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Bernstein's inequality

We proved last Thursday the following theorem.

Theorem. (Bernstein) Let $latex T(x)$ be a trigonometric polynomial of degree N. Then

$latex \displaystyle \max_x|T'(x)| \le 2\pi N\max_x |T(x)|$.

Bernstein's inequality provides an estimate for the derivative of $latex T(x)$ in terms of its degree and its maximum. The inequality is sharp, since the polynomial

$latex T(x) = \sin 2\pi Nx$

satisfies the equality.

We also proved the following corollary.

Corollary. Let $latex T(x)$ be a trigonometric polynomial of degree N. Then

$latex \displaystyle \max_x |T(x)| \le 2\pi N\int_0^1 |T(x)|dx$.

We observed that Fejér's kernel

$latex \displaystyle F_{N+1}(x) \sum_{n=-N}^N \Big(1 - \frac{|n|}{N+1}\Big)e^{2\pi inx} = \frac{1}{N+1}\Big(\frac{\sin \pi(N+1)x}{\sin \pi x}\Big)^2$

proves the inequality in the Corollary is essentially sharp, as its maximum is $latex N+1$ and its integral is 1. In fact, $latex F_{N+1}$ also proves that Bernstein's inequality is essentially sharp.

Exercise 1.

  1. Use Bernstein's inequality to prove that $latex |F_{N+1}'(x)|\le 2\pi N(N+1)$;

  2. Note that $latex F_{N+1}(-1/(N+1))=0$ and $latex F(0)=N+1$, and conclude that there exists some $latex x$ such that $latex F_{N+1}'(x)\ge(N+1)^2$.

In the Real Analysis course we prove that, for functions on [0,1], we have

$latex ||f||_1 \le ||f||_2,$

which follows from the Cauchy-Schwarz inequality, and that the quotient $latex ||f||_2/||f||_1$ may be arbitrarily large. However, for trigonometric polynomials, we have the following.

Exercise 2. Let $latex T(x)$ be a trigonometric polynomial of degree N. Then

$latex ||T||_2 \le \sqrt{2\pi N}||T||_1$.

Note that the Corollary above states that $latex ||T||_\infty \le 2\pi N||T||_1$.

In fact, Bernstein's inequality states that $latex ||T'||_\infty \le 2\pi N||T||_\infty$.

Exercise 3. Apply Parseval's identity to $latex T$ and $latex T'$ to show

$latex ||T'||_2 \le 2\pi N||T||_2$.

Let $latex c_k$ the coefficients used in the proof of Bernstein's inequality and $latex x_k=(2k-1)/(4N)$, $latex k=1, 2, \ldots, 2N.$ Then, by taking a proper traslation,

$latex \displaystyle T'(x) = \sum_{k=1}^N c_k T(x+x_k)$.

Exercise 4.

  1. Show that $latex \displaystyle \int_0^1|T'(x)|dx \le \Big(\sum_{k=1}^N|c_k|\Big)\int_0^1|T(x)|dx.$

  2. Conclude that $latex ||T'||_1 \le 2\pi N||T||_1$.