Proof of Master Theorem

Master Theorem：
$T(n)=aT\left(\frac{n}{b}\right)+f(n)$
where $a\ge 1,b>1$, $a,b$ is constant, $\frac{n}{b}$ can be either $\lfloor \frac{n}{b}\rfloor$ or $\lceil \frac{n}{b}\rceil$

1. $\exists ϵ>0s.t.f(n)=O(n^{{\log }_{b}a-ϵ})$, then $T(n)=\Theta (n^{{\log }_{b}a-ϵ})$
2. $f(n)=\Theta (n^{{\log }_{b}a}{\log }^{k}n)$, then $T(n)=\Theta (n^{{\log }_{b}a}{\log }^{k+1}n)$
3. $\exists ϵ>0s.t.f(n)=\Omega (n^{{\log }_{b}a+ϵ})$, and $af\left(\frac{n}{b}\right)\le cf(n)$ holds for some constant $c<1$ and all sufficiently large $n$ (usually called the regularity condition), then $T(n)=\Theta (f(n))$

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Proof:

$$T(n)=aT\left(\frac{n}{b}\right)+f(n)\iff T(n)=\Theta (n^{{\log }_{b}a})+\sum _{i=0}^{{\log }_{b}n-1}{a}^{i}f\left(\frac{n}{b^i}\right)$$

Obviously $T(n)$ depends on case of $f(n)$, now let
$$g(n)=\sum _{i=0}^{{\log }_{b}n-1}{a}^{i}f\left(\frac{n}{b^i}\right)$$

Case 1 : $\exists ϵ>0s.t.f(n)=O(n^{{\log }_{b}a-ϵ})$
$$g(n)=\sum _{i=0}^{{\log }_{b}n-1}{a}^{i}f\left(\frac{n}{b^i}\right)=O(\sum _{i=0}^{{\log }_{b}n-1}{a}^i{\left(\frac{n}{b^i}\right)}^{{\log }_{b}a-ϵ})$$
Simplification
$$\sum _{i=0}^{{\log }_{b}n-1}{a}^i{\left(\frac{n}{b^i}\right)}^{{\log }_{b}a-ϵ}$$
$$=n^{{\log }_{b}a-ϵ}\sum _{i=0}^{{\log }_{b}n-1}{b}^{ϵi}$$
$$=n^{{\log }_{b}a-ϵ}\frac{n^{ϵ}-1}{b^{ϵ}-1}$$
$$\Rightarrow g(n)=O(n^{{\log }_{b}a})$$
$$T(n)=\Theta (n^{{\log }_{b}a})+g(n)=\Theta (n^{{\log }_{b}a})$$
Case 2 : $f(n)=\Theta (n^{{\log }_{b}a}{\log }^{k}n)$
$$g(n)=\sum _{i=0}^{{\log }_{b}n-1}{a}^{i}f\left(\frac{n}{b^i}\right)=\Theta (\sum _{i=0}^{{\log }_{b}n-1}{a}^i{\left(\frac{n}{b^i}\right)}^{{\log }_{b}a}{\log }^{k}n)$$
$$=\Theta (n^{{\log }_{b}a}{\log }^{k}n\sum _{i=0}^{{\log }_{b}n-1}1)$$
$$=\Theta (n^{{\log }_{b}a}{\log }^{k}n{\log }_{b}n)$$
$$=\Theta (n^{{\log }_{b}a}{\log }^{k+1}n)$$

Case 3 : $\exists ϵ>0s.t.f(n)=\Omega (n^{{\log }_{b}a+ϵ})$, and $af\left(\frac{n}{b}\right)\le cf(n)$ holds for some constant $c<1$ and all sufficiently large $n$ (usually called the regularity condition)
$$af\left(\frac{n}{b}\right)\le cf(n)\iff f\left(\frac{n}{b}\right)\le \frac{c}{a}f(n)\iff f\left(\frac{n}{b^i}\right)\le {\left(\frac{c}{a}\right)}^{i}f(n)\iff a^{i}f\left(\frac{n}{b^i}\right)\le c^{i}f(n)$$

$$g(n)=\sum _{i=0}^{{\log }_{b}n-1}{a}^{i}f\left(\frac{n}{b^i}\right)\le \sum _{i=0}^{{\log }_{b}n-1}{c}^{i}f(n)\le \sum _{i\ge 0}{c}^{i}f(n)=\frac{1}{1-c}f(n)$$
$$\Rightarrow g(n)=O(f(n))$$
On the other hand
$$g(n)=\sum _{i=0}^{{\log }_{b}n-1}{a}^{i}f\left(\frac{n}{b^i}\right)\ge f(n)$$
$$\Rightarrow g(n)=\Omega (f(n))$$
Thus
$$g(n)=\Theta (f(n))$$
$$T(n)=\Theta (n^{{\log }_{b}n})+\Theta (f(n))=\Theta (f(n))$$