Prove that $TQBF \notin SPACE(n^{\frac{1}{3}})$
The argument is actually surprisingly delicate. This sketch follows Tompa's Introduction to computational complexity, Chapter 10. The space hierarchy theorem shows that there is some problem $L \in \mathrm{SPACE}(n^{1+\epsilon}) \setminus \mathrm{SPACE}(n)$ for every $\epsilon &gt; 0$. Since TQBF is PSPACE-complete, there is some logspace reduction from $L$ to TQBF. But we actually know more - using Savitch's theorem, we can come up with some reduction which blows up instances of $L$ of size $n$ to instances of TQBF of size $O(n^{2(1+\epsilon)} \log n)$ (at least according to Ryan Williams, page 1 at the bottom). If TQBF were solvable in space $O(n^\delta)$ then this would give an algorithm for $L$ in space $O(n^{2\delta(1+\epsilon)})$. Taking the limit $\epsilon \to 0$, this shows that $\delta \geq 1/2$.