L is properly contained in PSPACE

Both bounds meet the local work-space-constructibility contract. On unary input $1^n$, count its length in binary, using $O(\log (n+2))$ cells, and add $2$. The binary length and a power-of-two test compute $f(n)=\lceil {\log }_2(n+2)\rceil$ in the same space; outputting the binary count gives $g(n)=n$ within $O(n)$ cells. Also $f(n)\ge \lceil {\log }_2(n+2)\rceil$ tautologically, $f(n)=o(n)$, and $g(n)$ satisfies the hierarchy's address lower bound for all sufficiently large $n$. Thus The read-only-workspace space hierarchy theorem applies and yields a language in $\mathrm{D}\mathrm{W}\mathrm{O}\mathrm{R}\mathrm{K}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}(n)\setminus L$.

The bridge Read-only work space becomes all-tapes space after charging input puts that witness language in $\mathrm{D}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}(n+n)=\mathrm{D}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}(2n)$, hence in PSPACE. For any $L$ computation, the same bridge gives $\mathrm{D}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}(n+f(n))\subseteq \mathrm{D}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}(2n+O(1))\subseteq \mathrm{P}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}$, by the polynomial-space definition. The witness lies outside $L$ by step 1.1, so the inclusion is proper.