Show that $ inf_{c in mathbb{R}} |f - c|_{infty} = | f - frac{1}{2}(min f + max f) |_{infty}$
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I'm very unsure about what I did here, so if someone could proof check on the following, I would be very thankful. Let $I$ be a an interval of $mathbb{R}$ . For $f in mathcal{C}(I)$ , we say that $P in mathcal{P}_n$ ( $P$ is a polynomial of degree $n$ ) is the best approximation to $f$ if $| f- P |_{infty} = inf , { |f-P|_{infty} : P in mathcal{P}_n } $ . I want to show that the best approximation of $f in mathcal{C}(I)$ by a constant is $frac{1}{2}(inf_{x in I} f(x) + sup_{x in I} f(x))$ . $text{Proof.}$ Since $f$ is continuous and $I$ is compact, $inf f$ and $sup f$ are reached and we have that $inf f < infty$ and $sup f < infty$ . Hence our claim is equivalent to saying that $inf , { |f - c|_{infty}: c in mathbb{R}} = | f - frac{1}{2}(min f + max f) |_{infty}$ Note that we have...