# Extremal property of Chebyshev polynomials of second kind

Theorem

Let $$p(x)=x^n+\dots$$ monic polynomial of degree $$n$$ and $$U_n$$ n-th Chebyshev polynomial of second kind.Then it holds $$\|p\|\geq\|\dfrac{1}{2^{n}}U_n\|=\dfrac{1}{2^{n-1}}$$ where $$\|f\|=\int_1^{-1}|f(x)|dx$$.

I found this theorem but without proof.Can someone give me a reference or prove it here?

Note that $$U_n(x)=\frac {\sin (n+1)\theta}{\sin \theta}, x=\cos \theta$$, so in particular the degree of $$U_n$$ is $$n$$ and its leading coefficient is $$2^n$$ since $$\sin ((n+1)\theta)+\sin ((n-1)\theta)=2\cos \theta \sin (n\theta)$$, so $$U_n(x)=2xU_{n-1}(x)-U_{n-2}(x), n \ge 2, U_0(x)=1, U_1(x)=2x$$

Let $$s(x)=\frac{x}{|x|}, x \ne 0, s(0)=0$$ be the sign function for real $$x$$. We note that $$f(x)s(f(x))=|f(x)| \ge f(x)s(g(x))$$ for any pair of real functions $$f,g$$ as $$|f(x)| \ge \pm f(x)$$.

Lemma: $$\int_{-1}^1U_k(x)s(U_n(x))dx=0, k=0,..,n-1$$

Assuming the lemma is proved, it then implies (if $$p$$ has real coefficients, while otherwise, we apply it to $$\Re p$$ which is a monic polynomial of the same degree with real coefficients and $$|p(x)| \ge |\Re p(x)|$$) that

$$\int_{-1}^1(p(x)-2^{-n}U_n(x))s(U_n(x))=0$$ since $$p(x)-2^{-n}U_n(x)$$ has degree at most $$n-1$$ so it is a linear combination of $$U_0, U_1,..U_{n-1}$$

But now as $$|p(x)| =p(x)s(p(x)) \ge p(x)s(U_n(x))$$ as noted, we integrate and get:

$$\int_{-1}^1|p(x)|dx \ge \int_{-1}^1p(x)s(U_n(x))dx=\int_{-1}^12^{-n}U_n(x)s(U_n(x))dx=2^{-n}\int_{-1}^1|U_n(x)|dx$$ so we only need to prove the lemma and compute $$\int_{-1}^1|U_n(x)|dx$$ to finish

First the Lemma:

Remembering the definition and using $$x=\cos \theta, 0 \le \theta \le \pi, \sin \theta \ge 0$$:

$$\int_{-1}^1U_k(x)s(U_n(x))dx=\int_0^{\pi}(\sin (k+1)\theta)s(\sin (n+1)\theta)d\theta$$ and since the integrand is even, it is enough to prove $$\int_{-\pi}^{\pi}(\sin (k+1)\theta)s(\sin (n+1)\theta)d\theta=0$$ as that is twice our integral.

We use the usual Fourier trick and notice that if we let

$$I_k=\int_{-\pi}^{\pi}e^{ik\theta}s(\sin (n+1)\theta)d\theta=\int_{-\pi+a}^{\pi+a}e^{ik(\theta+a)}s(\sin (n+1)(\theta+a))d\theta=\int_{-\pi}^{\pi}e^{ik(\theta+a)}s(\sin (n+1)(\theta+a))d\theta =e^{ika}\int_{-\pi}^{\pi}e^{ik\theta}s(\sin (n+1)(\theta+a))d\theta$$

with first equality by substitution, second by periodicity.

But now for $$a=\frac{\pi}{n+1}$$ we clearly have $$s(\sin (n+1)(\theta+a))=-s(\sin (n+1)\theta)$$, so $$I_k=-e^{i\frac{k\pi}{n+1}}I_k$$, or $$I_k=0, 1 \le k \le n$$ and the lemma follows by noticing that our original integral is the imaginary part of $$I_k$$

For the last part:

$$\int_{-1}^1|U_n(x)|dx=\int_0^{\pi}|\sin (n+1)\theta|d\theta=\frac{1}{n+1}\int_0^{(n+1)\pi}|\sin \theta|d\theta=\int_0^{\pi}|\sin \theta|d\theta=2$$ so we are done (again we use substitution, periodicity and then $$|\sin \theta|=\sin \theta, 0 \le \theta \le \pi$$)

Note that the proof shows that we can have equality iff $$p=2^{-n}U_n$$