Decreasing positive function less than $x$? Is there a decreasing function $f$ defined on $(0,∞)$ such that $0<f(x)<x$?
I thought about it and couldn't come up with a conventional function. I was thinking that for all $x>0$ we can find (smallest) $n_x$ which satisfies $\frac{1}{n_x}<x$. If I can construct these $n_x$s such that $n_x>n_y$ whenever $x>y$ I will be done. But maybe I'm thinking too deep and the answer is much simpler than that. 
 A: I suppose that you require $f$ to satisfy $0 < f(x) < x$ for every $x \in \left( 0, \infty \right)$.
Then the answer is no. One way to see this is to note that for any fixed $x \in \left( 0, \infty \right)$ the estimate
$$ f(x) < f(n^{-1}) < n^{-1}$$
holds for every $n \in \mathbb{N}$ with $n > x^{-1}$.
But letting $n \to \infty$ then shows that $f(x) = 0$.
A: Pick $t \in (0, \infty)$ such that $t < \min(f(1), 1)$. If such $f$ exists, we have $f(t) \geq f(1)>t $ since $f$ is decreasing. On the other hand, we have $0<f(t) < t $. Thus we have $$0<f(t)<t<f(1)\leq f(t)$$
which is absurd.
A: How could it be?
Let $x > 0$. what is $f(x)$?.  As the function is decreasing, then  for any $\epsilon; 0 < \epsilon < x$ we must have $f(x) \le f(\epsilon)$.  But your condition that $0 < f(x) < x$ for all $x$ means $f(\epsilon) < \epsilon$.
So for any $x$ we must have $f(x)\le f(\epsilon) < \epsilon$ for any positive $\epsilon$.  But your condition is that $f(x) > 0$ so that means $f(x) < f(x)$ itself.
(Note:  Is a well-known that if $a < \epsilon$ for any positive $\epsilon$ that $a \le 0$.  Because if $a > 0$ then $a < a$ itself.)
