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solucao para um problema



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