Application of the Maximum Principle for the squared norm of the second fundamental form Firstly, I would like to say that I didn't do a PDE's course and I'm trying study by myself Mean Curvature Flow and consult references for PDEs theory when needed, so I would appreciate if someone helps me giving details. My doubt is about the application of a Maximum Principle on proposition $4.1$ which was extracted from Lectures on Mean Curvature Flow by Xi-Ping Zhu, but before I put the proposition here, I will put the evolution equation of proposition $2.2$ that will be used on the proof of proposition $4.1$.
$\frac{\partial |A|^2}{\partial t} = \triangle |A|^2 - 2 |\nabla A|^2 + 2 |A|^4.$

$\textbf{Proposition 4.1}$
$\max_\limits{X(\cdot, t)} |A|^2 \geq \frac{1}{2(\omega - t)}$ for $t \in [0, \omega).$
$\textbf{Proof:}$
Denote $U(t) = \max_\limits{X(\cdot, t)} |A|^2$. By applying the maximum principle to the evolution equation of $|A|^2$ in proposition $2.2$, we get
$\frac{d U}{d t} (t) \leq 2 \left( U(t) \right)^2$
and then
$U(t) \geq \frac{1}{2(\omega - t)}$. $\square$

What is the kind of maximum principle that was applied in order to obtain $\frac{d U}{d t} (t) \leq 2 \left( U(t) \right)^2$? I'm having difficult to apply the Maximum Principle, because I don't know what it will be the evolution equation for $U(t)$ since $U(t) = \max_\limits{X(\cdot, t)} |A|^2$ and not $U(t) = |A|^2$.
Thanks in advance!
$\textbf{EDIT:}$
I found on Proposition $2.3.5$ of this lecture notes what Maximum Principle is applied for the proposition $4.1$ of Zhu's book, but now I don't know why $|A|^2$ satisfy the hypothesis of the Maximum Principle given by the author of the lecture notes. Anyone know why $|A|^2$ satisfy the hypothesis of the Maximum Principle?
 A: 
Lemma: Let $f: M\times [0,T]\to \mathbb R$ be smooth, then 
  $$ U: [0,T]\to \mathbb R, \ \ \ U(t) = \max_{x\in M} f(x, t)$$
  is Lipschitz. If $U$ is differentiable at $t_0$, then
  $$ \frac{dU}{dt} (t_0) = \frac{\partial f}{\partial t} (x_0, t_0),$$
  where $x_0\in M$ is where $U(t_0) = f(x_0, t_0)$. 

Proof: Let $t_1<t_2 \in [0,T]$. Then there are $x_1, x_2$ so that $U(t_i) = f(x_i, t_i)$ for $i=1,2$. Then
\begin{align}
U(t_1) - U(t_2) &= f(x_1, t_1) - f(x_2, t_2) \\
&\le f(x_1, t_1) - f(x_1, t_2) \\
&\le \left | \frac{\partial f}{\partial t}\right| |t_1-t_2|\le C|t_1-t_2|.
\end{align}
Similar for $U(t_2) - U(t_1)$. Thus $U$ is Lipschitz. 
Argue similarly, let $U$ be differentiable at $t_0$. Then for $t > t_0$, 
\begin{align}
\frac{U(t) - U(t_0)}{t-t_0} &= \frac{f(x_t, t) - f(x_0, t_0)}{t-t_0}\\
&\ge \frac{f(x_0, t) - f(x_0, t_0)}{t-t_0}\\
\Rightarrow \frac{dU}{dt} (t_0) &\ge \frac{\partial f}{\partial t} (x_0, t_0).
\end{align}
Considering $t<t_0$ you have the opposite inequality. 

(Detailed) proof of Proposition 4.1: for all $t_0$ so that $U$ is differentiable at $t_0$, we have by the Lemma
  \begin{align}
\frac{dU}{dt}(t_0) &= \frac{\partial |A|^2}{\partial t} (x_0, t_0) \\
&\le \Delta |A|^2 (x_0, t_0) - 2|\nabla A|^2 (x_0, t_0) + 2|A|^4 (x_0, t_0) \\
&\le 2|A|^4 (x_0, t_0) = 2U^2(t_0)
\end{align}
  note that (weak) maximum principle is used in the last inequality, where one has $\Delta |A|^2 (x_0, t_0)\le 0$ since $|A|^2(\cdot, t_0)$ is maximized at $x_0$. Thus the inequality $\frac{dU}{dt} \le 2U$ is only satisfied almost everywhere, but this is sufficient to conclude proposition by integrating. 

