= Solution
Let $1$ be the constant function and define $h=L1\in C([0,1])$. Positivity gives $h\ge0$. Fix any $x$ and any continuous $f$. The function $g=f-f(x)1$ vanishes at $x$, so the zero-preserving assumption gives $Lg(x)=0$. By linearity this is $Lf(x)-f(x)L1(x)=0$. Since $x$ was arbitrary,
$$
\boxed{Lf=hf\quad\text{for every }f\in C([0,1]),\qquad h=L1\ge0.}
$$
This proves the <zero-preserving linear operator on continuous functions> representation. Zero preservation and linearity determine the <multiplication operator>; positivity is needed only for the multiplier's sign. The three numbered conditions in the source are hypotheses of this one request, rather than three separate subquestions.
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