Uniform contiunuity of $f$?

Question

If $g $ is uniformly continuous and $g(x) = (f(x))^2$,$f(x) \geq 0$, then is $f$ uniformly continuous?

So, $\forall \epsilon > 0 , $ there exists $\delta > 0$ such that $\forall x,y \in \Bbb{R}$ with $|x - y|<\delta$ $\implies$ $|f^2(x) - f^2(y)|<\epsilon$.

$|f(x) - f(y)| < \frac{\epsilon}{|f(x)+f(y)|}$

But how to proceed after this? , or is there any counterexample to this?

Answer 1

A composition of two uniformly continuous functions is uniformly continuous and $h(x) = \sqrt{x}$ is uniformly continuous, hence so is $f(x) = h(g(x))$.

Answer 2

Use the inequality $|\sqrt{u}-\sqrt{v}|\leq \sqrt{|u-v|}$ valid for $u,v\geq 0$, for $u=f(x)^2=g(x)$ and $v=f(y)^2=g(y)$.

This give that $$|f(x)-f(y)|\leq \sqrt{|g(x)-g(y)|}$$ and it is easy to finish.