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Topic: From addition preservation to linearity
Replies: 3   Last Post: Sep 24, 2012 1:14 PM

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 David C. Ullrich Posts: 21,553 Registered: 12/6/04
Re: From addition preservation to linearity
Posted: Sep 24, 2012 12:15 PM

On Mon, 24 Sep 2012 01:41:01 -0700 (PDT), Rupert
<rupertmccallum@yahoo.com> wrote:

>On Sep 24, 1:14 am, José Carlos Santos <jcsan...@fc.up.pt> wrote:
>> Hi all,
>>
>> Can someone please tell me where can I find a reference for the fact
>> that any Lebesgue-measurable map from R^n into R^m which preserves sums
>> is necessarily a linear map?
>>
>> Best regards,
>>
>> Jose Carlos Santos

>
>I don't know a reference, but one of the crucial points is that if A
>is a subset of R^n of positive measure then A+A has nonempty interior.
>David Ullrich showed me a proof of this a while back but he didn't
>know a reference in the literature.

Hmm. It's unclear to me what the "this" refers to - I've posted the
simple proof of that fact about A + A, but I don't recall ever
saying anything about the OP's question. In fact it's not clear
to me how that follows, let's see:

Lemma 0. If A subset R^n and m(A) > 0 then A + A has
nonempty interior.

Pf Sketch: WLOG 0 < m(A) < infinity. Let h be the characteristic
function of A, and let g be the convolution of A with
itself. Then g >= 0, and

int g = (int h)^2 > 0.

Also g is continuous, so g > 0 on some nonempty
open set B. But B subset A + A. QED.

Hmm. Now suppose f : R^n -> R^m is measurable
and

f(x+y) = f(x) + f(y).

There exists L so that if A = {x : |f(x)| < L} then
m(L) > 0. Hence A + A has nonempty interior O.

But x in O implies that |f(x)| < 2L.

Fix x in O and let V = O - x = {y : x + y in O}.
Then V is a neighborhood of the origin, and
z in in V implies z = y - x with y in O, hence

|f(z)| < 2L + |f(x)| = c.

So f is bounded in O.

Now since f(nx) = n f(x) for any positive integer n, it follows
that

|f(z)| < c/n for z in O/n,

and hence f is continuous at the origin. Since f is additive
this shows that f is continuous.

And now, since f(qx) = q f(x) for any rational q and f is
continuous, it follows that f(rx) = r f(x) for any real r, so
f is linear. QED.

Huh. I didn't know I knew that, thanks.

>
>I think this might be relevant
>
>@article{Kestelman51,
> author="Hyman Kestelman",
> title="Automorphisms of the field of complex
>numbers",
> journal="Proc. London Math. Soc.",
> issue=2,
> number=53,
> pages="1-12",
> year=1951}

Date Subject Author
9/23/12 Jose Carlos Santos
9/24/12 Rupert
9/24/12 David C. Ullrich
9/24/12 Jose Carlos Santos