For what $\alpha$ (and $C$) does $|f|^p - |g|^p \le C|f-g|^\alpha$ hold? ($f,g:\mathbb{R} \to \mathbb{R}$ continuous functions and $p>1$) Let $f,g:\mathbb{R} \to \mathbb{R}$ be continuous functions and $p>1$. 
For what $\alpha$ and constant $C>0$  does the inequality $$|f(x)|^p - |g(x)|^p \le C|(f-g)(x)|^\alpha$$ hold for every $x$?

If it is not true in general, what additional assumptions do we need to make it true?
We may assume in addition that $f,g$ are bounded.
 A: Lemma: If $s,t\in [-R,R],$ then
$$\tag 1 |t|^p - |s|^p \le pR^{p-1}|t-s|.$$
Proof: WLOG, $0\le |s|<|t|\le R.$ Now $|t|-|s|\le |t-s|.$ So by the MVT,
$$|t|^p - |s|^p = pc^{p-1}(|t|-|s|)\le pc^{p-1}(|t-s|), $$
where $|s|<c<|t|.$ Since $c<R,$ the lemma is proved.
Now assume $f,g$ are both bounded. Set $R=\max(\|f\|_\infty,\|g\|_\infty).$ Thinking of $t=f(x),s=g(x),$ we see the lemma implies
$$|f(x)|^p - |g(x)|^p \le pR^{p-1}|f(x)-g(x)|,\,\, x \in \mathbb R.$$
Thus the desired inequality holds with $C = pR^{p-1},\alpha = 1.$
So we're done in the case $f,g$ are both bounded.
The case of unbounded functions can lead to failure of the inequality. An easy case to look at is $f(x)=1+x,g(x) = x, p=2.$ Then for $x>0,$ $|f(x)|^2-|g(x)|^2 = 2x + 1,$ which is unbounded. But $|f(x)-g(x)|^\alpha = 1^\alpha = 1$ for all $x,$ so there are no $C,\alpha$ that work in this case.
However, the inequality doesn't always fail for unbounded $f,g.$ Just take $f=2g$ for example, where $g$ is unbounded. The inequality holds here if $C= 1/(4^p-1), \alpha = p.$
We get further examples looking at the case $0<g <f/2$ and $f$ is unbounded. Then for any $p>0,$
$$f^p-g^p \le f^p = 2^p(f/2)^p \le 2^p(f-g)^p.$$
So in this case the inequality is valid, with $C = 2^p,\alpha = p.$
A: By the Mean Value Theorem for $x^p$, we have
$$|f|^p - |g|^p = (|f|-|g|) p c^{p-1},$$
for some $c$ between $|f|$ and $|g|$. Since $c$ does not depend on $|f|-|g|$, $\alpha=1$.
Also, it's easy to see that if $|f|,|g|$ are not bounded, neither is $c$. Thus, $C$ does not exist. 
So additional assumption here is $|f|$ and $|g|$ are bounded. And $C$ is easy to see in this case.
