Zero-preserving linear operator on continuous functions
= Zero-preserving linear operator on continuous functions
{title2=$Z_f\subseteq Z_{Lf}\implies Lf=(L1)f$}
If a linear operator on real continuous functions preserves every zero, evaluate $f-f(x)1$ at $x$. The zero-preserving hypothesis forces $Lf(x)=f(x)L1(x)$, so the operator is multiplication by the continuous function $L1$. Positivity further makes that multiplier nonnegative; linearity and zero preservation alone already give the multiplication representation.