Re: Translation invariant subspace of L^1 is an ideal
- From: rusty <mr.rusty@xxxxxxx>
- Date: Wed, 22 Aug 2007 08:49:41 +0200
jane wrote:
jane wrote:
How can i prove that every closed subspace of theBanach algebra L^1 (
convolution being multiplication) is translationinvariant if and only if
is an ideal.
Let U_t be the action of R on L^1(R) (I presume).
Then g * f = \int g(t) U_t(f) \, dt, so translation
invariance implies
invariance under left (or right) convolution.
I didn't get this point. Do you mean that this should imply 1 of the
direction, namely that knowing that it is translation invariant it must be
an ideal ?
I've thought about it, and probably i know the way to solve by now.
I think for this direction we can do something like this:
Suppose M is our subspace, f in M, g in L^1. Then it is enough to prove
that f*g is in M for all compactly supposrted g, since they are dense in
L^1 and our algebra is closed. Suppose supp(g) in [A,B]
Then f*g(x) = intA^B f(x-t) g(t)dt. But for small h > o it is clear that
|| int_a^{a+h} (f(x-t)-f((a+h)/2))g(t)dt ||_L^1 can be done
arbitrary small, since t -> U_t(f) is uniformly continuous mapping R ->
L^1 and taking small h > 0 we know that |x-t ((a+h)/2)| < h/2 we will
have small
|| f_t - f((a+h)/2) ||_L^1.
Thus, subdividing [A,B] into finitely many intervals [a,a+h] with small
enough h > 0 we will get that
|| f*g - sum_a f(x - (a+h)/2)*int_a^{a+h}g(t)dt || < e and
sum_a f(x - (a+h)/2)*int_a^{a+h}g(t)dt is in M being the a finite linear
combination of elements from M.
Thus f*g is in M.
(1) Yes, Riemann integration applies to continuous functions f with
values in a Banach space E: the usual proofs based on uniform
continuity on compact intervals apply. The integrals are limits of
Riemann sums, as above, and if x* is in E*
x*( \int_[a,b] f(t) dt) = \int_[a,b] x*(f(t)) dt (*)
(2) When the Banach space is the dual of another Banach space,
the integral can be defined using the above equation for x*
in the predual, thus avoiding the need to develop a theory
from scratch. This will not work for L^1(R).
(3) In all cases, however, including this one, integration of continuous
functions from a compact space to E with respect to a
finite measure can be defined weakly as in (*) using convexity
theory. (See for example Rudin's functional analysis -
"integration of vector valued functions". You might also like
Abstract Harmonic Analysis by Loomis.)
(4) What is really going on here is convolution of measures.
U_t(f) is convolution of the point measure at t with the measure
f(x) dx, etc, etc.
Conversely, taking L^1 functions g_n with integral 1
and supported in [-1/n,
1/n], it is standard to check that g_n * f tends to
in the L^1 norm (it's
an "approximate identity" in L^1(R)). But then
U_t(g_n) * f = U_t (g_n *
f) tends to U_t(f).
--
rusty
Thanks, that part i've got.
--
rusty
.
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