| Continuing from .1, the question is whether there is a
second continuous f:R->R such that f(f(x))+f(x)=2x+b for
some b>0. From .1 f must be 1-1 and so either increasing
or decreasing. Also, f cannot be bounded above or below
as then f(f(x)) + f(x) would be bounded as well, and 2x+b
is not bounded. So f:R->R would be both 1-1 and onto, i.e.,
a bijection.
If g and h :R->R are continuous bijections, one increasing
and one decreasing, then their graphs must intersect, i.e.,
there is a c such that g(c) = h(c).
So if there is a decreasing solution f, then there must be
point where the graph of f crosses y=x. Let c be the point
where f(c) = c. Then f(f(c))+f(c) = f(c)+c = c+c = 2c which
cannot equal 2c+b because b>0. So if there is a continuous
f:R->R such that f(f(x))+f(x)=2x+b (b>0), then f must be an
increasing function.
Dan
|
| The answer in _Crux_ shows that f(x)=x+b/3 is the unique continuous
solution when a=2.
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