Lower bound for $a^kb^k$ in one-tape TM

For the language $$L= \{a^kb^k | k \geq 0 \}$$ How can i show there is no one-tape Turing Machine that can decide $$L$$ in less than $$O(n\log n)$$ time ?

• Have you tried to use the method of crossing sequences? Mar 22 at 18:40
• @YuvalFilmus If i define $L_n = \{ a^kb^k | n \geq k \geq 0 \}$ and consider crossing sequences between $n+1$ (to include at least one b) and $2n$ , is it good idea to think about ? Mar 22 at 18:59
• Yes, that's the first step. Mar 22 at 19:16