I am aware of the existence of the Calderon-Zygmund decomposition, as discussed here. But don't know how to actually compute the decomposition for a given function. For example:
Find a Calderon-Zygmund decomposition for $f: \mathbb{R}\to\mathbb{R}$ where $f(x)=|x|^{-1/3}$ at the height $\alpha = 1$.
Any hint regarding how to approach such problems will be helpful.
You want to find some intervals $I_k$ such that the average of $f$ on each interval is between $1$ and $2$, and on the rest of the line $|f|\le 1$. If the function was bounded by $1$ everywhere, we would not need any intervals. But here, $[-1,1]$ is a problematic set, so we need to cover that by $I_k$.
Using small intervals isn't practical here because the average will be too large when the interval is close to $0$. So how about we just use one interval $[-a,a]$ with $a\ge 1$? The average of $f$ on this interval is $$ \frac{1}{2a}\int_{-a}^a |x|^{-1/3}\,dx = \frac32 a^{-1/3} $$ So, $a=1$ works. The only "cube" we need is $[-1,1]$.
For a different example, consider $f(x)=|x|^{-0.9}$. Now $$ \frac{1}{2a}\int_{-a}^a |x|^{-0.9}\,dx = 10a^{-0.9} $$ so $a=1$ isn't good anymore. But $a=6$ is good enough: $10\cdot 6^{-0.9} \in (1, 2]$. This demonstrates that we don't always cut $f$ by the level set $|f|=\alpha$; the decomposition would hardly deserve a name if it was this simple.