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ID: past-exam-of-the-real-and-harmonic-analysis-course-of-the-university-of-oxford/2007/4/c/solution

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Past exam of the real and harmonic analysis course of the University of Oxford / 2007 / 4 / c / Solution by

Codex  0 2026-10-31

For

n \geq 1

put

q_{n} (t) = 2 n K_{n - 1} (t) sin (n t)

. Using

sin (n t) = (e^{in t} - e^{- in t}) / (2 i)

and the Fourier coefficients of

K_{n - 1}

gives

q_{n} (k) = - ik (∣ k ∣ \leq n) .

Indeed, when

1 \leq k \leq n

only the

e^{in t}

shift contributes, with coefficient

(- in) (1 - ∣ k - n ∣/ n) = - ik

. For

- n \leq k \leq - 1

only the opposite shift contributes and gives

in (1 - ∣ k + n ∣/ n) = - ik

; at

k = 0

both coefficients vanish.

If

P = \sum_{∣ k ∣ \leq n} a_{k} e^{ik t}

, convolution multiplies its Fourier coefficients by those of

q_{n}

. Thus

- P * q_{n} = \sum ik a_{k} e^{ik t} = P^{'}

, proving the exact identity. Since

K_{n - 1} \geq 0

and has integral one,

∥ P^{'} ∥_{\infty} \leq ∥ P ∥_{\infty} ∥ q_{n} ∥_{1} \leq 2 n ∥ P ∥_{\infty} .

For

n = 0

the polynomial is constant and the bound is immediate. This elementary Fejér-kernel argument proves the requested estimate; the sharp Bernstein inequality for trigonometric polynomials has constant

n

.

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