I need to show f(x)=x^(1/3) is uniformly continuous on (-1,1).
I can't find a delta st |x-y|<delta => |f(x)-f(y)|<epsilon..
any ideas?
You can try and find a delta but I am quite sure it's going to be hard, (been there done that), two aproaches that are easier are:
1) Show that it is a Lipschitz function (I'm not saying that it is, I don't know, haven't tried it)
2) Use extension theorem: a function is uniform continuous in (a,b) if and only if it can be defined in the end points a, b such that the extended function is continuous.
** Clearly the case of this function. The theorem is in the book Introduction to Real Analysis by Bartle.
It isn't Lipschitz since if $\displaystyle f$ is differentiable and Lipschitz then $\displaystyle f'$ is bounded. $\displaystyle f'(x)=\frac{1}{3\sqrt[3]{x^2}}$ which is unbounded as $\displaystyle x\to 0$
This only works for the exact reason that Plato said namely a continuous function on a compact metric space (in this case simply $\displaystyle [a,b]$) is automatically u.c.2) Use extension theorem: a function is uniform continuous in (a,b) if and only if it can be defined in the end points a, b such that the extended function is continuous.
** Clearly the case of this function. The theorem is in the book Introduction to Real Analysis by Bartle.