Is the set $F$ make the $M_{\lambda}(0)$ finite?

Question

Let $F$={$0$} $\bigcup$ {$\frac{1}{k}$:$k\ge2$} $\bigcup$ {$\frac{-1}{k}$:$k\ge2$} be a closed subset of $(-1,1)$.

For any $t\in(-1,1)$,define the distance function $\delta(t) = inf_{p\in F}${$|t-p|$}.

For any $\lambda \gt 0$,consider the Marcinkiewicz integral

$$ M_{\lambda}(x)=\int_{-1}^{1} \frac{\delta^{\lambda}(t)}{|x-t|^{1+\lambda}} dt$$

Is $M_{\lambda}(0)$ finite ?

I know that if $x \in F $ then $M_{\lambda}(x)$ is finite almost everywhere.

And if $x \in F$ such that $M_{\lambda}(x)$ is finite , then $F$ is "very dense" near x .

But we dont have the define of "very dense" , we have no way to know $x=0$ whether dense enough in $F$ , and I have no ideal how to calculate $M_{\lambda}(0)$

Answer 1

With your $F$, you can check that $\delta(t) \leq t^2$. Indeed, if $\frac{1}{k+1} \leq |t| \leq \frac{1}{k}$ for some $k \geq 2$ then

$$ \delta(t) \leq \frac{1}{2}\left(\frac{1}{k} - \frac{1}{k+1} \right) = \frac{1}{2k(k+1)} \leq t^2. $$

If $|t| \geq \frac{1}{2}$, then $\delta(t) = |t|-\frac{1}{2} \leq t^2$ and still the inequality is true. Thus

$$ M_{\lambda}(0) \leq \int_{-1}^{1} \frac{|t|^{2\lambda}}{|t|^{1+\lambda}} \, dt = \int_{-1}^{1} |t|^{\lambda-1} \, dt = \frac{2}{\lambda} < \infty. $$

Remark. Although not required to our proof, we can show that $\delta(t) \leq \frac{1}{2}t^2$ as we can see from the following graph:

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