Date: Sep 24, 2012 1:14 PM
Author: Jose Carlos Santos
Subject: Re: From addition preservation to linearity
On 24-09-2012 17:15, David C. Ullrich wrote:

> 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

You meant "of h" here.

> 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.

And here you meant "m(A) > 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.

Thanks. That was quite a help. I still would rather have a reference

that I could cite, but it's good to know how to prove it.

Best regards,

Jose Carlos Santos