Extend $f$ by zero outside $[0,1]$. Since the <Fourier transform> of $e^{-|x|}$ is $2/(1+\xi^2)>0$,
$$
\langle Kf,f\rangle
=\frac1{2\pi}\int_{\mathbb R}
\frac{2}{1+\xi^2}|\widehat f(\xi)|^2\,d\xi>0
$$
for $f\ne0$. Thus every eigenvalue is positive. From the relation in part (b),
$$
\boxed{\lambda=\frac2{1+\omega^2}>0}.
$$
The transcendental equation has its successive roots in intervals separated by the poles and zeros of $\tan\omega$, so $\omega_n$ grows linearly with $n$. Hence
$$
\boxed{\lambda_n=\frac2{1+\omega_n^2}=O(n^{-2})}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 326 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-326/2/c/solution