Maximum of the leading coefficient when $\text{deg}P=6$, $0\leq P(x)\leq1$ for $-1\leq x\leq 1$. Let $P(x)$ be a real polynomial of degree 6 with the following property:

For all $-1\leq x\leq 1$, we have $0\leq P(x)\leq 1$.

what is the maximum possible value of the leading coefficient of $P(x)$?
I obtained obvious answers when the degree is 1 or 2, but couldn't solve when the degree is 6.
I have a hunch that this has something to do with the Legendre polynomials, but I'm not quite sure. Any advice is welcome.
 A: Let $T(x)=\cos 6\theta$ $(x\in[-1,1])$ where $x=\cos\theta$. The leading coefficient of $T$ is $2^{6-1}$ The maximal difference to $0$ occurs at $\theta_k=k\pi/6,k=0,1,\ldots,6$. Let $x_k=\cos\theta_k$, when $k$ is even, $T(x_k)=1$, when $k$ is odd, $T(x_k)=-1$. Suppose $\deg R=6$ and the leading coefficient of $R$ is also $2^5$, if $\max|R|<1$, at $x_k$, we have $T(x_0)-R(x_0)>0$, $T(x_1)-R(x_1)<0$ ..., it changes sign at least $6$ times, contradicts with $\deg T-R<6$. Hence for all polynomial $Q$ s.t. $\deg Q=6$ and the leading coefficient is $2^{6-1}$,
the minimal of $\max|Q|$, which is $1$, occurs when $Q(x)=\cos6\arccos x$.
Now, denote $L(x)=\frac12(Q(x)+1)/\max|Q|$, we can see the maximal leading coefficient of $L$ is $2^4=16$.
A: Hint
The value you’re looking for is the same as the one of the problem ... For all $-1 \le x \le 1$, we have $-1/2 \le Q(x) \le 1/2$... as we’re just substracting $1/2$ to the polynomial which doesn’t change the leading coefficient.
Now the map $f(x) = \frac{1}{2^{6-1}}T_6(x)$ where $T_n$ is the n-th Chebyshev polynomials is the one having $1$ for leading coefficient and of minimal $\infty$-norm (see Wikipedia article for more details). And this norm is equal to $\Vert f \Vert_\infty =\frac{1}{2^{6-1}}$. In your problem you allow the norm to be equal to $1/2$.
Hence the value you’re looking for is equal to $\frac{1/2}{1/2^{6-1}}=2^4=16$.
