Does Multitape reduction to a one tape machine preserve space complexity? Suppose a Turing machine $M$ has a read-only input-tape and $k$ read-write work-tapes whose non-blank cells are each bounded by $f(|x|)$ where $|x|$ is the length of the input.

Is there some constant $C(k)$ and Turing machine $M^{'}$ with a read-only input tape such that $M^{'}$ accepts exactly when $M$ accepts, $M^{'}$ has only one read-write work tape, and the blank cells of that tape are bounded by $C f(|x|)$?

If there doesn't exist such a constant $C(k)$ that is independent of $M$, does there exist a constant $C(k, M)$?  In other words does reducing the number of tapes 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.