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