edu (John Baez) wrote in message news:<biuhrn$7ja$1...@glue.ucr.edu>...
>> What's an easy low-brow way to prove that polynomial
>> functions times the Gaussian exp(-x^2) are dense in L^2(R)?
The problem is harder than it looks. The property of the
function exp(-x^2) which is needed is that the corresponding
moment problem must have a unique solution. This looks like
an L_1 condition, but everything between 1 and 2 is equivalent
for this problem, and the usual nasc conditions can easily be
obtained by L_2 arguments.
This is easy, but lengthy, so I do not think it is what is being
desired. It can be done with only real analysis, but using
complex numbers.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
David C. Ullrich
>ba...@math.removethis.ucr.andthis.edu (John Baez) wrote in message news:<biuhrn$7ja$1...@glue.ucr.edu>...
>> What's an easy low-brow way to prove that polynomial
>> functions times the Gaussian exp(-x^2) are dense in L^2(R)?
>
>The way we were given in lectures was as follows. Since the
>trigonometric polynomials are uniformly dense in C[a,b] it is easy to
>see that the space of finite linear combinations of functions of the
>form:
>
>e^(iax)e^(-x^2/2) (a = 0,1,2,...)
>
>is uniformly dense in C_0(R) (the space of continuous functions on R
>tending to 0).
That's clear.
>From here it is possible to deduce that the space of
>functions of the form p(x)e^(-x^2/2) (where p is a polynomial) is
>uniformly dense in C_0(R).
How does one deduce that, exactly?
>Now given f in L^2(R) first approximate it in L^2 by continuous g with
>compact support (as in previous messages) then uniformly approximate
>g(x)e^(x^2/2) by functions of the form p(x)e^(-x^2/2) where p is a
>polynomial. It is easy to see this gives an L^2 approximation of g by
>functions of the form p(x)e^(-x^2) so we're done.
>
>Michael
************************
David C. Ullrich
Michael J. Doré
>On 3 Sep 2003 08:10:59 -0700, md...@cam.ac.uk (Michael J. Doré)
wrote:
>>The way we were given in lectures was as follows. Since the
>>trigonometric polynomials are uniformly dense in C[a,b] it is easy
to
>>see that the space of finite linear combinations of functions of the
>>form:
>>
>>e^(iax)e^(-x^2/2) (a = 0,1,2,...)
Oops, meant to say a is an integer rather than non-negative integer.
In fact the rest of the proof goes through fine if you only restrict a
to being in R.
>>is uniformly dense in C_0(R) (the space of continuous functions on R
>>tending to 0).
>
>That's clear.
>
>>From here it is possible to deduce that the space of
>>functions of the form p(x)e^(-x^2/2) (where p is a polynomial) is
>>uniformly dense in C_0(R).
>
>How does one deduce that, exactly?
Well it suffices to prove e^(iax)e^(-x^2/2) can be uniformly
approximated on R by p(x)e^(-x^2/2). Let p_n be the first n terms of
the Taylor series of e^(iax). By the triangle inequality:
|e^(iax)e^(-x^2/2)-p_n(x)e^(-x^2/2)| <= e^(|a||x| - x^2/2)
so this is < epsilon outside some compact interval I, for all n. But
p_n(x) -> e^(iax) uniformly on I so we're done.
Michael
David C. Ullrich
>On 04 Sep 2003, David C. Ullrich wrote:
>>On 3 Sep 2003 08:10:59 -0700, md...@cam.ac.uk (Michael J. Doré)
>wrote:
>
>>>The way we were given in lectures was as follows. Since the
>>>trigonometric polynomials are uniformly dense in C[a,b] it is easy
>to
>>>see that the space of finite linear combinations of functions of the
>>>form:
>>>
>>>e^(iax)e^(-x^2/2) (a = 0,1,2,...)
>
>Oops, meant to say a is an integer rather than non-negative integer.
>In fact the rest of the proof goes through fine if you only restrict a
>to being in R.
>
>>>is uniformly dense in C_0(R) (the space of continuous functions on R
>>>tending to 0).
>>
>>That's clear.
>>
>>>From here it is possible to deduce that the space of
>>>functions of the form p(x)e^(-x^2/2) (where p is a polynomial) is
>>>uniformly dense in C_0(R).
>>
>>How does one deduce that, exactly?
>
>Well it suffices to prove e^(iax)e^(-x^2/2) can be uniformly
>approximated on R by p(x)e^(-x^2/2). Let p_n be the first n terms of
>the Taylor series of e^(iax). By the triangle inequality:
>
>|e^(iax)e^(-x^2/2)-p_n(x)e^(-x^2/2)| <= e^(|a||x| - x^2/2)
Well duh. Took me a second to see why that inequality was
true, but yes, it's clear. This is the simple proof I was coming
to think didn't exist. Keen.
>so this is < epsilon outside some compact interval I, for all n. But
>p_n(x) -> e^(iax) uniformly on I so we're done.
