If $p$ is a polynomial approximating a continuous function $h$, why can we assume $\left\|p\right\|_\infty\le\left\|h\right\|_\infty$?

Question

Let $K\subseteq\mathbb R$ be compact and $h:K\to\mathbb R$ be continuous and $\varepsilon>0$. By the Stone-Weierstrass theorem, there is a polynomial $p:K\to\mathbb R$ with $\left\|h-p\right\|_\infty<\varepsilon$.

Why can we choose $p$ such that $\left\|p\right\|_\infty\le\left\|h\right\|_\infty$?

I'm not sure whether it's really crucial that $p$ is a polynomial. In general, if $K$ is a compact Hausdorff space, $\mathcal C$ is a subalgebra of $C(K)$ with $1\in\mathcal C$ and $f:K\to\mathbb R$ is continuous with $c:=\left\|f\right\|_\infty\ne0$, then there is a $\tilde g\in\mathcal C$ with $$\left\|\frac fc-\tilde g\right\|_\infty<\frac\varepsilon2.$$ Now $g:=c\tilde g\in\mathcal C$ and $\left\|f-g\right\|_\infty<\varepsilon$. But only if $\left\|\tilde g\right\|\le1$, we obtain $\left\|g\right\|_\infty\le c=\left\|f\right\|_\infty$ ...

Answer 1

$\|p\|\leq \epsilon +\|h\|$. Let $q=\frac {\|h\|} {\epsilon+\|h\|} p$. The $q$ is a polynomial and $\|q\| \leq \|h\|$. Now $\|q-h\| \leq \|q-p\|+\|p-h\|$. Can you finish?

Answer 2

Assuming $K$ is not empty and $h\ne 0.$

Let $e=\min (\|h\|/2, \varepsilon /2).$

For $x\in K$ let $\bar h(x)=h(x)$ if $|h(x)|< \|h\|-e\,;$ let $\bar h(x)=\|h\|-e$ if $h(x)\ge \|h\|-e\,;$ let $h(x)=e-\|h\|$ if $h(x)\le e-\|h\|.$

Now $\bar h$ is continuous. So take $p$ with $\|p-\bar h\|<e.$