$\forall p\in(1,\infty)$, show that under some conditions, it satisfies $|f(x)-f(y)|\leq c_p|f'(x)|^{\frac{1}{p}}|x-y|,\qquad\forall y\in[-1,1]$. 
For each $p\in(1,\infty)$, show that there exists constant $c=c_p$ satisfies: if $f\in C^2([-1,1]),\;f(-1)<f(1)$ and $\max_{[-1,1]}|f'(x)|\leq1$, then there exists $x\in[-1,1]$ such that $f'(x)>0$ and 
  $$
|f(x)-f(y)|\leq c_p|f'(x)|^{\frac{1}{p}}|x-y|,\qquad\forall y\in[-1,1].
$$

This is bonus question in a midterm examination of analysis, but within class time, I can't do this. And now I can't even start.
 A: Remark. I will interprete $f\in\mathcal C^2([a,b])$ as "there exists a $\bar f\in\mathcal C^2(]a-\varepsilon,b+\varepsilon[)$ such that $f$ and $\bar f$ coincide on $[a,b]$ for some $\varepsilon>0$."
First note that since $f(-1)<f(1)$ (and $f$ is continuous, differentiable), we know by the Mean value Theorem that
\begin{equation}\tag{1}\label{1}
 \text{There exists } x_0 \in [-1,1] \text{ such that } f'(x_0) = \frac{f(1)-f(-1)}{2}>0.
\end{equation}
We can use a Taylor expansion of second order ($f$ is twice continuously differentiable) to find that 
\begin{equation}\tag2\label2
 \forall\, x\in[-1,1]:\exists\,\xi\in[-1,1]:\forall\,y\in[-1,1]:f(y)-f(x)=(y-x)f'(x)+\frac{(y-x)^2}2f''(\xi)
\end{equation}
Since $f''$ is continuous on the compact $[a,b]$, we have $\sup_{x\in[a,b]}|f''(x)|=:{const}<\infty$.
By using this and \eqref{1} and \eqref{2} we find that there exists $x_0\in[a,b]$ such that $f'(x_0)>0$ and $|f(y)-f(x_0)|\le|y-x_0||f'(x_0)|+\frac{1}{2}|y-x_0|^2{const}\overset{f'(x_0)>0}\le c|y-x_0|\cdot\sqrt[p]{|f'(x_0)|}$ for some constant $c$ that depends only on $p$. This achieves a proof of your claim.
