Find all monotonic functions $u:\mathbb{R}\rightarrow\mathbb{R}$ which have the property that there exists a strictly monotonic function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that \[f(x+y)=f(x)u(x)+f(y) \] for all $x,y\in\mathbb{R}$. Vasile Pop
Problem
Source: Romanian TST 1998
Tags: function, algebra proposed, algebra, functional equation
pco
23.04.2011 23:17
WakeUp wrote: Find all monotonic functions $u:\mathbb{R}\rightarrow\mathbb{R}$ which have the property that there exists a strictly monotonic function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that \[f(x+y)=f(x)u(x)+f(y) \] for all $x,y\in\mathbb{R}$. Let $P(x,y)$ be the assertion $f(x+y)=f(x)u(x)+f(y)$ Subtracting $P(x,0)$ from $P(x,y)$, we get $f(x+y)=f(x)+f(y)-f(0)$ and so, since strictly increasing, $f(x)=ax+b$ with $a>0$ And so $x=(x+\frac ba)u(x)$ Setting $x=-\frac ba$, we get $b=0$ and so the solution : $u(x)=1$ $\forall x\ne 0$ and $u(0)=c$ any real
Dijkschneier
23.04.2011 23:28
Here is my solution :
Let u be a monotonic function which has that property.
P(x,y) : f(x+y) = f(x)u(x) + f(y)
P(0,0) ==> f(0)=f(0)u(0)+f(0)
==> f(0)=0 or u(0)=0
P(x,0) ==> f(x)=f(x)u(x)+f(0)
==> f(x)(1-u(x))=f(0)
If f(0)=0, then f(x)(1-u(x))=0, and because f is strictly monotonic, u(x) will always equal 1 except possibly in one point, and because it is monotonic, then we can conclude that it always equal 1.
Conversely, if $\forall x \in \mathbb{R} : u(x)=1$, then take f to be a monotonic solution to Cauchy's equation, to see that it works.
If $f(0)\neq 0$ and u(0)=0
Furthermore, we have by symetry : f(x)u(x)+f(y) = f(y)u(y)+f(x)
==> f(x)(u(x)-1) = f(y)(u(y)-1)
And so the function h(x)=f(x)(u(x)-1) is constant. (= c)
Suppose it is the allzero function. Then, because f is strictly monotonic, u(x) will always equals 1 except possibly in one point, and because it is monotonic, then we can conclude that it always equals 1.
But this is a contradiction with u(0)=0.
Now suppose that h it is not the allzero function. Then f(x) is never equal to zero, so it implies u(x)-1 = c/f(x), u(x)=c/f(x) + 1
Hence : f(x+y)=c+f(x)+f(y)
==> f(x+y) + c = (f(x)+c) + (f(y)+c)
==> g(x+y)=g(x)+g(y), where g(x)=f(x)+c (note that g is also monotonic)
==> g(x)=ax
==> f(x)=ax-c
But then, f(c/a)=0, and so we have a contradiction
Hence the unique solution :
$\forall x \in \mathbb{R} : u(x)=1$
Dijkschneier
23.04.2011 23:45
Dear pco, if you define $u(x)=1 \forall x\neq 0$ and $u(0)=c$, and $c\neq 1$, is u still monotonic ?
mavropnevma
24.04.2011 00:10
pco's solution emphasizes the weakness of this problem; the monotonicity of $u$ is never used, except at the very last moment, when having established that a function $u$ obeying the conditions must be $u(x)=1$ for all non-zero $x$, with arbitrary $u(0)$, hence in order to be monotonic needing having $u(0)$. No wonder Patrick forgot to enforce this last requirement
pco
24.04.2011 09:51
Thanks all for the remarks. Indeed, since it was useless for finding solutions (just to reduce the result), I really forgot the constraint $u(x)$ monotonic. Sorry for the error.
grupyorum
19.09.2018 00:20
There are two errors with this problem statement. The actual problem is, find all $u:\mathbb{R}\to\mathbb{R}$ (as opposed to monotonic $u$) such that, there exists a monotonic (strictly) $f(\cdot)$ such that $f(x+y)=f(x)u(y)+f(y)$ holds for every $x,y\in\mathbb{R}$.
Let $P(x,y)$ be the given assertion in the problem. Note that, $P(x-y,y)$ gives us $f(x)-f(y)=f(x-y)u(y)$. Now, if $u\equiv 0$ identically, we get that $f(x+y)=f(y)$, a contradiction. Now suppose that $x_0$ is such that $u(x_0)\neq 0$. Take $x=y=x_0$ above, and get, $f(0)=f(0)u(x_0)$, in particular, $f(0)=0$. Furthermore, taking $y=0$ in $f(x)-f(y)=f(x-y)u(y)$, we obtain $f(x)=f(x)u(0)$, and thus, $u(0)=1$.
We now look at $P(y,x)$ in the original assertion, and obtain,
$$
f(x+y)=f(x)u(y)+f(y)=f(y)u(x)+f(x) \implies f(x)(u(y)-1)=f(y)(u(x)-1).
$$In particular, since $f(0)=0$ and $f(\cdot)$ is strictly monotonic, we get that $f(x)\neq 0$ unless $x=0$; and therefore, we can safely obtain,
$$
\frac{u(x)-1}{f(x)}=\frac{u(y)-1}{f(y)}.
$$In particular letting $C$ to be the equality constant above, we get $u(x)-1=Cf(x)$, for every $x\neq 0$. Furthermore, for $x=0$, we have $u(0)-1=0=Cf(0)$, and thus, $u(x)-1=Cf(x)$ always holds.
Plugging this in, we get $u(x+y)=u(x)u(y)$. Finally, note that $u(\cdot)$ is monotonic, and after taking logs above we get $\log u(x+y)=\log u(x)+\log u(y)$, and letting $\log u = g$, we get $g(x+y)=g(x)+g(y)$. Since $g(\cdot)$ is monotonic, the only solution to this Cauchy equation is $g(x)=kx$, and thus, $u(x)=e^{kx}$ is obtained.
Using this, one can work backwards and recover the corresponding $f$. It is not hard at all to check that these $f(\cdot)$ found indeed enjoys the claim.
fukano_2
29.06.2020 05:43
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