Are the functions $g$ and $h$ uniformly continuous on $\mathbb{R}$

Question

I'm working on a question that states:

Let $f: \mathbb{R}^+ \to \mathbb{R}$ be a continuous functions.

(a) Is the function $g:\mathbb{R} \to \mathbb{R}: x \mapsto f(\frac{1}{1+x^2})$ uniformly continuous?

(b) Is the function $h:\mathbb{R} \to \mathbb{R}: x \mapsto g(x)^2$ uniformly continuous (here $g$ is the same function as in part (a))?

I am aware and can prove that the function $\frac{1}{1+x^2}$ is uniformly continuous and my I think that $g$ is uniformly continuous, but $h$ is not. I know that the composition of two uniformly continuous functions is uniformly continuous, but my issue is that $f$ is only continuous.

Any help on this problem is appreciated.

@aschepler Actually $x^{1/3}$ is uniformly continuous on $[0,\infty).$ — zhw., Mar 15 '21 at 17:22
I thought more commonly $\mathbb{R}^{+} = (0, \infty)$. (See also https://math.stackexchange.com/questions/1706/does-set-mathbbr-include-zero) I suspect this wouldn't change the answers, though. — aschepler, Mar 15 '21 at 18:05

score 2 · Accepted Answer · answered Mar 15 '21 at 17:52

For point a

The map $l : x \mapsto \frac{1}{1+x^2}$ is $2$-lipschitz on $\mathbb R$ and maps $\mathbb R$ onto $(0,1]$. As $f$ is supposed to be continuous on $\mathbb R_+$, it is uniformly continuous on the closed segment $[0,1]$. Therefore for any $\epsilon \gt 0$, it exists $\delta \gt 0$ such that $\vert f(x) - f(y) \vert \lt \epsilon$ for $\vert x - y \vert \lt \delta$ and $x,y \in [0,1]$. Now for $x,y \in \mathbb R$ and $\vert x - y \vert \lt \delta /2 $ we have $\vert l(x) - l(y) \vert \lt \delta$ (as $l$ is $2$-lipschitz) and $l(x) , l(y) \in (0,1] \subseteq [0,1]$ and finally $\vert (f \circ l)(x) - (f \circ l)(y) \vert \lt \epsilon$. This enables to conclude that $f \circ l$ is uniformly continuous as desired.

Point b

Just apply point (a) to $f^2$, the square of $f$.

Are the functions $g$ and $h$ uniformly continuous on $\mathbb{R}$

1 Answers1