"Free energy" moving the space elevator possible?

In this other question I have found the following:

A space elevator can climb at a low, constant vertical speed (albeit for a very long climb), and gets its orbital speed for free [borrowed from Earth's rotational speed of course].

So, does this this mean that to start the elevator, we invest some energy E, and then from some point elevator will start moving itself and give us E back, maybe even E'>E?

(if you think this is more for Physics SE I'm fine with that)

• Have you seen Tom Spilker's answer to that question, which goes into considerable detail on that point? Sep 11, 2018 at 8:26

Yes, but E' > E is dubious, at least in case of Earth. The point of reversal lies exactly at the GEO altitude, where the elevator moves at speed proper for that altitude and doesn't put any strain on the tether. As it climbs further, past GEO its 'horizontal' velocity exceeds orbital velocity of that altitude, and it's pulled away from Earth, needing to brake against the tether and putting a strain on it.
The problem here is that you'd need a lot of that tether extended past GEO to get E' > E and that tether is not cheap. It's much easier to extend, say, 10% of its length past GEO and put an asteroid or a big, sun-shaded chunk of ice on the end, to offset the mass of the tether pulling towards Earth.