Given $m\in\mathbb{N}$. Find all functions $f:\mathbb{R^{+}}\rightarrow\mathbb{R^{+}}$ such that $$f(f(x)+y)-f(x)=\left( \frac{f(y)}{y}-1\right)x+f^m(y)$$holds for all $x,y\in\mathbb{R^{+}}.$
($f^m(x) =$ $f$ applies $m$ times.)
Denote the assertion of the problem by $P(x, y)$.
Obviously $f \equiv \textbf{Id}$ is a solution. So, now assume $f$ is not the identity function. That means there exist some $y_0 \in \mathbb{R}^+$ such that $\frac{f(y_0)}{y_0} - 1 \neq 0$. We will establish some properties of $f$ first:
If $f(x_1) = f(x_2)$, comparing $P(x_1, y_0)$ and $P(x_2, y_0)$ gives us $\left(\frac{f(y_0)}{y_0}-1\right)(x_1-x_2) = 0 \implies x_1 = x_2 \implies f$ is injective.
Let $f(x) \le M$ for every $x \in \mathbb{R}^+$. From $P(x, y_0)$:
\[ x = \frac{|f(f(x)+y_0)-f(x)-y_0|}{|\frac{f(y_0)}{y_0}-1|}. \]The right-hand side is bounded by $\frac{y_0+2M}{|\frac{f(y_0)}{y_0}-1|}$, while the left-hand side is not. Therefore, $f$ is unbounded.
Let $F$ be the set of fix points of $f$, and let $\alpha \in F$. From $P(x, \alpha)$, we obtain $f(x) + \alpha \in F$. From $P(\alpha, f(y))$, we obtain:
\[ f(\alpha + f(y)) - \alpha = \left(\frac{f(f(y))}{f(y)}-1\right)\alpha + f^{m+1}(y) \implies f(\alpha + f(y)) = \frac{f(f(y))\alpha}{f(y)} + f^{m+1}(y). \]Therefore, $\frac{f(f(y))\alpha}{f(y)} + f^{m+1}(y) = f(\alpha + f(y)) = \alpha + f(y) \implies f(y)-f^{m+1}(y) = \alpha\left( \frac{f(f(y))}{f(y)} - 1 \right)$. Since we can change $\alpha$ to another fixed point (such as $\alpha + f(x)$ for any $x \in \mathbb{R}^+$), therefore the factors must be zero, i.e. $f(y) = f(f(y)) = f^{m+1}(y)$. Now, by subtracting $P(x, x)$ from $P(f(x), x)$, we obtain: $\left(\frac{f(x)}{x}-1\right)(f(x)-x) = 0 \implies f(x) = x$.
From $P(x, f(y))$, we obtain:
\[ f(f(x)+f(y)) - f(x) = \left(\frac{f(f(y))}{f(y)} - 1\right)x + f^{m+1}(y) \implies f(f(x)+f(y)) = f(x) + \left(\frac{f(f(y))}{f(y)} - 1\right)x + f^{m+1}(y). \]Interchanging $x$ and $y$ from the last equation gives us:
\[ f(x) + \left(\frac{f(f(y))}{f(y)} - 1\right)x + f^{m+1}(y) = f(y) + \left(\frac{f(f(x))}{f(x)} - 1\right)y + f^{m+1}(x) \text{. } (1) \]Denote the above equation by $Q(x,y)$. From $Q(x, 1)$ and $Q(x, 2)$, we obtain the following:
\[ Q(x, 1): f(x) + C_1x + D_1 = \frac{f(f(x))}{f(x)} + f^{m+1}(x) \]\[ Q(x, 2): f(x) + C_2x + D_2 = 2 \cdot \frac{f(f(x))}{f(x)} + f^{m+1}(x) \]for some constants $C_i$ and $D_i$. Subtracting them gives us $\frac{f(f(x))}{f(x)}$ is a linear function in $x$. Now we divide into two cases:
Case 01. $\frac{f(f(x))}{f(x)}$ is constant.
