Inequality of a non decreasing function

Question

I am trying to prove the following inequality

$f(2^kr)\leq \int_{2^kr}^{2^{k+1}r}\frac{f(t)}{t}dt$

The hypothesis is that $f$ is an almost non-decreasing function and $f:(0,\infty)\rightarrow(0,\infty)$. Any hints on how to approach that inequality?

Answer 1

Since $f$ is non-decreasing we have that:

$$\int_{2^kr}^{2^{k+1}r}f(2^kr)dt\leq \int_{2^kr}^{2^{k+1}r}f(t)dt$$

Now, let's divide both sides by $t$ inside the integral:

$$f(2^kr)\int_{2^kr}^{2^{k+1}r}\dfrac{1}{t}dt\leq \int_{2^kr}^{2^{k+1}r}\dfrac{f(t)}{t}dt$$

We can take out $f(2^kr)$ out of the integral because it does not depend on $t$. After solving the integral on the LHS we get:

$$\int_{2^kr}^{2^{k+1}r}\dfrac{1}{t}dt=\log(2^{k+1}r)-\log(2^kr)=(1+k)\log(2)+\log(r)-k\log(2)-\log(r)=\log(2)$$

which yields the inequality:

$$\log(2)f(2^kr)\leq \int_{2^kr}^{2^{k+1}r}\dfrac{f(t)}{t}dt$$

I know this isn't the inequality you were trying to show, but I couldn't find anything else.