Sum of a series involving integral

Question

Let $0<2a<b<\infty$ and $f:(0,\infty)\to(0,\infty)$ be a convenient function. Intuitively, I guess that

$$ \sum_{k=-\infty}^{-1}\int_{2^{k}a}^{2^{k}b}\frac{f(t)}{t}dt=\int_{0}^{b/2} \frac{f(t)}{t}dt. $$ But i can not prove that. Can anyone help me?

Answer 1

In fact, for any $a$ satisfying $0 < 2a < b < \infty$, there is no integrable function $f > 0$ which satisfies your equality.

We can say that $$\sum_{k=-\infty}^{-1} \int_{2^{k-1}b}^{2^kb} \frac{f(t)}{t}\;dt = \int_0^{b/2}\frac{f(t)}{t}\;dt.$$

Now, let $2a < b$. Then, $2^k a < 2^{k-1}b$, so $$\sum_{k=-\infty}^{-1} \int_{2^ka}^{2^kb} \frac{f(t)}{t}\;dt = \sum_{k=-\infty}^{-1}\left( \int_{2^{k-1}b}^{2^kb} \frac{f(t)}{t}\;dt + \int_{2^ka}^{2^{k-1}b} \frac{f(t)}{t}\;dt \right)$$ $$= \int_0^{b/2} \frac{f(t)}{t}\;dt + \sum_{k=-\infty}^{-1} \int_{2^ka}^{2^{k-1}b} \frac{f(t)}{t}\;dt$$

Thus, the inequality can only hold if $\sum_{k=-\infty}^{-1} \int_{2^ka}^{2^{k-1}b} \frac{f(t)}{t}\;dt = 0$. However, since $f > 0$, each term in the sum is positive, so this is not possible.