Distance from $x^n$ to lesser polynomials I am interested in the $L_1$ distance of $x^n$ to the $\mathbb R$-span of $\{1,x,\ldots,x^{n-1}\}$ over some interval. We can WLOG consider the interval $[0,1]$ (say) because scaling and shifting only affects the distance by a constant factor.

So far we have these basic results:
Theorem $x^n$ is not in the $\mathbb R$-span of $\{1,x,\ldots,x^{n-1}\}$
Proof Suppose it was: $x^n = \sum_{i=0}^{n-1} a_i x^i$. Differentiate $n$ times to get the contradiction $n! = 0$.
We can rephrase this theorem as saying that $\| x^n - p(x) \| > 0$ for all polynomials of degree $< n$.
Theorem There is no sequence of polynomials in the span, with limit $x^n$.
Proof We would have a sequence $p_n(x) = \sum_{i=0}^{n-1} a_{i,n} x^i$ where $x^n = \lim_{n \to \infty} p_n(x) = \sum_{i=0}^{n-1} (\lim_{n \to \infty} a_{i,n}) x^i$, but that is a contradiction.

Together with the previous result this implies there should be some positive $h(n)$ such that $\| x^n - p(x) \| > h$.
My question is about finding some $h$. How could we go about doing this? Are there relevant theorems?
My first idea was that we might use induction and find some way of saying: Since we can only get within $r$ of the derivative we can only get within $h(r)$ of the function itself.. but finding $h(r)$ seems just as hard as finding $h$ at all.
 A: To minimize
$$
\int_0^1\left|x^n-a_{n-1}x^{n-1}-a_{n-2}x^{n-2}-\dots-a_0\right|\;\mathrm{d}x\tag{1}
$$
we can differentiate with respect to each $a_k$ to get at the critical points
$$
\int_0^1\operatorname{sgn}(x^n-a_{n-1}x^{n-1}-a_{n-2}x^{n-2}-\dots-a_0)\;x^k\tag{2}\;\mathrm{d}x=0
$$
for $k=0\dots n-1$.
The minimal $P(x)=x^n-a_{n-1}x^{n-1}-a_{n-2}x^{n-2}-\dots-a_0$ must have $n$ distinct roots in $(0,1)$. If not, then there is a polynomial $Q(x)$ of degree $< n$ that has the same sign as $P(x)$ for all $x\in[0,1]$. Then for some $\epsilon>0$, $\int_0^1|P(x)-\epsilon Q(x)|\mathrm{d}x<\int_0^1|P(x)|\mathrm{d}x$.
Let $0< x_1< x_2<\dots< x_n<1$ be the roots of $P(x)$. By $(2)$ we have that
$$
\frac{1}{k+1}\left[(x_1^{k+1}-0)-(x_2^{k+1}-x_1^{k+1})+(x_3^{k+1}-x_2^{k+1})-\dots+(-1)^{n-1}(1-x_n^{k+1})\right]=0
$$
for each $k=0\dots n-1$, which is equivalent to
$$
x_n^k-x_{n-1}^k+x_{n-2}^k-\dots+(-1)^{n-1}x_1^k=\tfrac12\tag{3}
$$
for each $k=1\dots n$. Equation $(3)$ has roots
$$
x_j=\sin^2\left(\frac{\pi}{2}\frac{j}{n+1}\right)\tag{4}
$$
for $j=1\dots n$. That is, since $\displaystyle \operatorname{Re}\left(\frac{1}{e^{ix}+1}\right)=\frac12$,
$$
\begin{align}
&\sum_{j=1}^n(-1)^{n-j}\sin^{2k}\left(\frac{\pi}{2}\frac{j}{n+1}\right)\\
&=(-\tfrac14)^k\sum_{j=1}^n(-1)^{n-j}\sum_{m=0}^{2k}\binom{2k}{m}(-1)^m\exp\left(i\pi\frac{(k-m)j}{n+1}\right)\\
&=(-\tfrac14)^k(-1)^{n}\sum_{m=0}^{2k}\binom{2k}{m}(-1)^m\left(\frac{(-1)^{k-m+n}-1}{\exp\left(i\pi\frac{k-m}{n+1}\right)+1}-1\right)\\
&=(\tfrac14)^k\sum_{m=0}^{2k}\binom{2k}{m}(-1)^{k-m+n}\frac{(-1)^{k-m+n}-1}{\exp\left(i\pi\frac{k-m}{n+1}\right)+1}\\
&=(\tfrac14)^k\sum_{m=0}^{2k}\binom{2k}{m}\frac{1-(-1)^{k-m+n}}{\exp\left(i\pi\frac{k-m}{n+1}\right)+1}\\
&=(\tfrac14)^k\sum_{m=0}^{2k}\binom{2k}{m}\frac12\left(1-(-1)^{k-m+n}\right)\\
&=(\tfrac14)^k\sum_{m=0}^{2k}\binom{2k}{m}\frac12\\
&=\frac12\tag{5}
\end{align}
$$
Thus, the minimal monic polynomial is
$$
P(x)=\prod_{j=1}^n\left(x-\sin^2\left(\frac{\pi}{2}\frac{j}{n+1}\right)\right)\tag{6}
$$