Let $f(f(x)) = cf(x)$, for every $x \in \mathbb{R}^+$. If $c=1$, injectivity gives us $f(x) = x$. Now, assume $c \neq 1$. By induction, it's easy to see that $f^{n}(x) = c^{n-1}f(x)$. Plugging this to $(1)$ gives us:
\[ f(x) + x(c-1) + c^mf(y) = f(y) + y(c-1) + c^mf(x) \implies (c^m-1)f(x) - (c-1)x = (c^m-1)f(y) - (c-1)y, \]which means $(c^m-1)f(x) - (c-1)x$ is constant, i.e. $f(x) = \frac{(c-1)x+d}{c^m-1}$ for some other constant $d$. Plugging $x \mapsto f(x)$ to the last equation, we obtain:
\[ cf(x) = f(f(x)) = \frac{(c-1)f(x)+d}{c^m-1} \implies d = 0.\]This means $f(x) = ex$ for some constant $e$. Plugging to the original equation gives us $e = 1$, or $f(x) = x$. A contradiction since $f$ is not the identity function.
Case 02. $\frac{f(f(x))}{f(x)}$ is non-constant linear in $x$.
Let $f(f(x)) = f(x)(cx+d)$ for some constant $c \neq 0, d$. $c$ must be positive, and $d$ must be non-negative, otherwise for some positive real $x$, $cx + d \le 0$. Now, we will prove the following claim:
Claim. $d < 1$.
Proof. Assume for the sake of contradiction that $d \ge 1$. Since $f$ is unbounded, there exist $t$ such that $f(t)>1$. Denote $A = \{x | f(x) > 1 \}$. We know that $f(f(t)) = (ct+d)f(t) > ct+d > 1$ and $f(f(f(t)))) = (cf(t)+d)f(f(t)) > (c+d)f(f(t)) > 1$. By simple induction we obtain $f^{n}(t) = (cf^{n-2}(t) + d)f^{n-1}(t) > (c+d)^{n-2}f(f(t)) > 1$ for every $n \ge 3$ and $t \in A$. From $Q(f(t),1)$, but now with the unmasked constant, we obtain:
\begin{align*}
f(f(t)) + \left(\frac{f(f(1))}{f(1)}-1\right) f(t) + (f^{m+1}(1)-f(1)+1) = \frac{f(f(f(t)))}{f(f(t))} + f^{m+1}(f(t)) &> \frac{f(f(f(t)))}{f(f(t))} + (c+d)^{m}f(f(t)) \\
\implies f(f(t)) + (c+d-1)f(t) + D_1 &> (cf(t)+d) + (c+d)^{m}f(f(t)) \\
\implies (d-1)f(t) + D_1 &> d + ((c+d)^{m}-1)f(f(t)) = d + ((c+d)^{m}-1)(ct+d)f(t) \\
\implies D_1 &> d + (((c+d)^{m}-1)(ct+d) - (d-1))f(t),
\end{align*}for every $t \in A$. Since $f^n(t) > 1$, so $f^n(t) \in A$, and so by substituting $t \mapsto f^n(t)$ in the equation above, we obtain:
\[ D_1 > d + (((c+d)^{m}-1)(cf^n(t)+d) - (d-1))f^{n+1}(t) > (((c+d)^{m}-1)(cf^n(t)+d) - (d-1))(c+d)^{n-1}f(f(t)), \]which is false for large enough $n$. Therefore, the claim is proven. $\blacksquare$
From the claim above, we get $d < 1$. That means, $u = \frac{1-d}{c} \in \mathbb{R}^+$. We get $f(f(u)) = (cu+d)f(u) = f(u)$. Therefore, $f(u)$ is a fixed point. But by the third property of $f$ above, this means $f$ is an identity function.
