A problem on the boundedness of maximal operator by using linearization method We know that the maximal operator is bounded on $L^{p}(\mathbb{R}^{n})$ where $n\geq 1$ and $1<p<\infty$ and the proof would be contained in many classical harmonic analysis books. Here I find a much more different method to verify this when $n=1$ and $p=2$ (This method would be found in the book "Some Topics in Dyadic Harmonic Analysis" by Michael T.Lacey, pp.13-14).
The general ideal is that since $L^{2}(\mathbb{R})$ is a Hilbert space, therefore, the bounded linear operator $T$ on $L^{2}(\mathbb{R})$ has the property $\left \| T \right \|^{2}\leq \left \| TT^{\star} \right \|$. Thus the essential thing is to construct a proper bounded linear operator $T$ on $L^{2}(\mathbb{R})$ such that $Mf(x)\leq CTf(x)$ where $C>0$ is a fixed constant and $f(x)$ is positive. This operator has been given by the form in Lacey's book:
\begin{align*}
Tf(x)=\sum_{I\in \mathfrak{D}}\frac{\chi_{E(I)}}{|I|}\int_{I}f(y)dy
\end{align*}
where $\mathfrak{D}$ denotes the set of dyadic intervals on $\mathbb{R}$ and for each $I\in \mathfrak{D}$ associate $E(I)\subseteq I$ so that the sets $\{E(I): I\in \mathfrak{D}\}$ are disjoint subsets. Lacey points out that $Mf(x)\leq 2Tf(x)$ in his book.(I could not verify this even if taking $C$ instead of $2$.) 
To Bound the maximal operator, we calculate 
\begin{align*}
TT^{\star}f(x)=\sum_{I\in \mathfrak{D}}\sum_{J\in \mathfrak{D}}\chi_{E(I)}\frac{<\chi_{I},\chi_{J}>}{|I|}\frac{<\chi_{E(J)},f>}{|J|}\leq Tf(x)+T^{\star}f(x).
\end{align*}
Therefore, from the inequalities
\begin{align*}
\left \| Tf \right \|^{2}_{2}\leq \left \| T \right \|^{2}\left \| f \right \|^{2}_{2}\leq \left \| TT^{\star} \right \|\left \| f \right \|^{2}_{2}\leq 2\left \| T \right \|\left \| f \right \|^{2}_{2}
\end{align*} 
we have $\left \| T \right \|\leq 2$ and then $\left \| Mf \right \|_{2}\leq C\left \| Tf \right \|_{2}\leq 2C\left \| f \right \|_{2}$. 

However, for the key step, the description of the choice of $E(I)$ in Lacey's book seems not too clear (maybe to me) . Hence, $\mathbf{my \ question}$ is:
How to choose a desirable set $\{E(I): I\in \mathfrak{D}\}$ such that the pointwise inequality $Mf(x)\leq CT(x)$ is valid？
 A: I have just obtained an answer to this question but there would have some modifications that differ from Lacey's description. The detail is as follows (forgive me to pose it here):
First, we should proof a Lemma (it is just as an exercise in Lacey's book, p.4) : 
$\textbf{Lemma.}$ $\mbox{}$For any interval $I$, there is an interval $J\in \mathfrak{D}\cup \mathfrak{D'}$ with $I\subseteq J$ and $|J|\leq 8|I|$, where $\mathfrak{D'}$ is another choice of dyadic intervals defined by $\mathfrak{D'}=\left \{ \left [j2^{k},(j+1)2^{k}\right)+(-1)^{k}\frac{2^{k}}{3}: j,k\in \mathbb{Z}\right \}$. 
Proof of the Lemma: $\mbox{}$
Suppose the length of $I$ satisfies $2^{k}\leq |I|<2^{k+1}$ with $k\in \mathbb{Z}$. We should only separate it into two cases. Firstly, $I\subseteq J$ where $J\in\mathfrak{D}$ and $|J|=2^{k+1} $. Then it is easy to get $|J|\leq 8|I|$. Secondly, on the other hand, there must exists only one point $j_{0}2^{k+1}$ contained in $I$. We denote $J_{1}=\left[j_{0}2^{k+1},(j_{0}+1)2^{k+1}\right)$ and $J_{s}=J_{1}+(s-1)2^{k+1}$, $s\in \mathbb{N}$, in which $J_{1}$ is the right interval which has a nonempty intersection with $I$. Note that there exists only one interval $J'\in \mathfrak{D}$ such that $J'\supseteq J_{1}$ with $|J'|=2^{k+3}$. Then, if $J'\bigcap J_{4}=\varnothing$, we choose $J=J'\in \mathfrak{D}$ and the $J$ satisfies the Lemma. If $J'\bigcap J_{4}\neq \varnothing$ (hence $J_{4}\subseteq J'$), from the two inequalities $2^{k+1}<\frac{2^{k+3}}{3}$ and $\frac{2^{k+3}}{3}<2^{k+3}-2^{k+1}$, we have $J'-\frac{2^{k+3}}{3}\supseteq I$. Suppose that the interval $J''\in \mathfrak{D}$ is the left one which is adjacent to $J'$ with legth $|J''|=2^{k+3}$. Also we have $J''+\frac{2^{k+3}}{3}\supseteq I$. Hence, there must exists a $J\in \mathfrak{D'}$ with length $|J|=2^{k+3}$ such that $J\supseteq I$ and $|J|\leq 8|I|$. Therefore we complete the proof of the Lemma. 
Now we can construct the set $\left \{E(I)\right \}$ and the operator $T$:
For any fixed $x\in \mathbb{R}$, suppose that $Mf(x)<\infty$, there exists an interval $Q\ni x$ such that $Mf(x)\leq \frac{2}{|Q|} \int_{Q}f(y)dy$, (recall that $f\geq 0$). By Lemma, there exists an $I_{x}\in \mathfrak{D}\bigcup \mathfrak{D'}$ such that $I_{x}\supseteq Q$ and $|I_{x}|\leq 8|Q|$. Hence $Mf(x)\leq \frac{16}{|I_{x}|} \int_{I_{x}}f(y)dy$. We relabel $\left \{ I_{x}:x\in\mathbb{R} \right \}$ as $\left \{I_{j}:j\in\mathbb{N} \right \}$. For each $j$, we set $$\tilde{E}(I_{j})=\left \{ x\in I_{j}:Mf(x)\leq \frac{16}{|I_{j}|}\int_{I_{j}}f(y)dy \right \}$$
and let 
$$E(I_{j})=\tilde{E}(I_{j})\backslash \bigcup_{k=1}^{j-1}\tilde{E}(I_{k})$$ for $j>1$, and $E(I_{1})=\tilde{E}(I_{1})$.
We can easily get $\bigcup_{k=1}^{\infty}E(I_{k})=\mathbb{R}$ and $E(I_{j_{1}})\bigcap E(I_{j_{2}})= \varnothing$, if $j_{1}\neq j_{2}$. Analogously, denote $$Tf(x)=\sum_{j=1}^{\infty}\frac{\chi_{E(I_{j})}(x)}{|I_{j}|}\int_{I_{j}}f(y)dy.$$
The most important thing is that the similar inequality 
$$Tf(x)\leq Mf(x)\leq 16Tf(x)$$
is now still valid for $x\in \mathbb{R}$.

$\textbf{Supplement}$ $\mbox{}$ The operator $T$ defined in the above is a little different from the one in Lacey's book. However, the rest proof for the boundedness of maximal operator $M$ on $L^{2}(\mathbb{R})$ is still followed by the method of Lacey's. 
I have modified some mistakes in the proof of the Lemma. Now the proof will become more clear.
