Is DSPACE(n) = DSPACE(1.5n)? From space-hierarchy theorem it is  known that if $f$ is space-constructible then
DSPACE($2f(n)$) is not equal to DSPACE($f(n))$. 

Here, by DSPACE($f(n))$  I mean the class of all problems that can be solved in space $f(n)$ by a Turing machine with some fixed alphabet. This allows  to consider Space-hierarchy theorem with such accuracy. 

The standard argument gives multiplicative constant $2$: we need space $f(n)$ for constructing a calculation of some Turing machine by an universal one. Also we need $f(n)$ to solve a problem with halting. 

Question: Is DSPACE($f(n)$) equal to DSPACE($\frac{3}{2}f(n)$)?
It can be proved that DSPACE$(f(\frac 32 n))\ne $ DSPACE$(f(n))$ if $f$ grows at least linearly by using a simple variant of the standard padding argument.
For a language $L$, let $L'=\{x0^{|x|/2}\mid x\in L\}$.

Claim. $L\in$DSPACE$(f(n))$ if and only if $L'\in$DSPACE$(f(\frac 23n))$ if $f(n)\ge \frac 32n$.

(My first answer had several incorrect statements, thanks to Emil for spotting this.)

I will first show how to use the claim to prove the hierarchy.
Since $f$ grows at least linearly, we have DSPACE$(2f(n))\subset$DSPACE$(f(2n))$.
Take a language $L\in$DSPACE$(f(2n))\setminus$ DSPACE$(f(n))$.
Using the claim, $L'\in$DSPACE$(f(\frac 43 n))=$ DSPACE$(f(n))$, where the last equality is by the indirect assumption.
But then $L\in$DSPACE$(f(\frac 32 n))=$ DSPACE$(f(n))$, where the last equality is again by the indirect assumption, giving a contradiction.

Proof of the claim.
If $L'\in$DSPACE$(f(\frac 23 n))$, then to prove $L\in$DSPACE$(f(n))$, we just need to write $|x|/2$ 0's to the end of the input $x$ and simulate the machine that accepted $L'$.
Since $f(n)\ge \frac 32 n$, this won't increase the space we use.
(In fact, knowing how many 0's to write is not clear at all if $f$ is small and we cannot increase the alphabet size - instead, we can use another tape and write on that everything that would come after the end of $x$.)

The other direction is just this simple by replacing the 0's with *'s, if we are allowed to write *'s. (See the issues with this in my comment to the question.)
If we are not allowed to write stars, then we slightly modify the definition of $L'$ as $L'=\{x10^{|x|/2}\mid x\in L\}$.
Now, instead of writing stars, we keep the original input $x10^{|x|/2}$ and work with that.
But whenever we reach a 1, we go right until we hit another 1 to check whether it was the end-of-word 1 or not.
If we've found another 1, we just go back to our 1.
If we haven't, we still go back, but we'll know that it should be treated as a star - if we were to write on it, then we also write a 10 after it to have a new end-of-current-word marker.
(In fact, there's also a small catch in this part if $f$ is small - how can we check if the input is of the form $x10^{|x|/2}$? Without destroying the input, I can only solve this by using multiple heads for small $f$.)