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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?

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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 $p=2^{-n}U_n$

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