smallest constant $c$ for which: $|x^k - y^k| < c \cdot |x - y|$, with $x,y < 1/2$

Question

Let $k \in \{2,3,\dots\}, a \in [0,1/2)$ and $x,y \in [0,a)$.

I haven't really shown it but am $99\%$ certain the following holds following inequality holds: $$ |x^k - y^k| < |x - y|, $$ But I am looking for a constant $c < 1$ in function of $k$ and $a$ s.t.: $$ |x^k - y^k| < c|x - y|. $$

Answer 1

Hint: Think about mean value theorem applied for $f(x)=x^k$. I.e. $$\left|f(x)-f(y)\right|=\left|f'(\varepsilon)\right||x-y|$$ where $\varepsilon \in (x,y) \subset[0,a)$ (assuming $x<y$) of course.

Answer 2

$$x^k-y^k = (x-y)(x^{k-1}+x^{k-2}y+....+y^{k-1})<(x-y)k\Big({1\over 2}\Big) ^{k-1}$$ so $c\leq k\Big({1\over 2}\Big) ^{k-1}$