Recursion $T(n) = T(\log (n)) + O(\log(\log (n)))$ $T(n) = T(\log (n)) + O(\log(\log (n)))$

assuming $n =2 \ ^ {m}$ for $m \in N $ 

I need to prove by induction an upper bound.

I thought of doing the following:

$$T(2\ ^{m}  )= T(m) + O(log(m)) $$

define $S(m) := T(2 \ ^ {m})  $ we get : 
$$S(m) = S(log(m)) + O(log(m)) $$

I'm not sure how to continue.
You don't have upper bound cos $T(2^m)-T(m)\ge t&gt;0$  since you can continue doing infinite times $T(2^{2^{.^{.^{.^m}}}})$ and each next is for a non infinitely small number bigger then previous.
Hint:

Assume $T(n)=O(\log(\log(n)))$. Then the recurrence says

$$T(n)=O(\log(\log(n)))=O(\log(\log(\log(n))))+O(\log(\log(n)))=O(\log(\log(n))),$$ which is true.