>
>Michael
************************
David C. Ullrich
ArtflDodgr
md...@cam.ac.uk (Michael J. Doré) wrote:
> ba...@math.removethis.ucr.andthis.edu (John Baez) wrote in message
> news:<biuhrn$7ja$1...@glue.ucr.edu>...
> > What's an easy low-brow way to prove that polynomial
> > functions times the Gaussian exp(-x^2) are dense in L^2(R)?
>
> The way we were given in lectures was as follows. Since the
> trigonometric polynomials are uniformly dense in C[a,b] it is easy to
> see that the space of finite linear combinations of functions of the
> form:
>
> e^(iax)e^(-x^2/2) (a = 0,1,2,...)
>
> is uniformly dense in C_0(R) (the space of continuous functions on R
> tending to 0).
Please elaborate on this point. (It would seem that you need some
control on the number of complex exponentials involved in the linear
combination.)
--
A.
David C. Ullrich
wrote:
This part is easy (meaning it was immediately clear to me why it
worked...) The point is that given a continuous function f on a
compact interval I, with |f| <= K on I, we can find a trigonometric
polynomial g with |f-g| < epsilon on I and _also_ |g| <= K on all
of R. This follows if we just set g equal to the convolution of
(the periodization of) f with, say, a Fejer kernel. The fact that
|g| <= K globally takes care of the problem that you say it seems
we need control over the number of terms for - the point is that
the bound |g| <= K is independent of the number of terms in g,
independent of epsilon.
What I didn't see was how the next step went - going from a
trignometric polynomial g to a polynomial p. We need _some_
global estimate on |p| independent of epsilon (or if the estimate
depends on epsilon it's going to be a very fussy argument.)
I came perilously close to saying it didn't work, luckily just
asked how the argument went. The answer is very simple:
We can assume here that g(x) = exp(iax). We let p_n
be the n-th partial sum of the Taylor series for g. This
is something I never woulda thought of - a Taylor series
is not the best way to get a uniform approximation to
a function! But using the Taylor series gives us the
global bound we need: g - p_n is the sum of the
remaining terms in the Taylor series, so it's <= the
sum of the absolute value of all the terms, which
is exactly exp(|ax|). Very clever, I gotta write this
down (actually one reason I'm typing this is to try
to fix the thing in my memory.) Since exp(|ax|) is
o(exp(-x^2/2)) we're set.
So it seems clear to me that this _is_ the Simple
Proof that we've been looking for.
****************
What interests me is the question of what functions
work in place of exp(-x^2) here. Seems like the
functions that work in this argument are the same
as the ones that work in the Fourier transform
argument - we need G(x) exp(|ax|) -> 0 for all a
in both cases (regarding the argument above,
note that this is the same as saying
sqrt(G(x)) exp(|ax|) -> for all a.)
(I haven't thought about whether that hypothesis
is enough to give the convergence in the version
of the FT argument that you posted, but it's
enough for the version that I didn't post: Suppose
that f is orthogonal to all the P(x) G(x). Define
F(z) = int f(x) exp(ixz) G(x).
Then F is entire, and the hypothesis shows that
all the derivatives of F at the origin vanish, so F = 0,
hence f = 0.)
************************
David C. Ullrich
Michael J. Doré
Suppose f is in C_0(R). Given epsilon > 0 there is a compact interval
I outside which |f| < epsilon and e^(-x^2/2) < epsilon. Now find a
finite linear combination g of e^(iax)e^(-x^2/2) (a integer) such that
|f-g| < epsilon on I. We just need to check f and g are reasonably
close outside I.
We can write g(x) = e^(-x^2/2)h(x) where h(x) is a trigonometric
polynomial. It is easy to get a bound on h - WLOG I contains [-pi,pi]
so |e^(-x^2/2)h(x)-f(x)| < epsilon on [-pi,pi] so |e^(-x^2/2)h(x)| <=
|f(x)|+epsilon <= M+epsilon on [-pi,pi] where M = sup_R |f|. So |h(x)|
<= e^(pi^2/2)(M+epsilon) on [-pi,pi]. But h has period 2pi so this is
true on R.
Then outside I, |g(x)| = e^(-x^2/2)*|h(x)| < epsilon *
e^(pi^2/2)(M+epsilon) and |f| < epsilon so |f-g| < epsilon *
(e^(pi^2/2)(M+epsilon)+1) outside I, while |f-g| < epsilon on I.
Michael
Michael J. Doré
Hmmm, on second thoughts I seem to have messed this part up. Of course
you can't restrict a to being an integer since we want to uniformly
approximate on an interval I, which could have any length in general.
So what I should have said is:
Claim: The space of finite linear combinations of functions of the
form e^(iax)e^(-x^2/2) (a in R) is uniformly dense in C_0(R).
