Show $f(x)={x\over x+1}$ is uniformly continuous on $[0,\infty)$.

Question

Show $$f(x)={x\over x+1}$$ is uniformly continuous on $[0,\infty)$. The hints say that it should be easy to see it is lipschitz because the derivative is bounded. I tried and tried and I just can't show it. I would really appreciate your help.

Answer 1

Start using the MVT: $$|f(x)-f(y)| = |f'(\xi)| |x-y| $$ for some $\xi \in (x,y)$ under the assumption $x<y$ (otherwise $\xi\in(y,x)$). How can you bound $f'(\xi)$? Of course we need to use $x,y\in [0,\infty)$

Answer 2

An approach is just to write, on $[0,\infty)$: $$ |f(a)-f(b)|=\left| \frac{a}{a+1}- \frac{b}{b+1}\right|=\left| \frac{a(b+1)-b(a+1)}{(a+1)(b+1)}\right|=\left| \frac{a-b}{(a+1)(b+1)}\right|\leq|a-b| $$ and the function is lipschitzian.