For every monic polynomial $P$ of degree $n$ (with leading coefficient 1), it is well-known that $$\max _{x \in[-1,1]}|P(x)| \geq \frac{1}{2^{n-1}}.$$ A standard proof uses Chebyshev polynomials.
Is there a proof which does not use Chebyshev polynomials?
I have come up with a (somewhat strange) proof that uses only the first Chebyshev polynomials $$Q_2(x) = 2x^2 - 1$$ I will use the fact that this is a surjection from $[-1, 1] \to [-1, 1]$.
Define $$M(n) = 2^{n-1}\inf_P \max_{x \in [-1, 1]} |P(x)|.$$ where $P$ goes over monic polynomials of degree $n$.
The key steps of the proof are two facts.
Fact 1: $M(2n) \leq M(n)^2.$
Proof: If $P$ is a degree $n$ polynomial with $\max_{x \in [-1, 1]} |P(x)| = \alpha$, then the polynomial $$P'(x) = 2^{-1} \alpha^{2} Q_2(\alpha^{-1}P(x))$$ is of degree $2n$ and satisfies $\max_{x \in [-1, 1]} |P’(x)| = 2^{-1} \alpha^{2}$. Taking $\alpha$ to be sufficiently close to $M(n)$, we get $M(2n) \leq M(n)^2$.
Fact 2: $M(n) \leq M(2n).$
Proof: If $P'$ is a degree $2n$ polynomial with $\max_{x \in [-1, 1]} |P'(x)| = \alpha$, then $P'(x) + P'(-x)$ is an even polynomial. Thus we can write $$P'(x) + P'(-x) = 2^{-n+1} P(Q_2(x))$$ where $P$ is some monic polynomial of degree $n$. Thus we have $\max_{x \in [-1, 1]} |P(x)| \leq 2^{n}\alpha$. Taking $\alpha$ to be sufficiently close to $M(2n)$, we get $M(n) \leq M(2n)$.
Combining the two parts, we conclude that $M(n) \leq M(n)^2$, so $M(n)$ is either at least $1$ or equal to $0$. It suffices to show that $M(n)$ is not equal to $0$. This is easy: for any monic $P$ of degree $n$, one can find some $x \in [-1, 1]$ that is distance at least $1/n$ from each root, so $M(n) \geq (1/n)^n$.
