Relaxing hypotheses of Master Theorem


This question is related to Master Theorem on oscillating function.

Consider a recurrence of the form

$ T(n) = a T(n/b) + f(n)$

Master Theorem’s regularity condition excludes some cases (for example, when $ f(n)$ is oscillating).

Suppose, however, that $ f(n)=\Theta(g(n))$ for a function $ g(n)$ that does not violate the regularity condition, so that the Master Theorem is applicable if $ g(n)$ is used instead of $ f(n)$ . Consider then the following recurrence:

$ T'(n) = a T'(n/b) + g(n)$

and assume that the master theorem gives the solution $ T'(n)=\Theta(g(n))$ .

Can I then safely conclude that $ T(n)=\Theta(g(n))$ ? Or there are some reasonable conditions on $ g(n)$ we can add (I suppose that if $ g(n)$ is polinomially bounded then maybe Akra-Bazzi method will apply, even if one have to swich integration and $ \Theta$ , and I’m not sure this is sound)?

Notice that from $ f(n)=\Theta(g(n))$ we cannot deduce that $ f(n)$ is always less than or equal to $ g(n)$ or to a $ d\cdot g(n)$ for some fixed $ d$ . So I cannot prove by induction that $ T(n)<T'(n)$ for all $ n$ and, anyway, I only want to prove $ T(n)=\Theta(T'(n))$ .

To give some context, this is the case I want to apply the result to: consider the recurrence

$ T(n)=2\cdot T(n/2)+f(n)$

where I only know that $ f(n)=\Theta(n\sqrt{n})$ . Then I can unfold the recurrence obtaining \begin{equation*} \begin{split} T&(n)=2T\left(\frac{n}{2}\right)+\Theta(n\sqrt{n})= 2\left(2T\left(\frac{n}{4}\right)+\Theta\left(\frac{n}{2}\sqrt{\frac{n}{2}}\right)\right)+\Theta(n\sqrt{n}) =\ &=\ldots=2\left(2\left(2\ldots \left(2T\left(\frac{n}{2^h}\right)+\Theta\left(\frac{n}{2^{h-1}}\sqrt{\frac{n}{2^{h-1}}}\right)\right)\ldots\right)\right)+\Theta(n\sqrt{n}) \end{split} \end{equation*}

until $ 2^h=n$ , i.e. $ h=\log(n)$ . Then $ T(n)=\sum_{i=0}^{\log(n)-1} 2^i\cdot\Theta\left(\frac{n}{2^i}\sqrt{\frac{n}{2^i}}\right)$ .

Performing the calculations I get \begin{equation*} \begin{split} T(n)&=\sum_{i=0}^{\log(n)-1} 2^i\cdot\Theta\left(\frac{n}{2^i}\sqrt{\frac{n}{2^i}}\right)=\Theta\left(\sum_{i=0}^{\log(n)-1} 2^i\cdot\frac{n}{2^i}\sqrt{\frac{n}{2^i}}\right)=\ &=\Theta\left(n\sqrt{n}\cdot\sum_{i=0}^{\log(n)-1} \frac{1}{\sqrt{2^i}}\right) \end{split} \end{equation*} The series $ \sum_{i=0}^{+\infty} \frac{1}{\sqrt{2^i}}$ is convergent, so I can conclude (I hope correctly) that $ T(n)=\Theta(n\sqrt{n})$ .

On the other hand, I cannot directly apply Master Theorem, as from $ f(n)=\Theta(n\sqrt{n})$ I cannot conclude $ 2f\left(\frac{n}{2}\right)<cf(n)$ for some $ c<1$ . Indeed, if definitively we have $ c_1 n\sqrt{n}\leq f(n)\leq c_2 n\sqrt{n}$ , then

$ f(n)\geq c_1 n\sqrt{n}= 2\sqrt{2}c_1 \frac{n}{2}\sqrt{\frac{n}{2}}\geq\frac{2\sqrt{2}c_1}{c_2} f\left(\frac{n}{2}\right)=\frac{\sqrt{2}c_1}{c_2} 2f\left(\frac{n}{2}\right)$

and so

$ 2f\left(\frac{n}{2}\right)\leq \frac{c_2}{\sqrt{2}c_1} f(n)$

but $ \frac{c_2}{\sqrt{2}c_1}$ is not necessarily less than 1.