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Show that for each $f in C[0,1]$ there is an $n$-th degree polynomial $p(x)$ on $[0,1]$ such that $||f(x)-p(x)||_infty leq ||f(x)-q(x)||_infty$ for any other $n$-th degree polynomial $q(x)$.

This looks similar to the following

If $A subseteq (X,||cdot||)$ is compact and non-empty then for each $x in X$ there is some $y_0 in A$ such that
$$||x-y_0||=inf{||x-y||: y in A}$$

However, the set of $n$-th degree polynomial is finite-dimensional (hence closed), can we prove the it is compact?

Theorem (Distance between compact sets): Let $(X,d)$ be some metric space, $A,Bsubset X$ be nonempty and compact.
Then there are $xin A$ and $yin B$ with
$$d(x,y)=d(A,B).$$

Here, the distance between two sets $P,Qsubset X$ is defined as

$$d(P,Q) = inf_{xin P, yin Q}d(x,y).$$

Proof: Let $(x_n,y_n)_{ninBbb N}subset Atimes B$ be a minimizing sequence, say

$$d(A,B) leqslant d(x_n,y_n) < d(A,B)+frac1n.$$

Because $A$ is compact, there is a subsequence $(x_{n_k})_{kinmathbb N}$ that is convergent, say to $xin A$.

Because $B$ is compact, there is a subsequence $(y_{n_{k_j}})_{jinmathbb N}$ that is convergent, say to $yin B$.

Then, as $jtoinfty$ we see from

$$d(A,B) leqslant dleft(x_{n_{k_j}},y_{n_{k_j}}right) < d(A,B)+frac1{{n_{k_j}}}$$

that $d(x,y) = d(A,B)$. $square$

We will use the theorem with $A = {f}subset C[0,1]$.
For $B$, let $P_nsubset C[0,1]$ be the subset of $n$-degree polynomials.
Pick any $qin P_n$ and consider $B = overline{mathcal B(f,lVert qrVert)}cap P_n$.

It suffices to check that $B$ is compact.
To that end, notice that $P_n$ is a closed, finite-dimensional subspace of $C[0,1]$, so that $B$ is the intersection of two closed sets and hence closed.
Moreover, $B$ is by construction bounded, so it is a closed and bounded subset of a finite-dimensional normed space, and hence compact.

The theorem will produce some $p$ that minimizes distance to $f$ over $B$.
Can you see why $p$ minimizes distance over all of $P_n$? (Use the triangle inequality!)