On the diagonal,
$$
P^k(x,x)-\pi(x)
=\pi(x)\sum_{i\geq2}\lambda_i^kf_i(x)^2,
$$
so every summand is nonnegative. Let $m=\lceil t_{\mathrm{rel}}\rceil$. For every $i\geq2$,
$$
\lambda_i^{m+1}
\leq\lambda_2^{t_{\mathrm{rel}}}
=\left(1-\frac1{t_{\mathrm{rel}}}\right)^{t_{\mathrm{rel}}}
\leq e^{-1}.
$$
Hence
$$
\frac1{1-\lambda_i}
\leq\frac e{e-1}\frac{1-\lambda_i^{m+1}}{1-\lambda_i}
=\frac e{e-1}\sum_{k=0}^m\lambda_i^k.
$$
Multiply by $\pi(x)f_i(x)^2$ and sum over $i\geq2$ to obtain the required inequality.


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 215 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-215/1/b/solution