Let f : R → R be a given function. The following property will ensure that f is uniformly continuous:
for all $x$ and $y ∈ \mathbb{R}$, $|f(x) − f(y)| ≤ |x−y|^\frac{1}{2}$.
My attempt :
From definition a function $f$ is uniformly continuous if for a arbitrary $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that for any two points $x,y$ in $\mathbb{R}$,
$$ |x-y| \lt \delta \implies |f(x)-f(y)| \lt \epsilon
$$
Now let $\epsilon \gt 0$ be given. Then
$$
|x−y|^\frac{1}{2} \lt \epsilon \\
\implies
|x-y| \lt {\epsilon}^2 = \delta \\
\implies |f(x)-f(y)| \lt \epsilon
.$$
Therefore the function is uniformly continuous.
Best Answer
You surely are on the right track. Given $\epsilon > 0$ you have to find a $\delta > 0$ such that the implication $|x-y| \lt \delta \implies |f(x)-f(y)| \lt \epsilon$ holds.
So the correct order of arguments would be: Let $\epsilon \gt 0$ be given. Set $\delta = \epsilon^2$. Then $\delta > 0$ and $$ |x−y| \lt \delta \implies |f(x) − f(y)| ≤ |x−y|^{1/2} < \delta^{1/2} = \epsilon \, . $$