Bounded operator on a non-empty set Let $S$ be non-empty set, and let $X$ be the vector space of bounded functions on $S$, subject only to the condition that it be a Banach space when $X$ is supplied with the supremum-norm. Suppose $f:S \to \mathbb{F}$ is a function such that $fg \in X$ for all $g \in X$. Then the multiplication operator $M_f:X \to X$, defined by $M_f(g)=fg(g \in X$) is bounded.
I do not ask for the proof of this theorem. I just wonder if we assume $X\neq \lbrace0\rbrace$,
*) Is it true that $f$ is necessarily bounded? That would certainly explain why $M_f$ is bounded.
*)Is the theorem still true if $X$ is not required to be complete?
 A: Well, I don't think there's much to ponder here. Assume that $f$ is bounded. This inequality holds in total generality: 
$$
|f(s)g(s)|\le \sup_{s\in S}|f(s)|\sup_{\sigma\in S} |g(\sigma)|.$$ 
Using the $\|\cdot\|_\infty$ notation we have 
$$
\|M_f(g)\|_\infty \le \|f\|_{\infty}\|g\|_{\infty},$$ 
which means precisely that the linear operator $M_f\colon X\to X$ is bounded and that its operator norm is smaller or equal to $\|f\|_{\infty}$. 
For all of this to make sense we only need that $X$ be a normed space, which is always the case for any set $S$. (Actually, $X$ is always Banach, but that is not needed at this stage). Notice that, if $S\ne \varnothing$, you cannot have $X=\{0\}$. Indeed, if $s\in S$, you can define the functions 
$$\tag{1}
\lambda_s(\sigma)=\begin{cases} \lambda, &\sigma=s\\ 0, & \sigma \ne s,\end{cases}$$
where $\lambda\in \mathbb R$ is arbitrary. Since $\lambda_s\in X$, the space $X$ is strictly bigger than $\{0\}$. 
Finally, you also ask whether the boundedness of $f$ is a necessary condition for the boundedness of $M_f$. The answer is affirmative. If $f\colon S\to \mathbb R$ is an unbounded function then there exists a sequence $s_n\in S$ such that $|f(s_n)|\to \infty$. Then 
$$
\|M_f(1_{s_n})\|_\infty \to \infty, $$
(where $1_{s_n}$ is defined in (1) with $\lambda=1$) and $\|1_{s_n}\|_\infty=1$. So $M_f$ maps a bounded sequence to an unbounded one, which means that $M_f$ is not bounded.
