Prove that for any real numbers $a,b$ we have $\lvert \arctan a−\arctan b\rvert\leq \lvert a−b\rvert$.

Prove that for any real numbers $a,b$ we have $\lvert \arctan a−\arctan b\rvert\leq \lvert a−b\rvert$. This should have an application of the mean value theorem.

• What is the derivative of $\arctan x$? What does the mean value theorem say about $$\frac{\arctan{a}-\arctan{b}}{a-b}?$$? – Thomas Andrews Mar 26 '13 at 14:04
• I will give a bigger hint. Take your function to be $\arctan$ and consider it at $a$ and $b$. Mean value theorem tells us that there exists $c$ such that ... ? then use the hint above. – Lost1 Mar 26 '13 at 14:05

You want to prove that $\lvert \arctan a−\arctan b\rvert\leq \lvert a−b\rvert$. This is equivalent to proving that $$\frac{\lvert \arctan a−\arctan b\rvert}{\lvert a−b\rvert}\leq 1.$$ Now let $f(x) = \arctan(x)$. Then (make a note of why you can say this) by the Mean Value Theorem you have a $c\in (a,c)$ such that $$f'(c) = \frac{\lvert \arctan a−\arctan b\rvert}{\lvert a−b\rvert}.$$ All that is left for you to do is to prove that for any such $c$, $f'(c)\leq 1$.
Note that $$\left|\frac{\rm d}{{\rm d}x}\arctan(x)\right|~=~\left|\frac{1}{1+x^2}\right|~\leq~ 1$$ and this ensures that $\arctan (x)$ is Lipschitz-continuous with Lipschitz constant $1$...
Let $f(x)=\arctan x$, then by mean value theorem there exist $c\in (a,b)$ such that
$$|f(a)-f(b)|=f^{\prime}(c) |a-b|$$
$$f^{\prime}(x)=\frac{1}{1+x^2}\leq1$$
So $f^{\prime}(c)\leq1$ and $|\arctan a-\arctan b|\leq |a-b|$.