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

For $n\ge 1$ put $q_n(t)=2n{K}_{n-1}(t)\sin (nt)$. Using $\sin (nt)=(e^{int}-e^{-int})/(2i)$ and the Fourier coefficients of $K_{n-1}$ givesIndeed, when $1\le k\le n$ only the $e^{int}$ shift contributes, with coefficient $(-in)(1-|k-n|/n)=-ik$. For $-n\le k\le -1$ only the opposite shift contributes and gives $in(1-|k+n|/n)=-ik$; at $k=0$ both coefficients vanish.

If $P={\sum }_{|k|\le n}{a}_k{e}^{ikt}$, convolution multiplies its Fourier coefficients by those of $q_n$. Thus $-P\ast q_n=\sum ik{a}_k{e}^{ikt}=P^{\prime }$, proving the exact identity. Since $K_{n-1}\ge 0$ and has integral one,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 $n$.