Showing $\tan^{-1}(x)$ is Lipschitz

Question

I am trying to show $\tan^{-1}(x)$ is lipschitz for all $x\in\mathbb{R}$.

My attempt:

If $f$ is Lipshitz on $\mathbb{R}$, then $\exists L\in\mathbb{R}$ such that $\forall x,y\in\mathbb{R}$ $$|f(x)-f(y)|\leq L|x-y|.$$ Now consider the Mean Value Theorem ($f$ is continuous and differentiable on $\mathbb{R}$) $$\exists c\in\mathbb{R} \ \text{such that} \ \frac{|f(x)-f(y)|}{|x-y|}=|f'(c)|$$

Using the Mean Value Theorem we can see that, \begin{align} L&\geq\frac{|f(x)-f(y)|}{|x-y|} \\ &=|f'(c)| \\ &=|f'(0)| \\ &=1 \end{align}

I have a two questions:

1. what determines the value of $c$?

2. what if $x=y$? Doesn't this not fulfill the inital condition that $f$ is Lipschitz $\forall x,y\in\mathbb{R}$?

Answer 1

The present question may be resolved with the aid of the principle that a differentiable function is locally Lipschitz, and in fact globally Lipschitz if the derivative is globally bounded. To see this, let

$f \in \mathcal C^1(I, \Bbb R), \tag 1$

where $I \subset \Bbb R$ is an open interval. The then for $x, y \in I$, we have

$f(y) - f(x) = \displaystyle \int_x^y f'(s) \; ds \tag 2$

by the fundamental theorem of calculus; thus if $y \ge x$,

$\vert f(y) - f(x) \vert = \left \vert \displaystyle \int_x^y f'(s) \; ds \right \vert \le \displaystyle \int_x^y \vert f'(s) \vert \; ds \le M \vert y - x \vert, \tag 3$

provided

$\vert f(s) \vert \le M, \; s \in I; \tag 4$

if $x \ge y$ we obtain the same result by reversing the roles of $x$ and $y$ in (3); since we eventually take absolute values, in the end the result is independent of the relative order of $x$ and $y$; we therefore see that under such circumstances $f$ is Lipchitz continuous on $I$, with Lipschitz constant $M$; and if $I = \Bbb R$, $f$ is globally Lipschitz on all of $\Bbb R$.

We apply this observation to the question at hand:

An important and helpful aspect of solving this problem is keeping careful track of the domain and range of $\tan^{-1}x$; recall that, taking

$\tan: \left ( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right ) \to \Bbb R, \tag 5$

we have

$\tan^{-1}: \Bbb R \to \left ( -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right ); \tag 6$

since $\tan^{-1}$ is defined on all of $\Bbb R$, so is its derivative; we may estimate $(\tan^{-1} x)'$ as follows:

$y = \tan^{-1} x; \tag 7$

$x = \tan y; \tag 8$

$\dfrac{dx}{dy} = \sec^2 y = 1 + \tan^2 y; \tag 9$

$\left \vert \dfrac{dy}{dx} \right \vert = \dfrac{1}{1 + \tan^2 y} = \dfrac{1}{1 + x^2} \le 1, \; \forall x \in \Bbb R; \tag {10}$

in the light of (10), we see that in fact $\tan^{-1}x$ is globally Lipschitz on $\Bbb R$, with Lipschitz constant $1$, by virtue of the principle proved in the above.

As far as our OP Bell's two specific questions are concerned, it should be observed that $c$ will in fact depend on $x$ and $y$ if the mean value theorem is invoked, which we can't do directly in the case $x = y$; this is why I prefer the above argument based on the fundamental theorem.