Are innermost reductions perpetual in untyped λ-calculus?
I believe $tt$ with $t=(\lambda x. (\lambda y.1) (x x))$ will terminate using $L$ even if it has an infinite reduction.

The first reduction step is: $L(tt)=L(t)t=L(\lambda x.(\lambda y.1) (x x))t=(\lambda x.L((\lambda y.1) (x x)))t=(\lambda x.1)t$.

The first reduction step with $F_\infty$ is $F_\infty([(\lambda x.(\lambda y.1 (x x))) t]))=(\lambda y.1) (tt)$.