Does checking the middle bit of an input require a log(n) space pointer? (Reusing space)
You can do this using $O(1)$ space, using two pointers. Initially, the first pointer is at the left end of the input, and the second pointer is at its right end. Each round, advance both pointers toward the middle, until they meet. This requires no additional space at all!

(The latter, assuming we can enlarge the tape alphabet. Otherwise we can still make do with constant additional space, by implementing the pointers as spaces.)