No, that's not how Perl manages memory. I've dug into this code lately, so here's a slightly fuller explanation that only has a few oversimplifications (as far as I know).

When Perl compiles a subroutine, it attaches a couple of arrays to the subroutine. These are lexical pads. The first array holds several other arrays. The first contained array holds the names of all lexical variables. The second contained array holds the values of all lexical variables. (I believe subsequent contained arrays hold values on recursion, but I haven't dug that deeply.)

The compile-time effect of my is to create an entry in the pad, storing the appropriate name in the first contained array and making a space in the second contained array for the value of that lexical variable.

The run-time effect of an assignment to a lexical variable is to store the appropriate value in the second contained array. If there's a conditional, though, that assignment won't happen.

For reasons of optimization, I believe that Perl doesn't reset all elements in the contained arrays to undef when it leaves a subroutine. Thus if your assignment doesn't happen due to a false conditional, you could see an old value persist through invocations.

As far as I know, the memory won't change location; if you use up megabytes of heap memory elsewhere, there should be no effect on the lexical pads unless you exhaust all real and virtual memory and then you have bigger problems.