= Solution
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.
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