Sum of a series involving integral II

Question

Let $0<2a<b<\infty$ and $f:(0,\infty)\to(0,\infty)$ be a convenient function. I see an inequality in a paper such that

$$ \sum_{k=-\infty}^{-1}\int_{2^{k}a}^{2^{k}b}\frac{f(t)}{t}dt=\int_{0}^{b}\left(\sum_{k=-\infty}^{-1}\chi_{[2^{k}a,~2^{k}b]}(t)\right) \frac{f(t)}{t}dt\le C \int_{0}^{b} \frac{f(t)}{t}dt. $$ But i can not see how this equality $$ \sum_{k=-\infty}^{-1}\int_{2^{k}a}^{2^{k}b}\frac{f(t)}{t}dt=\int_{0}^{b}\left(\sum_{k=-\infty}^{-1}\chi_{[2^{k}a,~2^{k}b]}(t)\right) \frac{f(t)}{t}dt $$ and this inequality $$ \int_{0}^{b}\left(\sum_{k=-\infty}^{-1}\chi_{[2^{k}a,~2^{k}b]}(t)\right) \frac{f(t)}{t}dt\le C \int_{0}^{b} \frac{f(t)}{t}dt $$ holds. Can anyone help me to understand these facts?

Answer 1

For all $k\leq -1$, we have
\begin{align*} \int_{2^{k}a}^{2^{k}b}\dfrac{f(t)}{t}dt=\int_{0}^{b}\chi_{[2^{k}a,2^{k}b]}(t)\dfrac{f(t)}{t}dt \end{align*} Now note that $f$ is nonnegative function, by Monotone Convergence Theorem, \begin{align*} \sum_{k=-\infty}^{-1}\int_{2^{k}a}^{2^{k}b}\dfrac{f(t)}{t}dt&=\sum_{k=-\infty}^{-1}\int_{0}^{b}\chi_{[2^{k}a,2^{k}b]}(t)\dfrac{f(t)}{t}dt\\ &=\int_{0}^{b}\sum_{k=-\infty}^{-1}\chi_{[2^{k}a,2^{k}b]}(t)\dfrac{f(t)}{t}dt. \end{align*} Now let $n_{0}\geq 1$ be so large that $\dfrac{b}{2^{n_{0}}}<a$, then $I_{k}:=[2^{-k}a,2^{-k}b]$ for all $k\geq 1$ are such that $I_{k+n}\cap I_{k}=\emptyset$, $I_{k-n}\cap I_{k}=\emptyset$ for all $n\geq n_{0}$ with the convention that $I_{m}=\emptyset$ for $m\leq 0$, so if $t\in[0,b]$ is such that $t\in I_{k}$ for some $k$, then \begin{align*} \sum_{k=-\infty}^{-1}\chi_{[2^{k}a,2^{k}b]}(t)\leq \chi_{I_{k+n_{0}-1}}(t)+\cdots+\chi_{I_{k-n_{0}+1}}(t)\leq 2n_{0}. \end{align*}

More details: $I_{k-n}\cap I_{k}=\emptyset$ for all $n\geq n_{0}$. Assume that $k-n\geq 1$, then we have \begin{align*} \dfrac{b}{2^{k}}=\dfrac{b}{2^{k-n}\cdot 2^{n}}<\dfrac{a}{2^{k-n}}, \end{align*} so \begin{align*} I_{k-n}\cap I_{k}=\left[\dfrac{a}{2^{k}},\dfrac{b}{2^{k}}\right]\bigcap\left[\dfrac{a}{2^{k-n}},\dfrac{b}{2^{k-n}}\right]=\emptyset. \end{align*} Similar reasoning applies for $I_{k+n}\cap I_{k}$.

Now for a fixed $t$, find some $k$ such that $t\in I_{k}$. Since $I_{k-n}\cap I_{k}=I_{k+n}\cap I_{k}=\emptyset$ for all $n\geq n_{0}$, so $t\notin I_{k-n},I_{k+n}$ for all such $n$, then the sum \begin{align*} \sum_{l=1}^{\infty}\chi_{I_{l}}(t) \end{align*} reduces to at most of $\chi_{I_{k-n_{0}+1}},...,\chi_{k+n_{0}-1}$, and these summands sum up to at most $2n_{0}$.

Answer 2

Hint. There is a positive integer $N$ such that $b/2^N<a$. Then for any $n\geq N$, and any positive integer $k$ (I prefer to consider positive indices), $$\frac{b}{2^{n+k}}<\frac{a}{2^{k}}$$ which implies that $[a/2^{n+k},b/2^{n+k}]\cap[a/2^{k},b/2^{k}]=\emptyset$. Now, given a positive integer $n$, is the non-negative step function $$\sum_{k=1}^{n}\chi_{[a/2^{k},b/2^{k}]}(t)$$ bounded over $[0,b]$ by some constant $C$ which does not depend on $n$ (but it could be related to $N$)? Recall that given $t$, then $\sum_{k=1}^{n}\chi_{[a/2^{k},b/2^{k}]}(t)$ counts the number of intervals $[a/2^{k},b/2^{k}]$ which contains $t$.

If so then for any positive integer $n$, $$\sum_{k=1}^{n}\int_{a/2^{k}}^{b/2^{k}}\frac{f(t)}{t}dt= \sum_{k=1}^{n}\int_{0}^{b}\chi_{[a/2^{k},b/2^{k}]}(t)\frac{f(t)}{t}dt\\=\int_{0}^{b}\left(\sum_{k=1}^{n}\chi_{[a/2^{k},b/2^{k}]}(t)\right) \frac{f(t)}{t}dt\le C \int_{0}^{b} \frac{f(t)}{t}dt.$$

P.S. $0<2a<b$ implies that $$\frac{a}{2^{k+1}}<\frac{a}{2^{k}}<\frac{b}{2^{k+1}}<\frac{b}{2^{k}}$$ Hence $[a/2^{k+1},b/2^{k+1}]\cap[a/2^{k},b/2^{k}]\not=\emptyset$ and $$\bigcup_{k=1}^{\infty}[a/2^{k},b/2^{k}]=(0,b/2].$$ So $1\leq \sum_{k=1}^{n}\chi_{[a/2^{k},b/2^{k}]}(t)$ for any $t\in (0,b/2]$.