Therefore, the answer is only $\boxed{f(x) = x}$ for each $x \in \mathbb{R}^+$. $\square$
$P(x)=x^4+2x^3+x^2+x-1$, $P(0)<0$ and $P(\frac{1}{2})>0$
Suppose $\alpha$ is the max root in$(0,\frac{1}{2} )$($\approx 0.48402$ in fact)
set $k=0.49,k\in (\alpha ,\frac{1}{2}),k^4+2k^3+k^2+k>1$
and set $N$ is arbitary large positive number
Part1: $\forall x \in \mathbb{R^+}$, $f(x)\ge x$
Proofelse, $\exists f(t)<t$
Case1: $\exists \frac{f(t)}{t} <1-k$
$f(x)=f(f(x)+t)+(1-\frac{f(t)}{t})x-f^m(t)$
$\forall x>\frac{f^m(t)}{1-k-\frac{f(t)}{t}}$
$f(x)=f(f(x)+t)+kx+(1-k-\frac{f(t)}{t})x-f^m(t)$
$>f(f(x)+t)+kx$
We also obtain $f(x)>kx$
and $\forall x>\frac{f^m(t)}{k(1-k-\frac{f(t)}{t})} \Rightarrow f(x)+t>kx>\frac{f^m(t)}{1-k-\frac{f(t)}{t}}$
So $f(f(x)+t)>k(f(x)+t)>k^2x$
We have proved $f(x)>f(f(x)+t)+kx>(k^2+k)x$
Do the silimar thing once more :
$\forall x>\frac{f^m(t)}{k^2(1-k-\frac{f(t)}{t})},f(x)+t>\frac{f^m(t)}{k(1-k-\frac{f(t)}{t})}$
$f(f(x)+t)>(k^2+k)(f(x)+t)>(k^2+k)f(x)>(k^4+2k^3+k^2)x$
$f(x)>f(f(x)+t)+kx>(k^4+2k^3+k^2+k)x>x$
$\forall x>\frac{f^m(t)}{k^3(1-k-\frac{f(t)}{t})},f(x)+t>\frac{f^m(t)}{k^2(1-k-\frac{f(t)}{t})}$
$\therefore f(f(x)+t)-f(x)>(k^4+2k^3+k^2+k-1)f(x)>(k^4+2k^3+k^2+k-1)x$
$\therefore (\frac{f(t)}{t}-1)x+f^m(t)>(k^4+2k^3+k^2+k-1)x$
$f^m(t)>(k^4+2k^3+k^2+k-\frac{f(t)}{t})x$
Let$x>\frac{f^m(t)}{k^4+2k^3+k^2+k-\frac{f(t)}{t}}$ leads to contradiction.
Case2: $\forall x$ , $\frac{f(x)}{x}\ge 1-k$ and$1-k\le \frac{f(t)}{t}<1$
We will make it contradict $f(t)<t$ through the following approach , $\forall N>0$
a) $\exists x>N,f(x)<x$
b) $\forall d_2\in (0,1-\frac{f(t)}{t}),\exists x>N,\frac{f(x)}{x}<\frac{f(t)}{t}+d_2$
c) $\forall d_3\in (0,\frac{t}{f(t)}-1),\exists x>N,\frac{f(x)}{x}<2-\frac{t}{f(t)}+d_3$
d) $\forall n\in \mathbb{N^+},\forall d_4>0,\forall \frac{f(x)}{x}>\frac{n}{n+1}-d_4$
a) $\forall x>\frac{f^m(t)}{1-\frac{f(t)}{t}}$ ,
$f(f(x)+t)=f(x)-(1-\frac{f(t)}{t})x+f^m(t)<f(x)<f(x)+t$
and$f(x)+t>(1-k)x+t$
Additionally, let $x>\frac{N}{1-k}$
$\therefore f(x)+t>N ,f(f(x)+t)<f(x)+t$
i.e. $\exists x>N,f(x)<x$
b) Otherwise, $\forall d_2\in (0,1-\frac{f(t)}{t}),\exists N>0$ s.t. $\forall x>N,\frac{f(x)}{x}\ge \frac{f(t)}{t} +d_2$ .
$\forall x>\frac{N}{1-k},f(x)+t>N$
$(\frac{f(t)}{t}-1)x+f^m(t)=f(f(x)+t)-f(x)$
$\ge (\frac{f(t)}{t}+d_2)(f(x)+t)-f(x)$
$>(\frac{f(t)}{t}+d_2-1)f(x)$
$\forall x>max\{\frac{N}{1-k},\frac{f^m(t)}{d_2}\}$
$(\frac{f(t)}{t}+d_2-1)f(x)<(\frac{f(t)}{t}-1)x+f^m(t)$
$<(\frac{f(t)}{t}-1+d_2)x$
$\Rightarrow f(x)>x$ for every large $x$ , contradiction.
