Uniform continuity (in terms of Big-O notation)

Question

I believe the following statement is equivalent to $f$ being uniformly continuous:

$$ f(x + h) = f(x) + \mathcal{O}_{|h| \to 0}(h) $$

I am wondering whether this statement is equivalent to $f$ being uniformly continuous as well:

$$ f(x + h) = f(x) + \mathcal{O}_{|h| \to 0}(\sqrt{h}) $$

Or in general:

$$ f(x + h) = f(x) + \mathcal{O}_{|h| \to 0}(h^{\alpha}); \alpha \in (0, 1) $$

Intuitively I don't think the above statement holds, but I am struggling to find an counterexample.

Many thanks in advance!

Answer 1

I believe the following statement is equivalent to $f$ being uniformly continuous: $$ f(x + h) = f(x) + \mathcal{O}_{|h| \to 0}(h) $$

Unfortunately, your belief is incorrect. It is somewhat close to being correct though. Note that your statement implies that $$|f(x+h) - f(x)| \leqslant L\cdot|h|$$ This is precisely the definition of $L$-Lipschitz continuity. An important thing to note is that $L$-Lipschitz contunity $\implies$ Uniform continuity, but the reverse implication is NOT necessarily true. A simple counterexample is $f(x) = \sqrt{|x|}$. So, your statement implies uniform continuity, but it is surely not equivalent.

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Or in general: $$f(x + h) = f(x) + \mathcal{O}_{|h| \to 0}(h^{\alpha}); \alpha \in (0, 1)$$

Once again, for any fixed $\alpha \in (0,1)$, a simple counterexample will be $f(x) = \left(\sqrt{|x|}\right)^\alpha$, which is uniformly continuous. But you can show that (say, at $x = 0$), your statement does NOT hold. Thus, by no means your statement is equivalent to uniform continuity.