Let $f$ be a real-valued continuous function on $[0,1]$ which is not a polynomial. Prove that there does not exist a polynomial $p$ such that $||f-p||$=$\min_{q \in \mathbf{P}}||f-q||$, where $\mathbf{P}$ is the space of all polynomials.
Attempt: Suppose such a $p$ exists, say of degree $n$. Then $p$ is the best polynomial approximation of $f$ in $\mathbf{P_k}$ for all $k \geq n$, where $\mathbf{P_k}$ is the space of all polynomials of degree at most $k$.
Then by Chebyshev's criterion, $f-p$ attains its maximum magnitude at $k+2$ distinct points with alternating signs. As $k$ can be arbitrarily large, the continuous function $f-p$ oscillates infinitely often.
But I don't think that gives any contradiction. Any insights?
Following @person's comment:
By the Stone-Weierstrass theorem there is a sequence of polynomials $p_n$ which uniformly converges to your continuous function $f \colon [0,1] \to \mathbb R$. If such a "best-approximating" polynomial $p$ existed then $\|f - p\| > 0$, where $\|\cdot\|$ denotes the supremum-norm, because $f$ itself is not a polynomial. So the sequence of polynomials $p_n$ from the Stone-Weierstrass theorem would not have uniformly converged since we must have $\|p_n - f\| \geq \|p - f\| > 0$ for all $n$.