Discussing the uniform continuity of the following:

1. $\displaystyle f(x)=x^2, \ \ \ x \in (-1,1) $

Claim: f is uniformly continuous on (-1,1)

Proof.

Given $\displaystyle \epsilon > 0$, pick $\displaystyle \delta = \frac { \epsilon }{2} $, then for each $\displaystyle x,y \in (-1,1) $ with $\displaystyle |x-y|< \delta $, we would have $\displaystyle |x^2-y^2|=|x-y||x+y|< \delta |x+y| < \delta |1+1| = \epsilon $

Q.E.D.

2. $\displaystyle f(x)=e^{-x} \ \ \ x \in [0,\infty ) $

Claim: f is uniformly continuous.

Proof.

Given $\displaystyle \epsilon > 0$, pick $\displaystyle \delta >0$, then for each $\displaystyle x,y \in [0, \infty) $ with $\displaystyle |x-y|< \delta $, we would have $\displaystyle |e^{-x} - e^{-y}| \leq |e^0-e^0|=|1-1|=0 < \epsilon$

Q.E.D.

3. $\displaystyle f(x)=x^{ \frac {1}{3} } \ \ \ x \in [0, \infty ) $

Claim: f is not uniformly continuous.

So far, I know that $\displaystyle f'(x)= \frac {1}{3x^{ \frac {2}{3}}} $, which is not below above as x approaches to 0.

But is it possible to prove this by defintion?

Proof so far.

Pick $\displaystyle \epsilon =1 $, then $\displaystyle \forall \delta > 0 $ with $\displaystyle |x-y|< \delta $, I need to show that $\displaystyle |x^{ \frac {1}{3}} - y^ { \frac {1}{3} } | < \epsilon $ But I'm stuck in here, any hints?

Thanks!!!