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By Titu Andreescu, Iurie Boreico

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2 2 y Let a = 1. Setting z = leads to x f (x + y) = f (x) + f (y) for any x = 0 and any y. The same remains true holds if x = 0. It follows now by (3)that f (xy) = f (x)f (y). √ Then f (x + y) = f (x) + (f ( y))2 ≥ f (x) for y ≥ 0. Hence f is an additive increasing function and therefore f (x) = f (1)x = x. So, f ≡ 2, f ≡ 0 or f (x) ≡ x. It clear that all the three functions satisfy (3). Exercises Problem 109. (Bulgaria, 1994) Find all functions f : R → R such that xf (x) − yf (y) = (x − y)f (x + y) for all x, y ∈ R.

Thus g(x − y) = ±(g(x)f (y) − g(y)f (x)). Suppose g(x − y) = −(g(x)f (y) − g(y)f (x)) for some x, y. Let A be the set of such (x, y) and B be the set of (x, y) for which g(x − y) = g(x)f (y) − g(y)f (x). Then A, B are closed and A B = R2 thus there is a point (u, v) which belongs to both A, B. Then g(u − v) = (g(u)f (v) − g(v)f (u)) = −(g(u)f (v) − g(v)f (u)) possible only when g(u − v) = 0. If we interchange x with y we get that f is even. If we set y → −y we get f (x+y) = f (x)f (y)+g(x)g(−y) and analogously f (x+ y) = f (x)f (y) + g(−x)g(y).

So u1 would be rational, impossible. So one of them, say 11 meets an infinite number 60 of times. As we have infinitely many zeroes and ones, the blocks 01 must also occur an infinite number of times. So if the l + 1th , l + 2th digits of u1 are 1 we get 34 < {2l u1 } < 1 but if l + 1th , l + 2th digits of 1 are 0 and 1 we have 14 < {2l u1 } < 12 . So {2l u1 } has infinitely many u members in the disjoint intervals [ 41 ; 12 ] and [ 34 ; 1] thus cannot converge. We derive an analogous contradiction if 00 meets infinitely many times.

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Functional Equations Electronic Edition 2007 by Titu Andreescu, Iurie Boreico


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