Lower space bound on a turing machine accepting palindromes Let
$$
PAL = \lbrace x \in \lbrace 0, 1, \# \rbrace^* | x = rev(x) \rbrace
$$
How do I show that a turing machine deciding $PAL$ must use space $\Omega(\log n)$?

I have a feeling that I need to use crossing sequences when crossing the middle of the input tape, but I'm not sure how to relate that to space.
Hint: A Turing machine running in space $S$ runs in time at most $\exp S$.