This question is concerned with an exercise from the book Classical and Multilinear Harmonic Analysis by Muscalu and Schlag (cf. exercise 2.5. on page 38): The task is to find within a multiplicative constant $M \delta_0$. $\delta_0$ denotes the Dirac measure at $0$ and for a complex Borel measure $\mu\in \mathcal{M}(\mathbb{T})$ on the one dimensional torus $\mathbb{T}$(identified with the interval $[0,1]$), the associated maximal function is given by $(M \mu)(\theta)=\sup_{\theta\in I\subset \mathbb{T}} \frac{|\mu|(I)}{|I|}$. The supremum here is taken over all open intervals $I$ and $|I|$ in the denominator denotes the Lebesgue measure of $I$. Moreover, we are asked to show that the weak $L^1$-bound $|\{ \theta \in \mathbb{T} | (M \mu)(\theta) >\lambda \}| \le \frac{3}{\lambda} \left\|\mu\right\|$, where $\left\|\mu\right\|$ denotes the total variation of $\mu$ and $\lambda>0$, can not be improved in general, using this example. First of all I calculated $(M \delta_0)(\theta)=\begin{cases} \frac{1}{\theta}, \quad \theta<\frac12 \\ \frac{1}{1-\theta}, \quad \theta\geq\frac12 \end{cases}$. Is this correct so far? Then I intend to show that, for certain values of $\lambda$, we obtain equality in the aforementioned weak $L^1$-bound. But I don't know how to get the factor $3$ unfortunately.


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