The question:
Let $a>0, n\:\epsilon\: \mathbb{N} $ and let $f_n:[a,\infty)\rightarrow \mathbb{R}$ be defined by $f_n(x):=\sqrt[n]{x}$. Show that $f_n$ is Lipschitz continuous and determine the appropriate Lipschitz constants $L_n$.
Now I wrote down the basic statement:
$|\sqrt[n]{x}-\sqrt[n]{y}|≤L_n|x-y|$
But I can't come up with any idea of how to get a Lipschitz constant, especially for all n.
Edit: I'm not allowed to use derivates.
Best Answer
We can apply mean value theorem. Note that $\frac{\mathrm{d}}{\mathrm{d}x} x^\frac{1}{n} = \frac{1}{n}x^{\frac{1}{n} - 1}$, so: $$ \left|\frac{\sqrt[n]{x} - \sqrt[n]{y}}{x - y}\right| \leq \max_{x \in [a,\infty)}\left|\frac{1}{n}x^{\frac{1}{n} - 1}\right| = \frac{1}{n}a^{\frac{1}{n} - 1} $$ So we can choose $L_n = \frac{1}{n}a^{\frac{1}{n} - 1}$.
Note that the following theorem is used to conclude the inequality: Let $f : \Bbb{R} \supseteq D \to \Bbb{R}$ be a differentiable function. Then: $$ \left|\frac{f(x) - f(y)}{x - y}\right| \leq \sup_{x \in D}|f'(x)| $$ The proof is very simple. By MVT, there exists $c \in [x,y]$ such that $f(x) - f(y) = f'(c)(x - y)$, so $\left|\frac{f(x) - f(y)}{x - y}\right| = |f'(c)|$. Clearly $|f'(c)| \leq \sup_{x \in D}|f'(x)|$, so the result follows.