Inequality involving a convex function I am stuck, showing the following inequality in an easy way (using only inequalities or something):
Let $x\in [-a,a]$ for some $a>0$ and $p\in (1,2)$. I want to show that there then exists a constant $M>0$ such that for all $x$:
$$||x-1|^p-1|\leq M |x|^{c}.$$
(where $0<c \leq p$; i.e. $c=0.5$).
I am having a hard time showing this innocent looking inequality. The reason is that for $p\in (1,2)$, the function $f(x)=|x|^p$ is convex and homogeneous of degree $p>1$. As a result, $f(x)$ is not subadditive and we can't use the reverse triangle inequality anymore (from which the result would otherwise directly follow). 
On the other hand, applying the standard triangle inequality on the absolute value seems not to be sharp enough, since
$$f(x):=||x-1|^p-1| \leq |x-1|^p +1 \leq 2^{p-1}|x|^p + 2^{p-1}\cdot 1 +1 \leq M|x|^c + 3$$
which obviously does not imply the desired upper bound. 
=> Is there some kind of (reverse triangle) inequality for convex functions of this kind I am not aware of, or any other easy way to proof the assertion? $$$$
(To see that the inequality holds I can provide an heuristic proof here:
For some subset I can show that the inequality holds: For $x \in [0,2]$ the function $||x-1|^p-1|=1-|1-x|^p = f_1(x)$ is strictly concave and attains a unique maximum at $x=1$ with $f_1(1)=1$. Hence on the interval $[0,2]$ it is sufficient to show that there exists $M_1\geq 1$, such that 
$$f_1(x) \leq M_1 |x|.$$
Note that $M_1|x|\geq 1=\max_{x\in [0,2]} f(x)$ for $1\leq x\leq 2$ given $M_1\geq 1$. So, let $0\leq x \leq 1$ we then have by binomial inequality:
$$f_1(x)=1-|1-x|^p = 1-(1-x)^p \leq 1 - 1 + p x \leq 2\cdot x.$$
=> with $M_1\geq 2$ the inequality holds on the subinterval [0,2].
Moreover for $x\in [-c,0]$ the function $f(x)$ is strictly decreasing (and convex) with $f(0)=0$. While the function is strictly increasing on $x\in[0,c]$ so there will indead exist a $M\geq 0$ such that $f(x) \leq M|x|^{0.5}$)
 A: The function
$$f(x) = \left|1-\left|1-x\right|^p\right|,\quad p\in(1,2) $$
is non-negative and concave in a right neighbourhood of the origin, non-negative and convex in a left neighbourhood of the origin, hence there are no positive constants $M$ and $c\in(0,p]$ such that
$$ f(x)\leq M |x|^c$$
holds over a whole neighbourhood of zero.
A: Let $f(x)= ||x-1|^p-1|$ and $g(x) = M|x|^c$. Assume $p\in (1,2)$, $M>0$, $c \in (0,p]$, $a>0$. 
1) If $c>1$ then it will not work:
We have $f(0)=g(0)=0$.  But the right-derivative of $f$ at $x=0$ is $p>0$, while the right-derivative of $g$ at $x=0$ is $0$.  It follows that there is a $\delta>0$ such that $f(x)\geq g(x)$ for all $x \in [0, \delta]$.
2) If $c \in (0,1]$ then it will work for suitably large $M>0$.
The magnitudes of the left and right derivatives of $f$ are bounded over $[-a,a]$, so $f$ is $L$-Lipschitz continuous over $[-a,a]$ for some constant $L>0$, so that $|f(x)-f(y)|\leq L|x-y|$ for all $x,y \in [-a,a]$. In particular, since $f(0)=0$, we get: 
$$ |f(x)|\leq L|x|  \quad \forall x \in [-a,a] $$
It suffices to find an $M>0$ such that $L|x|\leq M|x|^c$ for all $x \in [-a,a]$.  
Case 1:  Suppose $c=1$.  Then choose $M=L$ and we are done. 
Case 2: Suppose $c \in (0,1)$. The functions $M|x|^c$ and $L|x|$ are symmetric about the origin. It suffices to find $M>0$ such that $Lx\leq Mx^c$ for all $x \in [0,a]$.  For $x>0$ we have $\frac{d}{dx} Mx^c = \frac{Mc}{x^{1-c}} \geq \frac{Mc}{a^{1-c}}$.  So $Mx^c \geq \left(\frac{Mc}{a^{1-c}}\right)x$ for all $x \in [0,a]$. Just choose $M$ so that $\frac{Mc}{a^{1-c}}=L$. $\Box$
