Polynomials are not complete in the supremum norm on a compact interval

By [L1], the polynomial algebra $P[a,b]$ is dense in $C([a,b])$, so in particular the continuous function $h(x)=\mathrm{\mid }x-c\mathrm{\mid }$ lies in its supremum-norm closure.

The function $h$ is not a polynomial on $[a,b]$: if a polynomial $p$ agreed with $h$, then on $[c,b]$ it would satisfy $p(x)=x-c$, and on $[a,c]$ it would satisfy $p(x)=c-x$. So the polynomial $p(x)-(x-c)$ would vanish on the interval $[c,b]$, hence be identically zero, and similarly $p(x)-(c-x)$ would be identically zero, forcing $x-c=c-x$ for all $x$, impossible because $a<b$.

Steps 1.1 and 1.2 show that $P[a,b]$ is dense and proper in the Banach space $C([a,b])$, so it is not complete.

Since $C([a,b])$ is Banach by [L2], and the natural inclusion of $P[a,b]$ into $C([a,b])$ is a linear isometry with dense image by [L1], $C([a,b])$ is a completion of $P[a,b]$.