c) Otherwise, $\forall d_3\in (0,\frac{t}{f(t)}-1),\exists N>0,\forall x>N,\frac{f(x)}{x}\ge 2-\frac{t}{f(t)}+d_3$
$\forall x>\frac{N}{1-k},f(x)+t>N$
$(\frac{f(t)}{t}-1)x+f^m(t)=f(f(x)+t)-f(x)$
$\ge (2-\frac{t}{f(t)}+d_3)(f(x)+t)-f(x)$
$>(1-\frac{t}{f(t)}+d_3)f(x)$
Wish$(\frac{f(t)}{t}-1)x+f^m(x)<(1-\frac{t}{f(t)}+d_3)(\frac{f(t)}{t}+d_2)x$
Which lead to$\frac{f(x)}{x}>\frac{f(t)}{t}+d_2$ , and contradiction with b).
And it$\Leftrightarrow (d_2+d_2d_3-\frac{d_2t}{f(t)}+\frac{d_3f(t)}{t})x>f^m(t)$
Thus , in b) pick $d_2<\frac{d_3(f(t))^2}{t^2},\forall x>max\{\frac{f^m(t)}{d_2+d_2d_3},\frac{N}{1-k}\}$
They respectively provides $(\frac{d_3f(t)}{t}-\frac{d_2t}{f(t)})x>0,(d_2+d_2d_3)x>f^m(t)$
the ineq desired is proven through adding them up.
d) Prove it by induction:
$n=1,\frac{f(t)}{t}>1-k>\frac{1}{2}>\frac{1}{2}-d_4$
when $n=k+1$ , if $\frac{f(x)}{x}\ge 1$ obviously true
else, $\forall \frac{f(x)}{x}<1,\forall d_3,\exists x_0$ s.t. $\frac{f(x_0)}{x_0}<2-\frac{x}{f(x)}+d_3$
and we know from hypothesis $\frac{f(x_0)}{x_0}>\frac{k}{k+1}-d_4$
$\therefore 2-\frac{x}{f(x)}>\frac{k}{k+1}-d_3-d_4$ for $\forall d_3,d_4$
$\therefore 2-\frac{x}{f(x)}\ge \frac{k}{k+1}$
$\therefore \frac{f(x)}{x}\ge \frac{k+1}{k+2}$
$\therefore \forall n\in \mathbb{N^+}, f(t)<t,\frac{f(t)}{t}\ge \frac{n}{n+1}$
Let $n>\frac{\frac{f(t)}{t}}{1-\frac{f(t)}{t}}$
$\frac{f(t)}{t}<\frac{n}{n+1}$ , contradiction.
$\therefore x\in \mathbb{R^+},f(x)\ge x$
If there exists $f(t)>t$
Part2: $\forall x\in \mathbb{R^+},f(x)>x$
Proofelse $\exists f(z)=z$
$\forall x,f(f(x)+z)=(\frac{f(z)}{z}-1)x+f(x)+f^m(z)=f(x)+z$
for $x>N,f(x)+z\ge x+z>N$
i.e. $\exists x>N,f(x)=x$
$f(f(x)+t)=f(x)+(\frac{f(t)}{t}-1)x+f^m(t)$
$\exists u>f^m(t)+t,f(u)=u$
Let $x=\frac{u-f^m(t)-t}{\frac{f(t)}{t}-1}$
$\therefore f(f(x)+t)+t=f(x)+u$
$f(u)=u\Rightarrow f(f(x)+u)=(\frac{f(u)}{u}-1)x+f(x)+f^m(u)=f(x)+u$
i.e. $ f(f(f(x)+t)+t)=f(f(x)+t)+t$
$t<f(t)<f^m(t)+(\frac{f(t)}{t}-1)(f(x)+t)$
$=f(f(f(x)+t)+t)-f(f(x)+t)=t$
Part3: $\forall x\in \mathbb{R^+},f(x)=x$
Proofa) denote $\frac{f(t)}{t}=r,\exists x>N,\frac{f(x)}{x}<\sqrt{r}+1-\frac{1}{\sqrt{r}}$
b) $\forall d>0,\exists x>N,\frac{f(x)}{x}<1+d$
c) $\forall d>0,\exists x>N,\frac{f^m(x)}{x}<1+d$
d) $\forall d,y>0,\forall x>\frac{f^m(y)}{d},\frac{f(f(x)+y)}{f(x)+y}<\frac{f(y)}{y}+d$
a) else, $\exists N,\forall x>N,\frac{f(x)}{x}\ge \sqrt{r}+1-\frac{1}{\sqrt{r}}$
$(r-1)x+f^m(t)=f(f(x)+t)-f(x)$
$\ge (\sqrt{r}+1-\frac{1}{\sqrt{r}})(f(x)+t)-f(x)>(\sqrt{r}-\frac{1}{\sqrt{r}})f(x)$
$>(\sqrt{r}-\frac{1}{\sqrt{r}})(\sqrt{r}+1-\frac{1}{\sqrt{r}})x$
$=(r+\frac{1}{r}+\sqrt{r}-\frac{1}{\sqrt{r}}-2)x$
$x<\frac{f^m(t)}{\frac{1}{r}+\sqrt{r}-\frac{1}{\sqrt{r}}-1}=\frac{f^m(t)}{(\sqrt{r}-1)(1-\frac{1}{r})}$ is a constant, contradiction.
