Prove that $\vert \tan^{-1}x-\tan^{-1}y \vert \le \vert x-y\vert\enspace \forall x,y \in ℝ$

Question

Prove that $$\vert \tan^{-1}x-\tan^{-1}y \vert ≤ \vert x-y\vert \quad \forall x,y \in \mathbb{R}. $$

What I did was to put $x \to x+h$ and $y \to x$ and then take $\lim_{h \to 0}$. This gives $$\lim_{h \to 0} \frac{\vert \tan^{-1}(x+h)-\tan^{-1}x\vert}{\vert h \vert} \le 1$$ which gives $$\frac{d \tan^{-1}x}{dx} = \frac{1}{1+x^2} \le 1.$$ But this proves it only for $x$ and another value $x+h$ very close to it and not $\forall x,y$. Please help.

Answer 1

The comments from Bumblebee and Maximilian Janisch are right. The mean value theorem states that given $[x,y]$, there is $c \in (x,y)$ such that $\arctan'(c) = \frac{\arctan(y) - \arctan(x)}{y - x}$. Since $\arctan'(x) = \frac{1}{1 + x^2}$, you have $\frac{1}{1 + c^2} = \frac{\arctan(y) - \arctan(x)}{y - x}$. Now, $\frac{1}{1 + c^2} \leq 1$, $\frac{\arctan(y) - \arctan(x)}{y - x} \leq 1$, which rearranges to $\arctan(y) - \arctan(x) \leq y - x$. The inequality still holds when you take the absolute value of both sides.

Answer 2

Hint. Use that for all $x\in\Bbb R$, $$\arctan'(x) = \frac 1{1+x^2}\le 1.$$

Answer 3

Use that $$\arctan(x)-\arctan(y)=\frac{1}{1+\xi^2}(x-y)$$ It is just the Mean Value Theorem.

Answer 4

The mean value theorem states that "there is a tangent which is parallel to a secant, if the function is continuous".

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The function is $f(u) = \tan^{-1}(u)$.

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The secant is between the points $(x,\tan^{-1}(x))$ and $(y,\tan^{-1}(y))$.

The slope $m$ of the secant is the slope of the line between these two points:

$$|m|=|(\tan^{-1}(y)-\tan^{-1}(x))/(y-x)|=|\tan^{-1}(y)-\tan^{-1}(x)|/|y-x|$$

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Assume the parallel tangent at $u =\xi$.

The slope of the tangent is

$$|m|=\left|\frac{d}{du} f(u)\vert_{u=\xi}\right| = \left|\frac{d}{du} \tan^{-1}(u)\vert_{u=\xi}\right| = \left|\frac{1}{1+u^2}\vert_{u=\xi}\right|= \left|\frac{1}{1+\xi^2}\right| = 1/|1+\xi^2|$$

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Having two expressions for $|m|$ we set them equal, because $|m| = |m|$ $$|\tan^{-1}(y)-\tan^{-1}(x)|/|y-x| = 1/|1+\xi^2| $$

This means

$$|\tan^{-1}(y)-\tan^{-1}(x)| = |y-x|/|1+\xi^2| $$

or (because $|a-b| = |b-a|$ for any two real numbers $a$ and $b$)

$$|\tan^{-1}(x)-\tan^{-1}(y)| = |x-y|/|1+\xi^2| $$

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If $\xi = 0$, then $$|\tan^{-1}(x)-\tan^{-1}(y)| = |x-y| $$

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If $\xi \neq 0$, then $|1+\xi^2| = 1+\xi^2 > 1$ and $$|\tan^{-1}(x)-\tan^{-1}(y)| < |x-y| $$

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In total, we proved that

$$|\tan^{-1}(x)-\tan^{-1}(y)| \leq |x-y| $$