Globally Lipschitz condition

Question

Show that the function $f(x)=\frac1{1+x^2}$ satisfies a global Lipschitz condition on $\mathbb R$.

By the definition we want to whow that $|f(x)-f(y)|\leqslant L|x-y|$. Here is my work so far. $$\left|\frac1{1+x^2}-\frac1{1+y^2}\right|=\left|\frac{y^2-x^2}{(1+y^2)(1+x^2)}\right|\leqslant L|x-y|$$ $$\implies\left|\frac{y+x}{(1+y^2)(1+x^2)}\right|\leqslant L.$$ It appears to be bounded, since the denominator becomes much larger than the numerator for large values of $x$ and $y$. Is this good enough or can I find a value for $L$? Also, is it also the same as showing that $f'(x)$ is globally bounded?

Thank you in advance for any help or constructive criticism. I am forever grateful.

Answer 1

An easier way is: $|f'(x)|=\left|\frac{2x}{(1+x^2)^2}\right|\leq 1$. By a theorem (I don't know the english name of this theorem, but if someone else can translate, in french is "théorème des accroissements finis") you have that $\forall x,y\in\mathbb R, |f(x)-f(y)|\leq |f'(c_{xy})||x-y|\leq |x-y|$ for a certain $c_{x,y}\in\mathbb R$.

Answer 2

$$\left| \frac{x+y}{(1+x^2)(1+y^2)}\right|=\frac{|x+y|}{(1+x^2)(1+y^2)}$$

By triangle inequality:

$$\frac{|x+y|}{(1+x^2)(1+y^2)} \leq \frac{|x|}{(1+x^2)(1+y^2)}+\frac{|y|}{(1+x^2)(1+y^2)}$$

Now $x^2+1 \geq 1$, so:

$$ \frac{|x|}{(1+x^2)(1+y^2)} \leq \frac{|x|}{1+x^2}$$

Next $\displaystyle \frac{|x|}{1+x^2} \leq 2$ be AM-GM inequality. The same with $\displaystyle \frac{|y|}{(1+x^2)(1+y^2)}$, so finally:

$$\left| \frac{x+y}{(1+x^2)(1+y^2)}\right| \leq 4$$