Let $\mathcal{PR}$ be the set of primitive recursive functions. Let $\mathcal{PR}(f)$ be $\mathcal{PR}$ which we have amplified by adding (a recursive) $f$ the in the set of initial functions. To make this a true extension, assume $f$ outgrows all primitive recursive functions.


Now, $\mathcal{PR}$ is "dense" in the sense that it has the Ritchie-Cobham property, i.e. every function computable in primitive recursive time is primitive recursive and vice versa.



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> Does $\mathcal{PR}(f)$ have the Ritchie-Cobham property?
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To me it seems natural the extension preserves the property and I would guess there is some "direct" argument for this, which I just fail to see. 



A "dense" extension of the set of primitive recursive functions
The answer is yes.


Suppose a function $g$ is computable by a procedure $p$ whose
computation running time is bounded by a function
$h\in\newcommand\PR{\text{PR}}\PR(f)$. I claim that $g\in\PR(f)$.


To see this, let's first argue that $\PR(f)$ is closed under the
bounded search operator $(x,z)\mapsto\mu y<z[R(x,y,z)]$, where $R$
is a relation whose characteristic function is in $\PR(f)$ and
where the $\mu$ operator returns the least $y<z$ such that
$R(x,y,z)$, if there is one, and otherwise returns $z$ as a signal
that the search failed. The proof is just the usual proof that the
primitive recursive functions are closed under bounded search,
since one can easily define this instance of the $\mu$ operator by
recursion, by using $R$. In other words, the usual argument is not
sensitive to the particular functions one starts with in the $\PR$
hierarchy.


It follows now that $g$ is in $\PR(f)$, since we can say that
$g(x)=y$ just in case there is a code for a computation of length
$h(x)$ showing that the output of $g$ on $x$ is $y$ according to
the fixed program $p$. This code can be checked using bounded
search, since we can bound how big the code will be if we know how
long the computation will be.


