[Solved]: Prove that $TQBF \notin SPACE(n^{\frac{1}{3}})$

Problem Detail: 

I would like some hints on how to approach this problem, I know for instance that $TQBF$ is $PSPACE$-$Complete$, so it can solved in poly space and any other $PSPACE$-$Complete$ problems can be log spaced reduced to $TQBF$. I believe that I need to employ the space hierarchy theorem in some way but I am not sure how, this is a homework question so I just want a hints. Thank you!

Asked By : InsigMath

Answered By : Yuval Filmus

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 $\epsilon > 0$. 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 $\delta \geq 1/2$.

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Question Source : http://cs.stackexchange.com/questions/18426