Find all functions $u:R\rightarrow{R}$ for which there exists a strictly monotonic function $f:R\rightarrow{R}$ such that $f(x+y)=f(x)u(y)+f(y)$ for all $x,y\in{\mathbb{R}}$
Problem
Source: Azerbaijan Math Olympiad Training
Tags: algebra, functional equation, TST
pco
15.12.2019 20:47
Kamran011 wrote: Find all functions $u:R\rightarrow{R}$ for which there exists a strictly monotonic function $f:R\rightarrow{R}$ such that $f(x+y)=f(x)u(y)+f(y)$ for all $x,y\in{\mathbb{R}}$ Let $P(x,y)$ be the assertion $f(x+y)=f(x)u(y)+f(y)$ $P(0,x)$ $\implies$ $f(0)u(x)=0$ and so $f(0)=0$ (else we get $f(x+y)=f(y)$ and so $f(x)$ is not strictly monotonic. $P(x,0)$ $\implies$ $f(x)(u(0)-1)=0$ and so $u(0)=1$ (else $f(x)$ is not strictly monotonic If $u(t)=0$ for some $t$, then $P(x,t)$ implies $f(x+t)=f(t)$, impossible since $f(x)$ strictly monotonic. So $u(x)\ne 0$ $\forall x$ Subtracting $P(x,y)$ from $P(y,x)$, we get New assertion $Q(x,y)$ : $f(x)(u(y)-1)=f(y)(u(x)-1)$ If $u(t)=1$ for some $t\ne 0$ (so that $f(t)\ne 0$), then $Q(x,t)$ $\implies$ $u(x)=1$ $\forall x$ which indeed is a solution (choose $f(x)\equiv x$ for example). If $u(x)=1$ $\iff$ $x=0$ : $Q(x,1)$ $\implies$ $f(x)=c(u(x)-1)$ for some $c=\frac{f(1)}{u(1)-1}\ne 0$ Plugging this in original equation, we get $u(x+y)=u(x)u(y)$ Since we previously got $u(x)\ne 0$ $\forall x$, this easily implies $u(x)>0$ $\forall x$ (set $x=y$ And so $\ln u(x)$ is additive. So $u(x)=e^{g(x)}$ where $g(x)$ is some additive function. If $g(x)$ is not linear, graph of $g(x)$ is dense in $\mathbb R^2$ and so graph of $f(x)$ is dense in $\mathbb R\times (-c,+\infty)$ (or $(-\infty,-c)$ if $c<0$, in contradiction with $f(x)$ stricly monotonic. So $g(x)$ is linear and we got $\boxed{u(x)=e^{ax}\forall x}$ which indeed is a solution, whatever is $a\in\mathbb R$, choosing for example $f(x)=e^{ax}-1$ if $a\ne 0$ or $f(x)=x$ if $a=0$
Pathological
15.12.2019 20:58
We claim that the answer is all functions $u(x) \equiv a^x$ for some $a > 0.$
First let's demonstrate that these work. For $u \equiv 1$, simply take $f \equiv x$.
For $u \equiv a^x$ for $a > 1$, take $f(x) \equiv a^x - 1.$ This works as $a^{x+y} - 1 = (a^x - 1) a^y + a^y -1.$
For $u \equiv a^x$ for $a < 1$, take $f(x) \equiv -(a^x - 1).$ This works analogously as the previous construction.
Let's now show that these are the only such functions which work.
Let $P(x, y)$ be the assertion of the problem. We have from $P(0, 0)$ that $f(0) u(0) = 0$. From $P(x, 0)$ we get $f(x) = f(x)u(0) + f(0),$ i.e. $f(x) (1 - u(0))$ is constant. As $f$ is nonconstant, this implies $u(0) = 1$, and $f(0) u(0) = 0$ gives $f(0) = 0.$ As a result of $f$'s monotonicity, $f$ is nonzero for nonzero inputs. We hence have for all $x \neq 0$ that:
$$u(y) = \frac{f(x+y) - f(y)}{f(x)}.$$
Let $Q(x, y)$ be the above assertion, for $x \neq 0, y \in \mathbb{R}.$ We have from $Q(x, y) + Q(x+y, y)$ that for all $x \neq 0, -y$:
$$u(x+y) + u(y) = \frac{f(2x+y) - f(y)}{f(x)}. \qquad (1)$$
We have by $Q(x, x)$ that:
$$\frac{f(2x)}{f(x)} = u(x) + 1,$$
so using this we can turn $(1)$ into:
$$u(x+y) + u(y) = u(y) (u(x) + 1),$$
or
$$u(x+y) = u(x) u(y), \forall x \in \mathbb{R} / \{0, -y\}.$$
From $Q(-y, y) \cdot Q(y, -y)$ we get that the above actually holds when $x = -y$ as well.
Since it clearly holds for $x = 0$ ($u(0) = 1$), we get:
$$u(x+y) = u(x)u(y), \forall x, y \in \mathbb{R}. \qquad (2)$$
We claim that this implies that $u$ is exponential, i.e. there is some $t$ such that $u(x) = t^x.$
By taking advantage of pseudo-symmetry, we get from $P(x, y) - P(y, x)$ that $f(x)(u(y) - 1) = f(y)(u(x) - 1),$ so $y = 1$ gives us $f(x) (u(1) - 1) = f(1) (u(x) - 1).$ If $u(1) = 1$, then since $f(1) > f(0) = 0$ we get $u \equiv 1,$ which works.
Else, $u$ is obtained from $f$ by a suitable transformation (translation + dilation), which implies that $u$ must also be either monotonically increasing or decreasing. In either case, it's clear that $u$ must be exponential.
Iora
04.12.2022 20:51
Same solution as post #2, posting anyways Let $P(x,y)$ be the assertion of f.e as usual. Let the strictly monotonic property be shown as $(\clubsuit)$. Clearly $u(x) \neq 0$ and $f$ is injective. \begin{align*} &P(0,x):f(0)u(x)=0 \stackrel{(\clubsuit)}{\Rightarrow} f(0)=0 \\ &P(x,0): f(x)=f(x)u(0) \stackrel{(\clubsuit)}{\Rightarrow} u(0)=1 \\ &P(-x,x): f(-x)u(x)+f(x)=0 \Rightarrow u(x) = \frac{ -f(x)}{f(-x)} \qquad &(\spadesuit) \\ &P(x,y)-P(y,x)= f(x)\big(u(y)-1 \big)= f(y)\big( u(x)-1 \big) \Rightarrow \frac{f(x)}{f(y)}= \frac{u(x)-1}{u(y)-1} &\qquad (1) \end{align*}Assume $u(a)=1$ for some $a \neq 0$. By $(\clubsuit)$ if $x \neq 0$ in $(\spadesuit)$ we have $f(x) \neq f(-x)$. if $-f(x)=f(-x)$, or that is, $f$ is odd function, analyzing $P(-x,x)$ once more yields $$ f(x) \big( -u(x)+1\big)=0$$Since $f$ is obviously not constant, we must have $u(x)=1$ which is a clear solution. Hence, for sake of finding another solutions, assume $u(a)=1$ only if $a=0$. Observe $(1)$ now. We have $f(x)= au(x)-a$ for some $a$ such that $ a \neq 0$. Put into the original equation and we have $$u(x+y)= u(x)u(y)$$ Which is a well known problem and the only solutions are $u(x)=e^{ax}$ for some parameter $a$ which indeed works, hence we are done.