Question
Let $f : \mathbb R_+ \to \mathbb R_+$ be a continuous, non-decreasing function with $f(x) \to \infty$ as $x \to \infty$.
Suppose we have a non-decreasing sequence $(m_n) \subset \mathbb N$ such that $m_n \to \infty$ and $m_n = o(\min\{f(n), \, n\})$ as $n \to \infty$.
Let $\tilde m(\cdot) : \mathbb R_+ \to \mathbb R_+$ be a continuous, strictly increasing function such that $m_n = \lfloor \tilde m(n) \rfloor$ for all $n \in \mathbb N$.
Does it already follow that $\tilde m(x) = o(f(x))$ as $x \to \infty$?
Edit: I've now constructed a $\tilde m(\cdot)$ that works, which is really good enough for my purposes. But it would still be nice to know if this property holds for all such $\tilde m(\cdot)$.
Thoughts
This seems obviously true, but perhaps it isn't the case after all. Here are a couple things I've tried:
$$ 0 \leq \frac{\tilde m(x)}{f(x)} = \frac{m_{\lfloor x \rfloor}}{f(x)} \cdot \frac{\tilde m(x)}{m_{\lfloor x \rfloor}} \leq \underbrace{ \frac{m_{\lfloor x \rfloor}}{f(\lfloor x \rfloor)}}_{\to 0} \cdot \frac{\tilde m(x)}{m_{\lfloor x \rfloor}}, $$ so if $\limsup_{x \to \infty} \frac{\tilde m(x)}{m_{\lfloor x \rfloor}} = \limsup_{x \to \infty} \frac{\tilde m(x)}{\lfloor \tilde m (\lfloor x \rfloor) \rfloor } < \infty$, then we're done. But I haven't managed to show this.
Similarly, we might try $$ 0 \leq \frac{\tilde m(x)}{f(x)} \leq \frac{\tilde m(x)}{f(\lfloor x \rfloor)} \leq \underbrace{\frac{m_{\lfloor x \rfloor}}{f(\lfloor x \rfloor)}}_{\to 0} + \frac{\tilde m(x) - m_{\lfloor x \rfloor}}{f(\lfloor x \rfloor)} $$ So it would also suffice to show $\limsup_{x \to \infty} \frac{\tilde m(x) - m_{\lfloor x \rfloor}}{f(\lfloor x \rfloor)} = \limsup_{x \to \infty} \frac{\tilde m(x) - \lfloor m(\lfloor x \rfloor ) \rfloor }{f(\lfloor x \rfloor)} < \infty$.
The "danger" seems to be that even though $m_n = o(\min\{f(n), n\})$, there might be intervals $[n-1,n]$ where $f$ stays constant on most of the interval the interval, say $[n-1, n-1/2^N]$, while $\tilde m$ shoots up to almost $\tilde m(n)$ very early on this interval. My intuition is that even this shouldn't throw $\tilde m(t)/f(t)$ out of whack too much (since $f$ should be asymptotically much bigger). But I'm not sure how to analytically show this.