| > Proposed by Gerd Baron, Technische Universit�t, Vienna, and Walther
> Janous, Ursulinengymnasium, Innsbruck, Austria.
>
> Determine all pairs (a,b) of nonnegative real numbers such that the
> functional equation
>
> f(f(x))+f(x) = ax+b
>
> has a unique continuous solution f:R->R.
The trial solution f(x) = cx + d leads to
f(f(x)) = c(cx + d) = c^2x + cd + d
and
f(f(x)) = (c^2x + cd + d) + (cx + d) = (c^2 + c)x + (c + 2)d
= ax + b
So you need to solve
c^2 + c = a
(c + 2)d = b
This has two solutions for all nonnegative a except a=2.
For example, a=0 leads to f(x) = b/2 and f(x) = -x + b.
The problem at a=2 is that the solutions become c = 1,-2.
c = 1 leads to
f(x) = x + d
f(f(x)) + f(x) = ((x + d) + d) + (x + d) = 2x + 3d
so f(x) = x + b/3 is one solution. But c = -2 leads to
f(x) = -2x + d
f(f(x)) + f(x) = (-2(-2x + d) + d) + (-2x + d)
= 4x - 2d + d - 2x + d
= 4x
So there are two linear solutions, therefore at least two
continuous solutions, for all (a,b) with a and b nonnegative
with the exception of the case a=2, b>0. On that open ray
there is at least one continuous solution.
Since the problem asks for those (a,b) (both nonnegative)
for which there is a unique continuous solution f:R->R,
the partial answer is that the points with a=2,b>0 are
still candidates but the rest of the points have been ruled
out.
Note also that if f(x) = f(y) (let r be the common value)
then f(f(x)) + f(x) = f(r) + r = f(f(y)) + f(y), and so
ax + b = ay + b, so a(x - y) = 0. Thus for all nonzero a
we have f(x) = f(y) implies x = y, i.e., for all nonzero a
any solution f must be injective (one-to-one). A continuous
injective f:R->R must be strictly increasing or strictly
decreasing. So that limits the search for a second continuous
solution for the cases a=2,b>0.
Dan
|
| 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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