$f(n) = o(n^c) \rightarrow \exists \epsilon > 0 \ s.t. f(n) = O(n^{c-\epsilon})$ I'm trying to prove that for arbitrary $c &gt; 0$, 

$f(n) = o(n^c) \rightarrow \exists \epsilon &gt; 0 \ s.t. f(n) = O(n^{c-\epsilon})$

Intuitively, this seems to be true to me (little-o implies some gap "between" $f(n)$ and $n^c$ that we can formally express as $\epsilon$). However, I'm having some trouble formalising this argument.

What I've done so far:

$f(n) = o(n^c) \rightarrow \exists c_0, n_0 &gt; 0 \ s.t. \forall n \geq n_0, 0 \leq f(n) &lt; c_0 n^c$ 

Suppose (for a contradiction) that no such $\epsilon$ fulfilling the above condition exists.

Then we have

$\forall \epsilon &gt; 0, \exists n \geq n_0 \ s.t. f(n) \geq c_0 n^{c-\epsilon}$

I'm not sure how to proceed from here.
This is false. Consider for example
$$ f(n) = \frac{n^c}{\log n}. $$