Eigenfunction expansion with Markov operators I'm having trouble comprehending the following.
http://www.stat.berkeley.edu/users/pitman/s205s03/lecture28.pdf in the proof of theorem 28.3.
It says that $H_t(x,y)=H_t\delta_y(x)$, where $\delta_y(x)=1$ if $x=y$ and $0$ if $x\neq y$.
I'm stuck with understanding this. Please help?
Also why is $E(H_t f)=Ef$ ? I particularly don't know why $K(c)=c$ if $c$ is a constant and $K$ is a Markov operator, because I thought $K$ could only operate on functions. Could we consider $c$ as a function too although it really is just a scalar?
EDIT
I know this is another late question but about the same paper, there's an equality I don't understand.  Near the end, it says
$$|h_t(x,y)-1|=\left|\sum_{z}(h_{t/2}(x,z)-1)(h_{t/2}(z,y)-1)\pi(z)\right|$$
On the RHS it looks like I should get $|\sum_{z} (h_{t/2}(x,z)h_{t/2}(z,y)-h_{t/2}(x,z)-h_{t/2}(z,y)+1) \pi(z)|.$
However I can't simplify that to be the LHS.
And lastly, the inequality $$\left\|H_t-E\right\|_{2\to\infty}\leq \left\|H_{t_1}\right\|_{2\to\infty} \left\|H_{t_2}-E\right\|_{2\to 2}$$ for $t=t_1+t_2.$
These have been two questions I have been unable to solve and they would help me greatly in understanding Markov bounds.  Thanks!
pr.probability 
 A: The first is an application of the definition of $H_t$ as an operator which works on functions. 
$$H_t f(x) = e^{-t} \sum_{i=0}^{\infty} \frac{t^i K^i f(x)}{i!} \; .$$
Apply it on the function $\delta_y$ to give
$$H_t \delta_y(x) = e^{-t} \sum_{i=0}^{\infty} \frac{t^i K^i \delta_y(x)}{i!} \; .$$
So, the crucial thing to work out is what is the action of $K$ on $\delta_y$. Again using the definition, this becomes
$$K \delta_y (x) = \sum_{z \in \mathcal{X}} K(x,z)\delta_y(z)$$
Since $\delta_y(z)=1$ when $z=y$ and $0$ elsewhere this simplifies to
$$K \delta_y (x) = K(x,y) \; .$$
Applying K $i$ times to $\delta_y(x)$ should give you the $i$^th power of the stochastic matrix $K(x,y)$ which is denoted in the text by $K^i(x,y)$. (This confused me at first, since I though they meant an ordinary power of the $(x,y)$-component.) So for instance, for power two, this means
$$K^2 \delta_y(x) \equiv K^2(x,y) = \sum_{z \in \mathcal{X}} K(x,z)K(z,y) \; .$$
Putting this all together, we get
$$H_t \delta_y(x) = e^{-t} \sum_{i=0}^{\infty} \frac{t^i K^i(x,y)}{i!} \; .$$
But this is exactly how $H_t(x,y)$ is defined. So up till there, much ado about nothing, just some definitions.
EDIT 1 Purely using the definition 
$$h_t(x,y)=\frac{H_t(x,y)}{\pi(y)}$$
one can show that
$$\begin{eqnarray}
h_t(x,y) & = &\frac{H_t(x,y)}{\pi(y)}=\frac{\sum_z H_{t/2}(x,z)H_{t/2}(z,y)}{\pi(y)} \\
& = &\sum_z \frac{H_{t/2}(x,z)}{\pi(z)}\frac{H_{t/2}(z,y)}{\pi(y)}\pi(z) = \sum_z h_{t/2}(x,z)h_{t/2}(z,y)\pi(z)\end{eqnarray}$$
