What are the asymptotic upper and lower bounds for T(n) = 2T(n/2) + n lg lg n?

Question

The recurrence relation

T(n) = 2T(n/2) + n lg lg n

(where lg is logarithm to base 2) can be solved using the master theorem but I am not very sure about the answer. I have found my answer but am not mentioning it here in order to prevent information cascades. Please help me find the big O and Ω for above.

Answer 1

None of the 3 cases in the master theorem apply for

T(n)=2 T(n/2) + n log(log n)

(With arbitrary base, it doesn't really matter)

Case 1: f(n)=n log(log n) is 'bigger' than n^{log2 2}=n¹

Case 2: f(n) does not fit n log^k(n)

Case 3: f(n) is smaller than n^1+e

U(n)=2 U(n/2) + n log n
L(n)=2 L(n/2) + n

You can show that: U(n) >= T(n) and L(n) <= T(n). So U gives a upper bound, and L a lower bound for T.

Applying the master theorem for U(n), gives

Case 2: f(n)=n log n=Θ(n¹ log¹ n) thus U(n)=Θ(n log² n)

Applying the master theorem for L(n), gives

Case 2: f(n)=n =Θ(n¹ log⁰ n) thus L(n)=Θ(n log n)

Because L(n)<=T(n)<=U(n) it follows that T(n)=O(n log² n) and T(n)=Ω(n log n)

Also, note that O(log₂n)=O((log n)/log 2)=O((log n) * c)=O(log n).