= Supremum bound for a complete function-algebra norm
{title2=$\|f\|_\infty\le\|f\|$}
For an algebra of complex-valued functions with a complete <algebra norm>, every point value $f(t)$ belongs to the <spectrum of an element> of $f$ in the artificial <unitization>. Indeed $(h,\lambda)\mapsto h(t)+\lambda$ is an algebraic unital <character of an algebra>, and it vanishes on $f-f(t)1$. The <spectrum> bound gives $|f(t)|\le\|f\|$ without first assuming that evaluation is continuous.
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