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Given that $(\psi_i)_0^{|\chi|-1}$ is an orthonormal basis in $l^2(\pi)$ of real eigenfunctions of $K$ where $K$ is a Markov operator, then why is $\sum_0^{|\chi|-1} \psi_i^2(x)\pi(x) = 1$?

If we were summing over $x$, then it would make sense since $\psi_i$ is of unit norm. This time though, we're summing over $i$ so I don't see why the above still holds. I think it's false, but I want to make sure.

And the source of this is here: http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.aoap/1034968224

(At the end of page 702.)

I'm also stuck trying to prove corollary 2.2 in the paper, which is $$\|h_t^x-1\|_2^2 \leq \frac{1}{\pi(x)}e^{-t}+\|k_x^{[t/2]}-1\|_2^2.$$ Help would be much appreciated!

UPDATE I GOT IT.

However, I would still appreciate someone explaining to me why $\left\|H_t\right\|_{2\to \infty}=\sup_x h_{2t}(x,x)>1$ for all finite $t>0$, yet the limit as $t\to \infty$ is $1.$ That's unclear to me. Also why is $\left\|H_t\right\|_{2\to q}$ decreasing with respect to $t.$ I'd especially appreciate an intuitive explanation.

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    What is $\pi(x)$? Could you explain your notation a bit?2011-03-09
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    It's probably the stationary distribution.2011-03-09
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    $\pi(x)$ represents the probability w.r.t the stationary distribution $\pi$ of being at state $x.$2011-03-09
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    Shouldn't $K$ be reversible for there to exist an orthonormal basis of eigenfunctions?2011-03-09
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    Yep, forgot to mention that $K$ is reversible (thus self-adjoint).2011-03-09

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