# Studying the uniform continuity of $f(x) =x^2$ on the set of real numbers.

I know this question has been asked before , and the function can be proven to not be uniformly continuous using contradiction. The problem is I don't get how come the following definition of uniform continuity doesn't apply in this case. The definition states that a function $f = f(x)$ is uniformly continious if : $\forall \epsilon > 0, \quad\exists \delta> 0$ such that $\forall x' , x'' \in \mathbb{R} : 0 < |x'-x''| < \delta \implies |f(x')-f(x'')|<\epsilon$

Then:

$|f(x')-f(x'')|=|x'^2 -x''^2| = |x'-x''||x'+x''| < \delta|x'+x''|$

$\delta|x'+x''| = \delta|x'-x'' + 2x''|<\delta(|x'-x''|+2|x''|)=\delta (\delta+2|x''|) = \delta^2 + 2\delta x''$

$\delta^2 + 2\delta x''<\epsilon \implies \delta^2 + 2\delta x''-\epsilon<0$

The solutions for this equation are:

$\delta_1=-|x''|-\sqrt{x''^2+\epsilon}$ (Not acceptable , $\delta_1 < 0$)

$\delta_2=-|x''|+\sqrt{x''^2+\epsilon}$ (Acceptable , $\delta_2 > 0$)

Thus if we choose $0<\delta<\delta_2$ we would have $\delta^2 + 2\delta x''-\epsilon<0$ and $|f(x')-f(x'')|<\epsilon$

$\forall \epsilon > 0, \quad\exists \delta \in [0,\delta_2]$ such that $\forall x' , x'' \in \mathbb{R} : 0 < |x'-x''| < \delta \implies |f(x')-f(x'')|<\epsilon$

Where have i gone wrong?

Thanks for the help!

Your $\delta$ depends on $\delta_2$ and $\delta_2$ depends on $x''$ and on $\varepsilon$. That makes no sense. By the definition of uniform continuity, you should get a $\delta$ which depends only on $\varepsilon$.