Where is the problem in this proof that deciders are no stronger than LBAs?
You seem to be mixing some concepts. Over an empty input alphabet, there are no decision problems, so as a model, Turing machines with empty input alphabets are not interesting.
It is true that the problem of deciding whether a given TM accepts the empty word is interesting (and undecidable), but it's not the same thing as having an empty input alphabet.
So the &quot;reductions&quot; you describe are not exactly reductions in the formal sense, just some translations between models.
As for a concrete problem with your argument: first, an LBA runs on a fixed portion of the tape which is prescribed by the input length. If you have an empty input, the LBA is not allowed to touch the tape, so it can't really do anything. An LBA doesn't just mean that the space is linear in the input. So you can't manually add the &quot;m&quot; input cells. This breaks your argument.
But also note that the definition of the function $f(x)=x+m$ is a bit strange, since on an empty input alphabet, you always have $x=0$.