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Re: From addition preservation to linearity
Posted:
Sep 24, 2012 1:14 PM


On 24092012 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



