Can someone show me how to calculate this inequality? When $2^{k}\geq(2R)^{-1}$, how do we show that
$$\sum_{2^j\le 1/2^k|y|}2^{j+k}|y|+\sum_{1/2^k|y|\le 2^j\le 2R/|y|}1+\sum_{2^j\ge 2R/|y|}(2^{j+k}|y|)^{-1}\leq C\left(\left|\log R\right|+\left|k\right|\right)$$
where $C>0$ is some absolute constant independent of $R>0$ and $k\in\mathbb{Z}$.
It's from L. Grafakos' Classical Fourier Analysis, Second Edition, p.378.
 A: Normal Human's comment is how to deal with the first and third sums. Let $j_{max}\in\mathbb{Z}$ be the largest integer $j$ such that $2^{j}\leq (2^{k}\left|y\right|)^{-1}$. Then
$$\sum_{2^{j}\leq(2^{k}\left|y\right|)^{-1}}2^{j+k}\left|y\right|=2^{k}\left|y\right|\sum_{j=-\infty}^{j_{max}}2^{j}=2^{j_{max}+k}\left|y\right|\sum_{j=0}^{\infty}2^{j}=2^{j_{max}+k+1}\left|y\right|\leq 2$$
Similarly, let $j_{min}\in\mathbb{Z}$ be the minimal integer $j$ such that $2^{j}\geq 2R/\left|y\right|$. Then
$$\sum_{2^{j}\geq 2R/\left|y\right|}(2^{j+k}\left|y\right|)^{-1}=(2^{k}\left|y\right|)^{-1}\sum_{j=j_{min}}^{\infty}2^{-j}=2^{-j_{min}}(2^{k}\left|y\right|)^{-1}\sum_{j=0}^{\infty}2^{-j}\leq 2(2^{k}\left|y\right|)^{-1}(\left|y\right|/2R)\leq 2$$
If $\left|k\right|\geq 1$, then $2\leq 2(\left|\log R\right|+\left|k\right|)$. If $k=0$, then the hypothesis $2^{k}\geq (2R)^{-1}$ implies that $R\geq 1/2$. Whence, $\left|\log R\right|+\left|k\right|\geq\left|\log 2\right|$. We conclude that both sums are $\leq (2/\log 2)(\left|\log R\right|+\left|k\right|)$.
In the second term, taking logarithms, we see that
$$-(k\log 2+\log\left|y\right|)\leq j\log 2\leq\log R+\log 2-\log\left|y\right|$$
So the $j$ such that $2^{j}$ is in the above range lie in an interval of length $$\dfrac{\log R+(k+1)\log 2}{\log 2}\leq (\log 2)^{-1}\left|\log R\right|+\left|k\right|+1$$
So the second sum is at most 
\begin{align*}
\lfloor{(\log 2)^{-1}\left|\log R\right|+\left|k\right|+1}\rfloor&\leq(\log 2)^{-1}\left(\left|\log R\right|+\left|k\right|+1\right)\\
&\leq(2/\log 2)\left(\left|\log R\right|+\left|k\right|\right)
\end{align*}
if $k\neq 0$. If $k\geq 1$, then we have the estimate
$${(\log 2)^{-1}\left(1+\dfrac{1}{\left|\log R\right|+\left|k\right|}\right)(\log\left|R\right|+\left|k\right|)}\leq(\log 2)^{-1}\left(1+1/\log 2\right)\left(\left|\log R\right|+\left|k\right|\right)$$
since $R\geq 1/2$.
