How much time to recognize palindromes in logarithmic space? It is well-known that palindromes can be recognized in linear time on $2$-tape Turing machines, but not on single-tape Turing machines (in which case the time needed is quadratic). The linear-time algorithm uses a copy of the input, and thus also uses a linear space. 

Can we recognize palindromes in linear time of a multitape Turing machine, using only  a logarithmic space? More generally, what kind of space-time trade-off is known for palindromes?
You can use the same argument used to prove the $\Omega(n^2)$ time bound on single tape.

Suppose that you have a TM with $S(n)$ space that recognize palindromes 
$\{x\,0^{\frac{n}{3}} x^R \mid |x|=n/3 \}$ (where $x^R$ is the reverse of $x$) in time $T(n)$.
When the (input) head crosses the middle $0^{n/3}$ it can carry only $S(n)$ bits of information. So it needs to make $\Omega(n / S(n))$ crosses, and each cross requires $n/3$ time.

So $T(n)S(n)=\Omega(n^2)$.