Uniform contiunuity of $f$?

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?


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

Leave a Reply

Your email address will not be published. Required fields are marked *