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There are already good answers from several perspectives regarding the "hardness" of $PSPACE$-complete problems, such as this: What is practical difference between NP and PSPACE-complete?

But what are the practical implications when we actually try to solve (decide) a $PSPACE$-complete problem on a deterministic Turing machine? If $PSPACE \neq NP$ (or $PSPACE \neq P$, should it make a difference) then obviously it will take super-polynomial (exponential?) time. But what about space? Is there some $k>1$ that we know of such that it will definitely take $\Omega(n^k)$ space? And do we know for sure that it won't take more than $O(n^l)$ space for some $l$?

(Or is the question itself stated badly for some reason?)

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Duke
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By definition, PSPACE consists of all languages decided by some Turing machine using polynomial space. So every language in PSPACE can be decided by some Turing machine using space $O(n^\ell)$ for some $\ell$. The space hierarchy theorem ensures moreover that for any $\ell$ there is a language in PSPACE which requires space $\Omega(n^\ell)$.

Any PSPACE-hard language requires space $\Omega(n^k)$ for some $k$, since otherwise any language in PSPACE would be decided by a Turing machine using subpolynomial space, and this contradicts the space hierarchy theorem.

For any $k > 0$ there is a PSPACE-hard language which can be decided in space $O(n^k)$. Indeed, take any PSPACE-complete language $L$. It can be decided using space $O(n^\ell)$ for some $\ell$. Consider now the language $L' = \{(x,0^{|x|^{\ell/k}}) : x \in L\}$, which is PSPACE-hard by reduction from $L$. An algorithm for $L$ can be converted to one for $L'$ which uses $\log n$ space to verify the input structure and $O(|x|^\ell)$ for the rest. Since $n \geq |x|^{\ell/k}$, $O(|x|^\ell) = O(n^{(k/\ell)\ell}) = O(n^k)$.

Yuval Filmus
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