There is no function $f:\Bbb R\to \Bbb R$ such that $f(y)\le a f(x)-b\ln{(y-x)},\forall xLet $a>1,b>0$. Prove that there is no function $f:\Bbb R\to\Bbb R$ such that
$$f(y)\le af(x)-b\ln{(y-x)},\forall x<y$$
Some of my thoughts：Let $y=x+1$,then we have
$$f(x+1)\le af(x)$$so we have
$$f(x+n)\le af(x+n-1)\le a^2f(x+n-2)\le\cdots \le a^{n}f(x)$$
 A: If we choose a constant $x$, we can see from the condition, that the values for $f(y)$ can be arbitrarily small as $y\to\infty$.
Thus, let us choose $x_1\in\mathbb R$ such that $f(x_1)<-c_1$ for a constant $c_1>0$.
We recursively define $x_{k+1}:= x_k+2^{-k}$.
Then from the condition we observe that
$$
f(x_{k+1}) \leq a f(x_k) - b \ln(2^{-k}) = a f(x_k) + k b \ln 2.
$$
Now it is time to choose our constant $c_1$.
First we define $\gamma:=2b \ln(2)/(a-1)$.
Then we define $c_1:= (b\ln(2)+\gamma)/(a-1)$.
Recursively, we define $c_{k+1}:= a c_k-kb\ln(2)$.
By induction, one can show that
$$
 kb\ln(2) +\gamma \leq (a-1) c_k
$$
holds.
It follows that
$$
 c_{k+1} \geq \gamma+ c_k
$$
holds.
In particular, we have $c_k\to\infty$.
Again, by induction it can be shown that
$$
 f(x_k)\leq - c_k
$$
holds.
Thus $f(x_k)\to-\infty$ as $k\to\infty$.
Finally, we consider $y:=x_1+2$.
Note that $y\geq x_k+1$ for all $k$.
Then, by the original main condition we have
$$
 f(y) \leq a f(x_k) - b \ln(y-x_k) \leq a f(x_k) \leq -a c_k \to -\infty
$$
Thus $f(y)$ has to be arbitrarily small, which is a contradiction to being real valued.
