Product of $f(x)\cdot f(x)$ when the domain is empty

Question

Is this in some way a meaningful mathematical expression ? Would the product be equal to just $f(x)$ on its own ?

Answer 1

Keep in mind that "$\cdot$" is a function restricted to some domain, let's call it $D$. So, for every $a,b \in D$ we can evaluate the function $\cdot$ at these values and in return get a value $\cdot(a,b)$ which we also denote with $a \cdot b$.

Now let $f$ be a function with empty domain. Then $f = \emptyset$ and thus there are no $x,y$ with $(x,y) \in f$ or in other words, there are no $x,y$ such that $f(x) = y$. In order to evaluate $\cdot(f(x),f(x))$, however, we need some $y$ such that $f(x) = y$. In fact $\cdot(f(x),f(x)) = \cdot(y,y) = y \cdot y$ for precisely some such (and hence unique) $y$.

To summarize this argument: If $f$ doesn't take values, we can't use these values as parameters in other functions. So, no, $f(x) \cdot f(x)$ doesn't make sense for the empty function $f$. In fact, $f(x)$ doesn't make sense.