# Comparing two condition

Let $\rho:(0,\infty)\to (0,\infty)$ be a measurable function.

The condition 1 for the function $\rho$ is: There exist constants $C_1>0$ and $0<2k_1<k_2<\infty$ such that $$\sup_{r/2<s\le r}\rho(s)\leq C_1 \int_{k_1 r}^{k_2 r}\frac{\rho(s)}{s}ds,\qquad r>0.$$

The condition 2 for the function $\rho$ is: There exist constant $C_2>0$ such that $$\frac{1}{C_2}\le \frac{\rho(r)}{\rho(s)}\le C_2$$ whenever $r$ and $s$ satisfy $$r,s>0 \text{ and }\frac{1}{2}\le \frac{r}{s}\le 2.$$ The claim is "conditon 1 is weaker than condition 2, i.e., condition 2 implies condition 1."

I can not prove this claim. Can anybody help me to prove this claim?

## 1 Answer

Condition 2 implies that

$$\rho(s) \leqslant C_2 \rho(r)$$

for $\frac{r}{2} \leqslant s \leqslant r$, from which

$$\sup_{\frac{r}{2} \leqslant s \leqslant r} \rho(s) \leqslant C_2 \rho(r)$$

follows immediately. On the other hand, choosing $k_2 = 1$ and $0 < k_1 < \frac{1}{2}$ yields

\begin{align} \int_{k_1r}^{k_2r} \frac{\rho(s)}{s}\,ds &> \int_{r/2}^r \frac{\rho(s)}{s}\,ds \\ &\geqslant \frac{\rho(r)}{C_2}\int_{r/2}^r \frac{ds}{s} \\ &= \frac{\log 2}{C_2}\rho(r) \end{align}

using the inequality

$$\rho(s) \geqslant \frac{\rho(r)}{C_2}$$

for $\frac{r}{2} \leqslant s \leqslant r$ implied by condition 2. Putting both pieces together, we have

$$\sup_{\frac{r}{2} \leqslant s \leqslant r} \rho(s) \leqslant C_1 \int_{k_1r}^{k_2r} \frac{\rho(s)}{s}\,ds$$

with

$$C_1 = \frac{C_2^2}{\log 2}$$

and $0 < 2k_1 < k_2 = 1 < \infty$, that is, we have condition 1.