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Let $m > 0$ and $ a:[0,1] \rightarrow \mathbb R$ be a function with $a(\epsilon) \rightarrow 0$.

Then $ \epsilon^m a(\epsilon) \rightarrow 0 $ for $ \epsilon \rightarrow 0$.

But is $ \epsilon^m a(\epsilon) \in o(\epsilon^m) $, with O-Notation used? Put into other words, if we multiply a zero sequence which converges by order $m$ with another zero sequence, is the decay strictly faster than by order $m$?

You may assume $a$ to be uniformly continuous if that helps.

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Yes. If you have two functions $a,b : [0,1]\to\mathbb{R} $ satisfying $a(\epsilon)\in o(f(\epsilon))$ and $b(\epsilon)\in O(g(\epsilon))$ for $\epsilon \to 0$ then $a(\epsilon)b(\epsilon) \in o(f(\epsilon)g(\epsilon))$ wich specializes to $\epsilon^ma(\epsilon)\in o(\epsilon^m)$ in your case, since clearly $a(\epsilon)\in o(1)$.