It seems upsetting that the only tetrations that could satisfy this are non-analytic. I'm not prone to believe this, only because it doesn't look nice...
I'm wondering if we can look at it the following way
\( \exp_b^{d}(x) = \exp_b^{c}(\exp_b^{\delta}(x)) \)
Then the question boils into whether, for all \( \delta, \delta' >0 \) then
\( \exp^c_{b+\delta}(x) = o(\exp_b^{c+\delta'}(x)) \)
I think we can show this is true when \( c \in \mathbb{N} \). By induction, first, for \( c = 0 \) the induction step is obvious. Namely \( x = o (\exp_b^{\delta'}(x)) \). Suppose the result holds for \( c = n \) then (taking \( < \) to mean asymptotically less than):
\( \exp_{b+\delta}^{n+1}(x) =\exp^{n}_{b+\delta}(\exp_{b+\delta}(x))<\exp_{b}^{n+\delta'/2}(\exp_{b+\delta}(x)) < \exp_{b}^{n+\delta'/2}(\exp_b^{1+\delta'/2}(x)) = \exp_b^{n+\delta'+1}(x) \)
Sadly, I can't think of anyway to generalize this to non-integral \( c \)....
I'm thinking, a nice way to look at it from here is to look at root functions of the \( \exp \) function. But then we'd need an implication \( f^{\circ n}(x) = o(g^{\circ n}(x)) \Rightarrow f(x) = o(g(x)) \), which looks like it could be true for monotonically growing unbounded functions. But that's probably too easy, we'd probably need a nice condition on the root functions for that to be true.
EDIT:
It appears I made a fruitful mistake in the above proof. The base induction step would have to be (1) \( \exp_{b+\delta}(x) < \exp_b^{1+\delta'}(x) \), not the obvious one \( x < \exp^{\delta'}(x) \). This is the base step I should have used. The proof then says if this base step (1) holds the result holds for all natural \( c \), namely \( \exp_{b+\delta}^n(x) < \exp_b^{n+\delta'}(x) \). And I think with some finesse we can show that this implies it's true for root functions of \( \exp_b \), which should leave for a proof where \( c \in \mathbb{Q} \). Then perhaps a density argument may work on non rational \( c \). I'll work on this more later, but I think maybe we can reduce this entire problem into the condition that if for all \( \delta,\delta'>0 \) we have \( \exp_{b+\delta}(x) < \exp_b^{1+\delta'}(x) \) then it follows that \( \exp_{b+\delta}^c(x) < \exp_b^{c+\delta'}(x) \).
...We'll probably have to assume that \( \exp_b^c(x) \) is monotone non-decreasing in \( x \) and unbounded, or at least, eventually monotone non-decreasing.
I'm wondering if we can look at it the following way
\( \exp_b^{d}(x) = \exp_b^{c}(\exp_b^{\delta}(x)) \)
Then the question boils into whether, for all \( \delta, \delta' >0 \) then
\( \exp^c_{b+\delta}(x) = o(\exp_b^{c+\delta'}(x)) \)
I think we can show this is true when \( c \in \mathbb{N} \). By induction, first, for \( c = 0 \) the induction step is obvious. Namely \( x = o (\exp_b^{\delta'}(x)) \). Suppose the result holds for \( c = n \) then (taking \( < \) to mean asymptotically less than):
\( \exp_{b+\delta}^{n+1}(x) =\exp^{n}_{b+\delta}(\exp_{b+\delta}(x))<\exp_{b}^{n+\delta'/2}(\exp_{b+\delta}(x)) < \exp_{b}^{n+\delta'/2}(\exp_b^{1+\delta'/2}(x)) = \exp_b^{n+\delta'+1}(x) \)
Sadly, I can't think of anyway to generalize this to non-integral \( c \)....
I'm thinking, a nice way to look at it from here is to look at root functions of the \( \exp \) function. But then we'd need an implication \( f^{\circ n}(x) = o(g^{\circ n}(x)) \Rightarrow f(x) = o(g(x)) \), which looks like it could be true for monotonically growing unbounded functions. But that's probably too easy, we'd probably need a nice condition on the root functions for that to be true.
EDIT:
It appears I made a fruitful mistake in the above proof. The base induction step would have to be (1) \( \exp_{b+\delta}(x) < \exp_b^{1+\delta'}(x) \), not the obvious one \( x < \exp^{\delta'}(x) \). This is the base step I should have used. The proof then says if this base step (1) holds the result holds for all natural \( c \), namely \( \exp_{b+\delta}^n(x) < \exp_b^{n+\delta'}(x) \). And I think with some finesse we can show that this implies it's true for root functions of \( \exp_b \), which should leave for a proof where \( c \in \mathbb{Q} \). Then perhaps a density argument may work on non rational \( c \). I'll work on this more later, but I think maybe we can reduce this entire problem into the condition that if for all \( \delta,\delta'>0 \) we have \( \exp_{b+\delta}(x) < \exp_b^{1+\delta'}(x) \) then it follows that \( \exp_{b+\delta}^c(x) < \exp_b^{c+\delta'}(x) \).
...We'll probably have to assume that \( \exp_b^c(x) \) is monotone non-decreasing in \( x \) and unbounded, or at least, eventually monotone non-decreasing.