So I'm doing a little self-study in Kirkwood'sIntroduction to Analysis, and I need some clarification on something. Exercise 8 (a) in chapter 5-1 is this:

Suppose $\displaystyle |f(x+h)-f(x)| \le Kh^{\alpha}$ for some constant $\displaystyle K$ and $\displaystyle \alpha > 0$. Show that f is continuous.

This doesn't make any sense if $\displaystyle h < 0$, since $\displaystyle \alpha = {1\over 2}$ makes $\displaystyle Kh^{\alpha}$ complex. However, the problem never states $\displaystyle h \ge 0$. I don't think that's implied either, since it looks like the result can be proven if we revise it to say $\displaystyle |f(x+h)-f(x)| \le K|h|^{\alpha}$.

I don't need, nor do I want help with the proof, only in clarifying this apparent typo. Thoughts?