Does Multitape reduction to a one tape machine preserve space complexity?
You can take $C(k)=1$.

Simulate the $(k+2)$-tape machine (input, output, $k$ work tapes) by storing the $i$th character of all the simulated tapes in the $i$th character of the single tape. This is a standard construction, which also requires you to store the head positions using the characters on the tape. Now, you can do the whole computation without using any extra space.

In a little more detail, suppose you're simulating a Turing machine with tape alphabet&nbsp;$\Sigma$. Let $\Sigma'$ be the alphabet containing a character&nbsp;$a'$ for each character $a\in\Sigma$. We'll simulate using alphabet $(\Sigma\cup\Sigma')^{k+2}$, so each character on the simulating tape is a tuple $(c_1, \dots, c_{k+2})$. If $c_j\in\Sigma$, that means that the corresponding cell on the $j$th tape of the simulated machine contains that character; if $c_j=a'\in\Sigma'$, that means the cell contains character&nbsp;$a$ and that tape's head is in this position.



Note that it only makes to talk about space $\Omega(n)$ on single-tape Turing machines, since you now have to count the space used by the input.