# $A\in LSPACE \Longrightarrow CYCLE(A)\in LSPACE$

Let $$A$$ be a language and define $$CYCLE(A)= \{ yx | xy \in A \}$$ I need to prove, or disprove, $$A\in LSPACE \Longrightarrow CYCLE(A)\in LSPACE$$.

First I tried to prove $$CYCLE(A) \le_{L} A$$ which if true means that if $$A\in LSPACE$$ then $$CYCLE(A)\in L$$, hence I tried to define a log space transducer $$M$$ that compute a log space reduction from $$CYCLE(A)$$ to $$A$$:

M="for input $$\langle w \rangle$$:

1. for every symbol $$s$$ in $$w$$:
1. $$y = w[:s]$$, $$x = w(s:]$$
2. build $$xy$$ on the working tape
3. run the Turing machine that decides $$A$$ on $$xy$$
4. if it accepts $$xy$$ write $$xy$$ to output and accept"

But this isn't good, as while $$A$$ is known to be in $$LSPACE$$, and therefore its Turing machine runs in O($$log{n}$$) space, the building of $$xy$$ each time will take $$O(n)$$ space.

I then tried to build a Turing machine that decides $$CYCLE(A)$$ in O($$log{n}$$) but the idea seems to be exactly as the transducer: find a possible split to $$yx$$, build $$xy$$ and run $$A$$'s Turing machine on $$xy$$ and if it accepts, accept the input. But again building $$xy$$ will take O($$n$$) space.

At this point I shifted to try to disprove. I need to find language $$A$$ such that $$A\in LSPACE$$ but $$CYCLE(A)\notin L$$, but the more I think about it the more I lean towards the hypothesis being true, yet I'm unable to find how to prove it.