f is uniformly continuous if for all $x$ and $y ∈ \mathbb{R}$, $|f(x) − f(y)| ≤ |x−y|^\frac{1}{2}$.

Question

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.

Answer 1

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 $$ |x−y| \lt \delta \implies |f(x) − f(y)| ≤ |x−y|^{1/2} < \delta^{1/2} = \epsilon \, . $$

Answer 2

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 $|f(x) - f(y)| < \epsilon$.

Answer 3

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.