Separating time classes A student of mine recently ask the following question: 

Assume $$DTIME(f(n)) \subsetneq DTIME(g(n)).$$ Must there exist an $h(n)$ such that $$DTIME(f(n)) \subsetneq DTIME(h(n)) \subsetneq DTIME(g(n))?$$

This could probably be shown true by constructing an $h(n)$ if $f,g$ are time-constructible. But in general, I feel that this should be false similar to $DSPACE(o(\log(\log(n)))) = DSPACE(1)$.
If $DTIME(f(n))$ is defined as the class of all languages decidable in $O(f(n))$ time by a two tape Turing machine, then I suspect that the answer is no.  In other words, I think that there does not always exist a strictly intermediate time complexity class.

Note: This answer might not be exactly what you are looking for because I'm considering non-computable functions and I don't include every detail of the argument.  But, I felt that it is a good start.  Please feel free to ask any questions.  Maybe I can fill in these details further at some point or maybe this will lead to a better answer from an interested reader.

Consider functions of the form $f : \mathbb{N} \rightarrow \mathbb{N}$.  We refer to these functions as natural number functions.


  Claim 1: I claim that we can construct a very slow growing non-decreasing natural number (non-computable) function
  $\varepsilon(n)$ such that:
  
  (1) $\varepsilon(n)$ is non-decreasing
  
  (2) $\varepsilon(n) = \omega(1)$
  
  (3) For all unbounded computable $f: \mathbb{N} \rightarrow \mathbb{N}$, the set
  $\{ n \; \vert \; \varepsilon(n) \leq f(n) \}$ is infinite.


We construct $\varepsilon(n)$ as a slow growing non-decreasing step function. Let us enumerate all unbounded computable functions $\{ f_i \}_{i \in \mathbb{N}}$. We want to construct $\varepsilon(n)$ in such a way that for every $i$ and every $j \leq i$, $min\{ k \; \vert \; \varepsilon(k) \geq i\} \geq min\{ k \; \vert \; f_{j}(k) \geq i \}$. In other words, we wait to map $\varepsilon(n)$ to $i$ until the first $i$ functions in the enumeration have mapped to a value greater than or equal to $i$ at least once.  Then, $\varepsilon(n)$ continues to map to $i$ until the first $i+1$ functions in the enumeration have mapped to a value greater than or equal to $i+1$ at least once and at this point it starts mapping to $i+1$. If we continue this iterative process for constructing $\varepsilon(n)$, then for any given unbounded computable function, although $\varepsilon(n)$ might not always be smaller, it will infinitely often be at least as small.

Note: I just provided some intuition behind claim 1, I did not provide a detailed proof.  Please feel free to join in on discussion below.

Because $\varepsilon(n)$ is such a slow growing function, we have the following:


  Claim 2: For all computable natural number functions $f(n)$ and $h(n)$, if $h(n) = \Omega(\frac{f(n)}{\varepsilon(n)})$ and $h(n) = O(f(n))$, then $h(n) = \Theta(f(n))$.


For claim 2, if there existed a computable function $h(n)$ between $\frac{f(n)}{\varepsilon(n)}$ and $f(n)$ such that $h(n) \neq \Theta(f(n))$, then we would be able to compute an unbounded natural number function that grows more slowely than $\varepsilon(n)$ which isn't possible.  

Let me explain some relevant details.  Suppose for sake of contradiction that such a function $h(n)$ existed.  Then, $\lfloor \frac{f(n)}{h(n)} \rfloor$ is unbounded.

Note: The preceding function is computable because $f(n)$ and $h(n)$ are computable.

Since $h(n) = \Omega(\frac{f(n)}{\varepsilon(n)})$, we have $\lfloor \frac{f(n)}{h(n)} \rfloor = O(\varepsilon(n))$.  It follows that there is some constant $\alpha$ such that for all $n$ sufficiently large, $\lfloor \alpha \frac{f(n)}{h(n)} \rfloor &lt; \varepsilon(n)$.  Since this function is unbounded and computable we may apply Claim 1 to get that $\varepsilon(n) \leq \lfloor \alpha \frac{f(n)}{h(n)} \rfloor$ infinitely often which contradicts the previous statement.


  Claim 3: For a time constructible function $f(n)$, we have that $DTIME(\frac{f(n)}{\varepsilon(n)}) \subsetneq DTIME(f(n))$, yet there does not exist $h(n)$ such that $\frac{f(n)}{\varepsilon(n)} \leq h(n) \leq f(n)$ and $DTIME(\frac{f(n)}{\varepsilon(n)}) \subsetneq DTIME(h(n)) \subsetneq DTIME(f(n))$.


In order to just show that, $DTIME(\frac{f(n)}{\varepsilon(n)}) \subsetneq DTIME(f(n))$ we need to use a stronger time hierarchy theorem and this is where we use the assumption that the number of tapes is fixed (we said two tapes above).  See "The tight deterministic time hierarchy" by Martin Furer.

Since there are no computable natural number functions between $\frac{f(n)}{\varepsilon(n)}$ and $f(n)$ other than those that are $\Theta(f(n))$, we have that for every function $h(n)$ such that $\frac{f(n)}{\varepsilon(n)} \leq h(n) \leq f(n)$ and $h(n) \neq \Theta(f(n))$, $DTIME(\frac{f(n)}{\varepsilon(n)}) = DTIME(h(n))$.