b) else $\forall d>0,\exists N>0,\forall x>N,\frac{f(x)}{x}\ge 1+d$
$\exists t_1>N,\frac{f(t_1)}{t_1}<\sqrt{r}+1-\frac{1}{\sqrt{r}}$
$(\frac{f(t_1)}{t_1}-1)<\frac{r-1}{\sqrt{r}}<(\frac{f(t)}{t}-1)\frac{1}{1+d}$
Repeating this will give us $\exists t_n,\frac{f(t_n)}{t_n}-1<\frac{r-1}{(\sqrt{1+d})^n}$
Pick $n$ s.t. $d(\sqrt{1+d})^n>r-1$ , contradiction.
c) else $\forall d>0,\exists N>0,\forall x>N,\frac{f^m(x)}{x}\ge 1+d$
for a fixed $x_0$ , $\forall d'>0,\exists a>N+f(x_0),\frac{f(a)}{a}<1+d'$
$f(f(x_0)+a-f(x_0))-f(x_0)=(\frac{f(a-f(x_0))}{a-f(x_0)}-1)x_0+f^m(a-f(x_0))$
$>0+(1+d)(a-f(x_0))$
$(1+d)(a-f(x_0))<f(a)-f(x_0)<(1+d')a-f(x_0)$
$\Leftrightarrow (d-d')(a-f(x_0))<d'f(x_0)$
Let $d'<d,a>\frac{d'f(x_0)}{d-d'}$ , contradiction.
d) else $\exists x>\frac{f^m(y)}{d}, \frac{f(f(x)+y)}{f(x)+y}\ge \frac{f(y)}{y}+d$
Since$\frac{f(f(x)+y)}{f(x)+y}=\frac{f(x)+(\frac{f(y)}{y}-1)x+f^m(y)}{f(x)+y}$
$f(x)+(\frac{f(y)}{y}-1)x+f^m(y)\ge (f(x)+y)(\frac{f(y)}{y}+d)$
$\therefore (\frac{f(y)}{y}-1)x+f^m(y)\ge (\frac{f(y)}{y}+d-1)f(x)>(\frac{f(y)}{y}+d-1)x$
$\Rightarrow f^m(y)>dx$ , contraditcion.
In the end, we prove that if $f(t)>t,f(f(x)+t)-f(x)=(\frac{f(x_0)}{x_0}-1)x+f^m(t)$ cannot hold.
