Bounding $(x^n-y^n)/(x-y)$?

Question

Let $x,y \in [0,1]$ it follows then from the binomial theorem that for integers $n \ge 1:$

$$\sup_{x,y} \left\lvert \frac{x^n-y^n}{x-y} \right\rvert \le n.$$

Is this also true if $q \in [1,\infty)$:

$$\sup_{x,y} \left\lvert \frac{x^q-y^q}{x-y} \right\rvert \le q?$$

Answer 1

Lagrange mean value theorem implies $$|x^q-y^q|=|q\xi^{q-1}(x-y)|,$$ where $\xi$ is between $x$ and $y$.

Answer 2

For a positive integer $n$, we have $$\left|\frac{x^n-y^n}{x-y}\right|=\left|\sum_{r=0}^n\,x^r\,y^{n-r}\right|\leq \sum_{r=0}^n\,|x|^r\,|y|^{n-r}\leq n\,.$$ The equality does not hold (as $x\neq y$), but the inequality is sharp (by taking $y=1$ and $x\to 1^-$). Thus, $$\sup_{\substack{{x,y\in[0,1]}\\{x\neq y}}}\,\left|\frac{x^n-y^n}{x-y}\right|=n\,.$$

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Now, for a real number $q\geq 1$, we have by Bernoulli's Inequality that $$(1+t)^q\geq 1+qt\text{ for all }t\geq-1\,.\tag{*}$$ For (*) to be an equality, either $q=1$ or $t=0$. The case $q=1$ is obvious. From now on, we assume that $q>1$.

Without loss of generality, suppose that $x>y$. Set $t:=\dfrac{x-y}{y}\in(0,\infty)$ in (*). Then, the inequality becomes $$\frac{x^q}{y^q}=\left(1+\frac{x-y}{y}\right)^q> 1+q\,\left(\dfrac{x-y}{y}\right) \,.$$ Thus, $$\frac{x^q-y^q}{x-y}>q\,y^{q-1}\,.$$ Similarly, taking $t:=\dfrac{y-x}{x}\in(-1,0)$ in (*) yields $$\frac{y^q}{x^q}=\left(1+\frac{y-x}{x}\right)^q> 1+q\,\left(\dfrac{y-x}{x}\right)\,,$$ so that $$\frac{x^q-y^q}{x-y}< q\,x^{q-1}\,.$$

Ergo, $$q\,y^{q-1}<\frac{x^q-y^q}{x-y}< qx^{q-1}\,.$$ Since $x\leq 1$, we have $$\frac{x^q-y^q}{x-y}<q\,.$$ However, this inequality is sharp by taking $x:=1$ and $y\to 1^-$. That is, $$\sup_{\substack{{x,y\in[0,1]}\\{x\neq y}}}\,\left|\frac{x^q-y^q}{x-y}\right|=q\text{ for all }q\geq 1\,.$$