(Hardy-Littlewood Theorem) : Let $ u(r,\theta)$ be the Poisson integral of $ \varphi \in L^{p}$, $ 1<P \leqslant \infty$ , and let $ U(\theta)=\sup_{r<1}|u(r,\theta)| $.
Then $ U \in L^{p}$, and there is a constant $A_{p} $ depending only on $p$ such that $ \Vert U\Vert_{P} \leqslant A_{p} \Vert \varphi \Vert_{P}$.
The proof of theorem is easy. But my question is why for $p=1$ Theorem is False. I try to use the modified Poisson kernel $ u(r,\theta)= \dfrac{R^2-r^2}{R^2-2R r \cos\theta+r^2}$ for $R>1$. I use max and min $ u(r,\theta)$ but I can't solve the problem.
Your idea to use $$u(r,\theta)= \dfrac{R^2-r^2}{R^2-2R r \cos\theta+r^2}$$ is a good one. Let's take $|\theta|$ small but greater than $R-1$; also let $r=1-|\theta|$. Then the denominator can be estimated as $$R^2-2R r \cos\theta+r^2 \le 10|\theta|^2$$ since it's the squared distance between points with polar coordinates $(R,0)$ and $(r,\theta)$. (I use $10$ to say there's some factor there, but it's fixed.)
On the other hand, $R^2-r^2=(R-r)(R+r)\ge |\theta|$. Conclusion: $$ U(\theta) \ge\frac{1}{10 |\theta|},\quad |\theta|>R-1 $$ and this shows that the $L^1$ norm of $U$ blows up as $R\to 1$, while the $L^1$ norm of $u(1,\cdot)$ remains $1$.