Re: A few basic functional analysis questions

In article
<7f9e76ec-c4c2-48b5-8121-4c81d917ccda@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Rotwang <sg552@xxxxxxxxxxxxx> wrote:

Thanks for your reply, but there is something I don't understand...

On 16 Sep, 21:36, W^3 <aderamey.a...@xxxxxxxxxxx> wrote:
In article
<0626338b-97de-425a-a73a-929bf134b...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,

 Rotwang <sg...@xxxxxxxxxxxxx> wrote:
[...]

Suppose that a function f has the property that, for every continuous
function g with compact support, \int f g = 0. How does one prove that
f = 0 almost everywhere?

Define F = |f|/f on the set where f is nonzero, F = 0 elsewhere. Let B
be an open ball. A corollary of Lusin's theorem shows there exists
continuous f_n with compact support in B, with |f_n| <= 1 everywhere,
such that f_n -> F a.e.

I'm not familiar with Lusin's theorem, but the statements I've found
in various places only refer to functions defined on |R^1, and the
proof I've found seems to depend on some facts that I don't see how to
generalise. Can you point me at a version of the theorem which is
applicable here?


Now f is integrable on B, so dominated
convergence gives  int_B |f| = int_B f*F = lim int_B f*f_n = 0. This
implies f = 0 a.e. on B, hence a.e. on R^n.

How does one prove that the space of continuous functions with compact
support is dense in L^1? Also, in L^2?

L^2 functions with compact support are dense in L^2, so it's enough to
prove density in L^2(B), where B is a ball. Let M be the
L^2(B)-closure of the subspace of continuous functions with compact
support in B. If M is not dense in L^2(B), there is an f in L^2(B)
orthogonal to M. Now use the first result.

I can follow this, thanks.


(L^1 follows from this.)

Right, since if f in L^1 is non-negative then sqrt(f) is in L^2, and
if g is in L^2 with ||sqrt(f) - g||_2 < e then the Hölder inequality
implies that

||f - g^2||_1 = \int |sqrt(f) - g|*|sqrt(f) + g|
<= e * (2||sqrt(f)||_2 + e).

We can choose continuous g with compact support such that the RHS is
arbitrarily small, and the general case of complex f follows easily
(have I missed an easier way)?

Another way is to use 1. L^2(B) is dense in L^1(B), and 2. the L^1(B)
norm is <= sqrt(m(B)) * L^2(B) norm.

The following may be a better way to proceed for the questions you
have here. Briefly, simple functions are dense in every L^p, 1 <= p <
oo, right? So it's enough to show that if m(E) < oo and eps > 0, then
there is a continuous function g with compact support such that ||g -
X_E||_p < eps. But we know there is a compact subset K of E with m(E \
K) as small as we like. Let V_r = {x : d(x, K) < r}. Choose a
continuous g : R^n -> [0, 1] with g = 1 on K, supp g in V_r. If r is
small you will have ||g - X_E||_p < eps as desired.

That answers your questions for L^1, L^2. For your first question,
restrict to B as before and consider the F I defined in my first post.
Then there exists a sequence of continuous functions with compact
support g_n with |g_n| <= 1 everywhere, such that g_n -> F a.e. Why?
We know we can get convergence in L^2, hence there is a subsequence,
let's still call it g_n, that converges a. e. to F. Now let h(z) = z,
|z| <= 1, h(z) = z/|z| otherwise (z is complex). Then h o g_n is the
desired sequence, which leads to the result you want.
.

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