Abstract?for $\frac{f(a)}{a}\rightarrow 1,\frac{f(f(x)+a)}{f(x)+a}=\frac{f(x)+(\frac{f(a)}{a}-1)x+f^m(a)}{f(a)+a}\rightarrow 1$
and in $f(f(x)+t)-f(x)=(\frac{f(t)}{t}-1)x+f^m(t)$
If $f(x)+t$ can be shown as $f(?)+a$ , we obtain $\frac{f(f(x)+t)}{f(x)+t}\rightarrow 1$
then $t\approx (\frac{f(t)}{t}-1)x+f^m(t)$ is linear about $x$ which could lead to contradition
$\forall d_1,N_1>0,\exists a>N_1,\frac{f(a)}{a}<1+d_1$
let $x=f(p)+a$ undetermined , $y=t$ , then we have
$f(f(f(p)+a)+t)=f(f(p)+a)+(\frac{f(t)}{t}-1)(f(p)+a)+f^m(t) \cdots (1)$
for the first term, let $x=p,y=a$
$f(f(p)+a)+t=f(p)+(\frac{f(a)}{a}-1)p+f^m(a)+t$
$\forall d_2>0,\exists b>f^m(a)+t,\frac{f^{\max\{m,1\}}(b)}{b}<1+d_2$
Let $p=\frac{b-f^m(a)-t}{\frac{f(a)}{a}-1}$ , $\therefore f(f(p)+a)+t=f(p)+b$
$\frac{p}{f^m(b)}>\frac{p}{b(1+d_2)}=\frac{b-f^m(a)-t}{b(1+d_2)(\frac{f(a)}{a}-1)}$
Let $d_2 +1<\frac{d_1}{\frac{f(a)}{a}-1} \Rightarrow \frac{p}{f^m(b)}>\frac{b-f^m(a)-t}{d_1b}$
$\forall d_3<\frac{1}{d_1}$ , let $b>\frac{f^m(a)+t}{1-d_1d_3} \Rightarrow \frac{p}{f^m(b)}>d_3$
and $b>(\frac{f(a)}{a}-1)d_3f^m(a)+f^m(a)+t$
$\Rightarrow \frac{p}{f^m(a)}=\frac{b-f^m(a)-t}{f^m(a)(\frac{f(a)}{a}-1)}>d_3$
$\forall x>d_3f^m(y),\frac{f(f(x)+y)}{f(x)+y}<\frac{f(y)}{y}+\frac{1}{d_3}$
We have $\frac{f(f(p)+a)}{f(p)+a}=\frac{f(f(p)+a)}{f(p)+a}<\frac{f(a)}{a}+\frac{1}{d_3}$
and $\frac{f(f(f(p)+a)+t)}{f(f(p)+a)+t}=\frac{f(f(p)+b)}{f(p)+b}<\frac{f(b)}{b}+\frac{1}{d_3}$
$\therefore (\frac{f(t)}{t}-1)(f(p)+a)=f(f(f(p)+a)+t)-f(f(p)+a)-f^m(t)$
$<(\frac{f(b)}{b}+\frac{1}{d_3})(f(f(p)+a)+t)-f(f(p)+a)-t$
$\le (\frac{f^{\max\{m,1\}}(b)}{b}+\frac{1}{d_3})(f(f(p)+a)+t)-f(f(p)+a)-t$
$<(1+d_2+\frac{1}{d_3})(f(f(p)+a)+t)-f(f(p)+a)-t$
$=(d_2+\frac{1}{d_3})f(f(p)+a)+(d_2+\frac{1}{d_3})t$
$<(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3})(f(p)+a)+(d_2+\frac{1}{d_3})t$
$\therefore (\frac{f(t)}{t}-1)(f(p)+a)<(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3})(f(p)+a)+(d_2+\frac{1}{d_3})t$
We require $(1+2d_1)d_1<\frac{f(t)}{t}-1,(d_2+d_1)(1+2d_1)<\frac{f(t)}{t}-1,$
$d_1<\frac12 ,2<d_3<\frac{1}{d_1} ,(d_2+\frac{1}{d_3})(1+d_1+\frac{1}{d_3})<\frac{f(t)}{t}-1$
$((\frac{f(t)}{t}-1)-(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3}))(f(p)+a)<(d_2+\frac{1}{d_3})t<t$
Let $b>\frac{(\frac{f(a)}{a}-1)t}{\frac{f(t)}{t}-1-(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3})}+f^m(a)+t$
$\Rightarrow p=\frac{b-f^m(a)-t}{\frac{f(a)}{a}-1}>\frac{t}{\frac{f(t)}{t}-1-(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3})}$
$\therefore t<((\frac{f(t)}{t}-1)-(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3}))p$
$<((\frac{f(t)}{t}-1)-(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3}))(f(p)+a)$
$<t$ , contradiction.
a little ridiculous_ (´_`」 ∠) _