f is uniformly continuous if for all $x$ and $y ∈ \mathbb{R}$, $|f(x) − f(y)| ≤ |x−y|^\frac{1}{2}$. 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.
 A: 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
$$
 |x−y| \lt \delta  \implies |f(x) − f(y)| ≤ |x−y|^{1/2} < \delta^{1/2} = \epsilon \, .
$$
A: You have the right idea, but the logic isn't quite there.
Fix $\epsilon > 0$. Now you need to show there exists a $\delta > 0$ such that if $| x - y| < \delta$, then $|f(x) - f(y) | < \epsilon$.
As you say, a good choice is $\delta = \epsilon^2 > 0$. Then
$$
\begin{align}
&\quad\quad\quad|x - y| < \delta
\\&\implies | x- y| < \epsilon^2
\\&\implies |x-y|^{\frac{1}{2}} < \epsilon
\\&\implies |f(x) - f(y)| \leq |x-y|^{\frac{1}{2}} < \epsilon
\\&\implies |f(x) - f(y)| < \epsilon
\end{align}
$$
This shows that there exists a $\delta > 0$ (namely $\epsilon^2$) such that if $|x - y| < \delta$, then $|f(x) - f(y)| < \epsilon$.
A: 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 let $$\delta = {\epsilon}^2.$$ Then
$$ |x-y| \lt {\epsilon}^2 = \delta \\ 
\implies |x−y|^\frac{1}{2} \lt \epsilon \\ 
\implies |f(x)-f(y)| \lt \epsilon
.$$
Therefore the function is uniformly continuous.
