Theorem still true if $X$ is not complete

Question

Consider the following theorem:

Let $S$ be a non-empty set and let $\{0\} \neq X$ be a vector space of bounded functions on $S$, with the condition that $S$ is a Banach space when $X$ is supplied with the supremum-norm. Suppose $f : S → \mathbb{F}$ is a function such that $fg \in X$ for all $g \in X$. Then the multiplication operator $M_f : X → X$, defined by $M_f (g) = fg$ $(g ∈ X)$, is bounded.

I want to know,

• Is $f$ necessarily bounded?
• Is this theorem still true if $X$ is not complete.

Any insights are much appreciated.

Answer 1

• No. Otherwise the theorem would be trivial ($\|M_f\|\leqslant\sup|f|$).
• No. Take $S=[0,+\infty)$,$$X=\{g\colon S\longrightarrow\mathbb{R}\,|\,g(x)=0\text{ if }x\gg0\},$$and $f(x)=x$. Then $g\in X\implies fg\in X$. However, $M_f$ is unbounded. Just take, for each $n\in\mathbb N$, $g_n=\chi_{[0,n]}$. Then $(\forall n\in\mathbb{N}):\|g_n\|_\infty=1$, but $\bigl\|M_f(g_n)\bigr\|=n$.