Proof: Either the way David Ullrich outlined, or let f be in C_0(R)
and epsilon > 0. Take K > 0 such that |f(x)| < epsilon for |x| > K/2
and e^((K/2)^2/2) <= epsilon*e^(K^2/2). Take h to be a finite linear
combination of e^(iax) where a is a multiple of pi/K such that
|f-he^(-x^2/2)| < epsilon on [-K,K]. So |h| <= (|f|+epsilon)e^(x^2/2)
on [-K,K]. Then on [-K/2,K/2] we have |h| <= (M+epsilon)e^((K/2)^2/2)
<= (M+epsilon)*epsilon*e^(K^2/2) (where M = sup |f|) and for K/2 < |x|
<= K we have |h| <= 2epsilon e^(K^2/2). So |h| <=
max(M+epsilon,2)epsilon e^(K^2/2) <= (M+2+epsilon)epsilon e^(K^2/2) on
[-K,K]. But h has period 2K so this so this holds on R - so for |x| >
K we have |h e^(-x^2/2)| <= (M+2+epsilon)epsilon e^(K^2/2-x^2/2) <=
(M+2+epsilon)*epsilon and |f - he^(-x^2/2)| <=
epsilon+(M+2+epsilon)*epsilon. While for |x| <= K, |f-he^(-x^2/2)| <
epsilon. So |f-he^(-x^2/2)| <= (M+3+epsilon)*epsilon on R, which can
be made arbitrarily small, QED.
The rest of the proof then goes through. Hope that's all correct now.
Michael
John Baez
wrote:
> ba...@math.removethis.ucr.andthis.edu (John Baez) wrote:
>> What's an easy low-brow way to prove that polynomial
>> functions times the Gaussian exp(-x^2) are dense in L^2(R)?
>Suppose that f is in L^2(R) and is orthogonal to all such products.
>Expanding exp(zx) in a power series and using Fubini's theorem, check
>that
>
>int_R f(x) exp(zx) exp(-x^2) dx = 0,
>
>first for all complex z of sufficiently small modulus, then for all
>complex z by analytic continuation. In particular,
>
>int_R f(x) exp(-itx) exp(-x^2) dx = 0,
>
>for all real t. As f(x)exp(-x^2) is clearly an element of L^1(R),
>this last display means that the Fourier transform of f vanishes.
>Thus f = 0, almost everywhere.
This is nice; it's really the proof of the L^2 Stone-Weierstrass
theorem stripped down to handle just this one case. Unfortunately,
the one-to-one-ness of the Fourier transform is exactly what I was
trying to avoid using. But maybe it's not worth avoiding. Everyone
needs to learn it sometime.
At this point I've almost decided to:
1) first remind the kids about the Stone-Weierstrass theorem
2) then use that to prove the functions exp(inx) are an orthonormal
basis of L^2[0,2pi]
3) then use that to prove the Fourier transform is unitary on L^2(R)
4) then use your argument to show that polynomials times exp(-x^2)
are dense in L^2(R)
John Baez
David C. Ullrich <ull...@math.okstate.edu> wrote:
>On Mon, 01 Sep 2003 18:32:19 -0400, Nemo <nob...@no.net> wrote:
>>John Baez wrote:
>> > What's an easy low-brow way to prove that polynomial functions times
>> > the Gaussian exp(-x^2) are dense in L^2(R)?
>>For each compact interval I, polynomials times exp(-x^2) are dense in
>>the continuous functions on I with the sup norm; you can appeal to
>>Stone-Weierstrass, or its special case for polynomials. And continuous
>>functions with compact support are dense in L^2(R).
>>
>>Is this low-brow enough?
>Possibly a little too low-brow, since it doesn't _quite_ give a
>proof. Say f is in L^2. You can choose g continuous with
>compact support so ||f-g||_2 < epsilon. And now you show
>that one can approximate g _on_ the support of g by a
>function of the required type, but you actually have to
>approximate g in L^2(R).
Right; this is the approach I wanted to use, but I couldn't
see an easy way to fill in that gap. It suffices to
show that given a continuous function F vanishing outside
some interval I, and given epsilon > 0, we can find a polynomial
P such that
1) |F-P| < epsilon in the interval I
2) The function P(x) exp(-x^2) has L^2 norm < epsilon
outside the interval I
It's not obvious that these two goals don't run into conflict:
we want to make P approximate F well inside the interval but not
get too big outside the interval - except at points where |x| is so
large that exp(-x^2) is really small.
A good lowbrow proof would accomplish all this without too
much fuss, but right now I'm leaning towards proving the
L^2 Stone-Weierstrass theorem. It's more general, and very
easy once you know the Fourier transform is unitary.
On the other hand, if I could prove that polynomials times
exp(-x^2) were dense in L^2(R), I could give an *incredibly*
slick proof that the Fourier transform is unitary! Sigh.
David C. Ullrich
Possibly you missed it in the dense thread, but a more elementary
proof has appeared. See Michael Dore's post (and his replies to
various questions about what he suggested - what I thought was
a gap at first is filled by a clever estimation of a certain error
term.)
************************
David C. Ullrich