Uniform continuity : stuck at an intermediate step

Question

$$f(x)= \frac{x}{1+x^2}, \, x \in \mathbb{R}$$

I have proved till the following step. I can't understand how do I proceed further. Please help $$f(x)\le|(x-y)||1-xy|$$

Answer 1

For all $x \in \mathbb{R},$ we have

$$f'(x) = \frac{1 - x^2}{(1+x^2)^2} $$

For $|x| \leqslant 1$ we have

$$|f'(x)| = \frac{1 - x^2}{(1+x^2)^2} \leqslant \frac{1}{(1+x^2)^2} \leqslant 1,$$

and for $|x| > 1$ we have $|f'(x)| \to 0$ as $x \to \pm \infty$ and local maxima at $x = \pm \sqrt{3}$ where $f'(\pm \sqrt{3}) = 1/8.$

Since the derivative is bounded, $f$ is uniformly continuous.

Alternatively, following your approach

$$\left|\frac{x}{1+x^2} - \frac{y}{1+y^2}\right| = |x-y|\frac{|1 - xy|}{1 + x^2 + y^2 +x^2y^2} \leqslant |x-y|\frac{1 + |xy|}{1 + x^2 + y^2 +x^2y^2}. $$

I'll leave it to you to show that the RHS is bounded by $C |x - y|$ where $C$ is a constant.

Hint:

If $|xy| < 1$ then

$$\frac{1 + |xy|}{1 + x^2 + y^2 +x^2y^2} < \frac{2}{1 + x^2 + y^2 +x^2y^2}$$

If $|xy| > 1$ then

$$\frac{1 + |xy|}{1 + x^2 + y^2 +x^2y^2} < \frac{2|x||y|}{x^2 + y^2}$$