non tangential maximal function and Hardy-Littlewood maximal function I'm studying harmonic analysis and found that we can bound non-tangential maximal function by Hardy-Littlewood maximal function. Most books don't give the proof of it. How can I see that? Is there a proof of it using dyadic decomposition?

It appears the OP is asking the following question. Given $\phi$ a bounded positive integrable function on $\mathbb{R}^n$ that is radial and radially decreasing. For $t> 0$ let $\phi_t = t^{-n}\phi(x/t)$. Define the maximal operator
$$ M_\phi f(x) = \sup \{ |\phi_t\ast f(y)| : |x-y| < t\} $$
Why does the following hold:
$$ M_\phi f(x) \leq C \mathcal{M} f(x) $$
where $\mathcal{M}$ is the Hardy-Littlewood maximal operator. 
 A: Why insist on dyadic decomposition of cubes when you don't need it? 
Observe that the maximal function $\mathcal{M}f = \mathcal{M} |f|$ by definition. And observe that since $\phi$ is positive, $|\phi_t\ast f| \leq \phi_t \ast |f|$. Hence we can assume without loss of generality that $f$ is positive. 
Let $\lambda_\phi(s)$ for $s \geq 0$ be the set $\{ \phi(x) \geq s\}$. We have that $\phi(x) = \int_0^P \chi_{\lambda_{\phi}(s)}(x) \mathrm{d}s$, where $P = \sup \phi$. Note that by assumption $\lambda_\phi(s)$ for a decreasing family of balls around the origin. Let $\lambda^*\phi(s)$ be the ball whose radius is 1 more than the radius of $\lambda_\phi(s)$. Now, let $|y-x| < 1$. We have that
$$ \int_{\lambda_\phi(s)} f(y-z) \mathrm{d}z \leq \int_{\lambda^*_\phi(s)} f(x - z) \mathrm{d}z $$
since $f$ is positive and $x + \lambda^*_\phi(s) \supset y + \lambda_\phi(s)$. 
Let $P'$ be the smallest number such that $\lambda_\phi(P')$ has radius at most 1. By assumption (that $\phi$ is a bounded integrable function) we have that $\lambda_\phi(P')$ has positive radius $R'$.
$$\begin{align} \phi\ast f(y) &= \int_0^{P'} \int_{\lambda_\phi(s)} f(y-z) \mathrm{d}z \mathrm{d}s + \int_{P'}^P \int_{\lambda_\phi(s)} f(y-z) \mathrm{d}z \mathrm{d}s \\ 
&\leq \int_0^{P'} \int_{\lambda^*_\phi(s)} f(x-z) \mathrm{d}z \mathrm{d}s + \int_{P'}^P\int_{\lambda^*_\phi(P')} f(x-z) \mathrm{d}z \mathrm{d}s \\
& \leq \int_0^{P'} |\lambda_\phi^*(s)| \mathcal{M}f(x) \mathrm{d}s + |P' - P||\lambda_\phi^*(P')|\mathcal{M}f(x) 
\end{align}$$
Now using that for $s \leq P'$ we have that $|\lambda_\phi^*(s)| \leq c^n |\lambda_\phi(s)|$ where $c = 1 + 1/R'$ we have
$$ \leq  \mathcal{M}f(x)\left( c^n \int_0^P |\lambda_\phi(s)| \mathrm{d}s + |P-P'| |\lambda_\phi^*(P')|\right)\tag{*}$$
The first term inside the parenthesis gives $\int \phi(z) \mathrm{d}z$ from the distributional function characterisation of Lebesgue spaces. The second term is quite obviously a finite constant depending on $\phi$. 

Now, if we replace $\phi$ by $\phi_t$ in the above argument, then $P \mapsto t^{-n} P$. We will need to consider $|y-x| < t$ and we let $\lambda^*_{\phi_t}(s)$ to have radius $t$ more than its counterpart without the star. We also let $P'$ be such that the corresponding ball has radius at least $t$: by the scaling property of $\phi_t$ we see that $P' \mapsto t^{-n} P'$, and $\lambda_\phi(P') \to t \lambda_\phi(P')$. So the above analysis goes to show that the constant derived above (the term inside the parentheses in (*)) does not depend on the scaling $t$. Hence we get the desired inequality. 
