I am trying to prove that any deterministic 1-tape Turing Machine which recognizes the language $L = \lbrace{0^n1^n | n \geq 0 \rbrace}$ requires $\Omega(\text{log }n)$ space.

I believe this can be done using a crossing sequence argument. I have been trying to imitate the $DSPACE(O(1)) = REG$ proof from wikipedia.

What I have tried is:

Suppose $L \in DSPACE(S(n))$, for some $S(n) = o(\text{log } n)$ and let $M$ be an $S(n)$ space bounded TM recognizing $L$. Since $L$ is not regular, $L \notin DSPACE(O(1))$. Therefore, given $k \in \mathbb{N}$, let $x \in L$ be a string of minimal length that requires more than $k$ worktape cells.

Let $C$ be the set of configurations of $M$ on $x$. That is, $C$ is the set of tuples of the form

(state, work tape head position, work tape contents).

Then $|C| \leq |Q_M| \times S(n) \times 2^{S(n)} \leq 2^{cS(n)} = o(n)$, where $c$ is some suitable constant.

The crossing sequence at $i^{\text{th}}$ cell boundary is the sequence of such configurations occurring as the input head moves across that boundary. Each term of a crossing sequence can be any of the $|C|$ elements from $C$.

Also, length of any crossing sequence cannot be more than $|C|$; for otherwise, some configuration will repeat and $M$ will enter into an infinte loop.

Therefore, number of crossing sequences of $M$ on $x$ $\leq |C|^{|C|} \leq 2^{cS(n)2^{cS(n)}} $.

The problem is that this doesn't give the required bound. So, a cleverer argumet is needed.