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

Question

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.