Best uniform polynomial approximants exist

Write $p_c(x)={\sum }_{j=0}^n{c}_j{x}^j$ for $c=(c_0,\dots ,c_n)\in {\mathbb{R}}^{n+1}$ and set $F(c)=\mathrm{\parallel }f-p_c{\mathrm{\parallel }}_{\mathrm{\infty },I}$. For $0\le j\le n$, compactness of $I$ and [F2] give $M_j:={\max }_{x\in I}\mathrm{\mid }x{\mathrm{\mid }}^j<\mathrm{\infty }$. For every $x\in I$, the triangle inequality gives $\mathrm{\mid }f(x)-p_c(x)\mathrm{\mid }\le \mathrm{\mid }f(x)-p_d(x)\mathrm{\mid }+\mathrm{\mid }{p}_d(x)-p_c(x)\mathrm{\mid }$; taking suprema and then exchanging $c,d$ yields $\mathrm{\mid }F(c)-F(d)\mathrm{\mid }\le \mathrm{\parallel }{p}_c-p_d{\mathrm{\parallel }}_{\mathrm{\infty },I}\le {\sum }_{j=0}^n\mathrm{\mid }{c}_j-d_j\mathrm{\mid }{M}_j$. Thus $F$ is continuous. The sublevel set $K:=\{c:F(c)\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },I}\}$ is closed and contains $0$.

If $c\in K$, then $\mathrm{\parallel }{p}_c{\mathrm{\parallel }}_{\mathrm{\infty },I}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },I}+F(c)\le 2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },I}$. For $n=0$, take ${\tau }_0=a$; for $n\ge 1$, take distinct nodes ${\tau }_j=a+j(b-a)\mathrm{/}n$, $0\le j\le n$. Each $\mathrm{\mid }{p}_c({\tau }_j)\mathrm{\mid }\le 2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },I}$, so [F4] bounds every coefficient $c_k$ uniformly on $K$. Hence $K$ is bounded.

By [F3], the nonempty closed bounded set $K$ is compact, so the continuous function $F$ attains a minimum $\alpha$ on $K$. Since $0\in K$, $\alpha \le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },I}$; every coefficient vector outside $K$ has $F(c)>\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty },I}\ge \alpha$. Therefore $\alpha$ is the minimum over all of ${\mathbb{R}}^{n+1}$, which by [F1] is $E_n(f;I)$. The attaining polynomial is the required $p_{\ast }$.