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Saudac,o~es,
Ai' segue a soluc,a~o do prof. Rousseau para um dos
problemas
da lista.
Dear Luis:
Perhaps I have overlooked something, but it seems at first glance that the only solutions of the functional equation x f(x) = [x] f({x}) + {x} f([x]) are f(x) = c where c is a constant. Let f(0) = c. First, let us prove that f(n) = c for every integer n. If n is any nonzero integer, then [n] = n and [n] = 0, so n f(n) = n f(0) + 0 f(n), and thus f(n) = f(0) = c. Next, let us prove that f(x) = c for x > 0. First, if 0 < x < 1 then x f(x) = 0 f(x) + x f(0), so f(x) = f(0) = c. If n < x < n+1 where n > 0, then {x} = x-n is between 0 and 1, so x f(x) = n f(x-n) + (x-n) f(n) = n c + (x-n) c = xc and f(x) = c. The proof that f(x) = c for all x < 0 goes in a similar way. Then -n < x < -(n-1) for some positive integer n and {x} = x+n is between 0 and 1, so f(x+n) = c. Thus x f(x) = (-n) f(x+n) + (x+n) f(-n) = (-n) c + (x+n) c = xc and f(x) = c. Hence the only solutions are the constant ones. Cecil |