Recently I'm studying something about Calderon-Zygmund decomposition, as in the following link https://en.wikipedia.org/wiki/Singular_integral_operators_of_convolution_type#Calder%C3%B3n-Zygmund_decomposition.
But I don't know how to perform the decomposition for a specific function.
Let $f$ be a non-negative integrable on $[-π,π]$, perform the Calderon-Zygmund decomposition for
- $f(x)= \delta_0 (x)$ at the level $\alpha =1$.
- $f(x)= \delta_0 (x)+\delta_1 (x)$ at the level $\alpha =1$.
where $\delta_a$ equals $1$ at $a$ and $0$ otherwise, and on any interval containing $a$ the integral of $\delta_a$ is $1$.
Any help to solve the problem will be appreciated!
Thanks in advance.
Here's one interpretation of how to carry out the decomposition for $f(x) = \delta_0(x)$, with $\alpha = 1$. I'll leave the other part as an exercise.
The starting average value of $f$, over the full interval $[-\pi, \pi]$, is $(2\pi)^{-1} < \alpha$, so we proceed to subdivide the interval. On the child intervals, $[-\pi, 0]$ and $[0, \pi]$, $f$ has the average value $\pi^{-1} < \alpha$. So we subdivide again, obtaining $[-\pi, -\pi/2]$, $[-\pi/2, 0]$, $[0, \pi/2]$, and $[\pi/2, \pi]$. This time the average values are $0$, $2/\pi$, $2/\pi$, and $0$, still all less than $\alpha$. Subdividing a third time (I'm going to stop listing the intervals) yields averages of $0$, $0$, $0$, $4/\pi$, $4/\pi$, $0$, $0$, and $0$. The middle two intervals, $[-\pi/4, 0]$ and $[0, \pi/4]$, now have averages greater than $\alpha$, so we set those intervals aside. But the other intervals continue to be subdivided forever, with average values never exceeding $\alpha$ (in fact the averages are all $0$).
The definition of $g$ is now potentially a bit hazy. Going back to Stein's exposition in Singular Integrals and Differentiability Properties of Functions (p. 31), we have that $g$ is defined to be equal to $f$ off of the intervals $[-\pi/4, 0]$ and $[0, \pi/4]$. So $g = 0$ on $[-\pi, -\pi/4)$ and $(\pi/4, \pi]$. On the interiors of the set-aside intervals, $g$ is defined to be the average value of $f$ on those intervals. So $g = 4/\pi$ on $(-\pi/4, 0)$ and $(0, \pi/4)$. Stein doesn't specify what should happen at $-\pi/4$, $0$, or $\pi/4$, because for ordinary functions that doesn't matter: that remainder has Lebesgue measure $0$, and for ordinary functions there's no mass on such a null set. But now, because we're dealing with a delta function, there is mass at one of these points. So at the moment it's a bit hazy.
Moving on temporarily, we can safely say that $b$ is $0$ on $[-\pi, -\pi/4)$ and $(\pi/4, \pi]$. Further, though (again per Stein's exposition), the average value of $b$ on each interval $[-\pi/4, 0]$ and $[0, \pi/4]$ is supposed to be $0$. Now, we need to have $b = -4/\pi$ on $(-\pi/4, 0)$ and $(0, \pi/4)$ so that $f = g + b$ there. But then to make the average values work out to be $0$, we also need to include $\delta_0$ as a part of $b$ (and so not as a part of $g$).
With that, we've essentially fully defined $g$ and $b$. There are still a few individual points where we haven't provided specific function values, but these should be immaterial for practical purposes as all of $f$'s mass is accounted for.