Show that $f(x) \le 2 \sqrt{(\varepsilon)} f(\frac{1}{\varepsilon} y)$ I'm trying to find a general condition under which $f(x) \le 2 \sqrt{(\varepsilon)} f(\frac{1}{\varepsilon} y)$  holds for $x<y$ and $\varepsilon > 0$. Perhaps a condition on the derivative?!
Any help would be great. 
 A: Let us assume continuity on $f$.
We have condition 
$$f(x)\leq 2\sqrt\varepsilon\, f\left(\frac y\varepsilon\right),\quad \forall y > x,\forall \varepsilon > 0\tag{1}$$ 
and by letting $\varepsilon = \frac 1 t$, we get equivalent condition
$$f(x)\sqrt t \leq 2f(ty),\quad \forall y>x,\forall t>0.\tag{2}$$
Let $(x_n)$ be strictly decreasing sequence such that $x_n\to x$. If we substitute $y = x_n$ into $(2)$ and take limit, by continuity of $f$ we get 
$$f(x)\sqrt t\leq 2f(tx),\quad\forall t>0.\tag{3}$$
Take $t = 1$ to get $f(x)\leq 2f(x)$, which implies that $f(x)\geq 0$, for all $x\in\Bbb R$. Now, let $x = 0$ to get $f(0)\sqrt t \leq 2f(0)$, for all $t>0$, which implies $f(0) = 0$. Now, for any $x<0$, let $y = 0$ in $(2)$ to get $f(x)\leq 0$. This means that $f(x) = 0$, for all $x\leq 0$.
Hence, it is not loss of generality to assume $f\colon\Bbb R_{\geq 0}\to \Bbb R_{\geq 0}$, $f(0) = 0$.
Now, if there exists $C\geq 0$ such that $$C\sqrt x\leq f(x)\leq 2C\sqrt x,\tag{4}$$ then condition $(1)$ is satisfied:
$$2\sqrt\varepsilon\,f\left(\frac y\varepsilon\right)\geq 2C\sqrt y\geq 2C\sqrt x\geq f(x),\quad \forall y>x,\forall \varepsilon >0.$$
We will show that condition $(4)$ is necessary as well. For $x>0$, define $g(x)=\frac{f(x)}{\sqrt x}$. Since for $x = 0$, condition $(3)$ is immediately satisfied, we have that $(3)$ is equivalent to 
$$g(x)\leq 2g(tx),\quad \forall x,t>0,$$ 
which is equivalent to $$g(x)\leq 2g(y),\quad \forall x,y>0,\tag{5}$$ 
or by symmetry $$\frac 12g(y)\leq g(x)\leq 2g(y),\quad\forall x,y>0.$$ 
Take any $y_0>0$ and let $C_0 = g(y_0)$.  We have $$\frac 12C_0\leq g(x)\leq 2C_0,\quad \forall x>0.$$ In particular, $g$ is bounded, and thus it makes sense to consider $S=\sup \{g(x)\mid x>0\}$ and $I=\inf \{g(x)\mid x>0\}$. If we show that $S\leq 2I$, we are done, since then we would have $I\leq g(x)\leq S = 2I$ and $I\geq 0$, i.e. $(4)$ would be true.
Choose sequences $(x_n)$ and $(y_n)$ such that $g(x_n)\to S$ and $g(y_n)\to I$. From $(5)$ we have $g(x_n)\leq 2g(y_n)$ and taking limit, we get $S\leq 2I$.
We conclude that continuous real functions on $\Bbb R$ that satisfy $(1)$ are precisely those that satisfy:

$(i)$ $f(x) = 0$, $x<0$
$(ii)$ $(\exists C\geq 0)\ C\sqrt x \leq f(x)\leq 2C\sqrt x$, $x\geq 0$

