```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 withYou 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
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