# Dirichlet forms

So I'm seeing a Dirichlet form written as

$$\mathscr{E}(f,f) = \frac{1}{2} \sum_{x,y} |f(x)-f(y)|^2 K(x,y)\pi(x)$$

where $K(x,y)$ is the probability of taking a step to state $y$ from $x$.

And the conventional way of writing it seems to be

$$\mathscr{E}(f) = \int \left|\triangledown f\right|^2 d\mu.$$

How come the two expressions are equivalent. I just don't see how $K(x,y)\pi(x)$ could be analogous to $d\mu$ like $(f(x)-f(y))^2$ is analogous to $|\triangledown f|^2.$ If it were $\pi$, I could understand, but why are they using $K(x,y)\pi(x)?$

I hope this isn't a stupid question (I feel like it is).

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You may want to look at Qiaochu Yuan's answer to this question on the discrete Laplacian. He describes the discrete Lapacian on a regular lattice. You appear to be looking at what amounts to a weighted graph. In that case, you need to know whether $x$ and $y$ are edge connected and how 'strongly'. You need $K(x,y)$ for that. –  Eric Nitardy Mar 23 '11 at 20:16
Did you get something out of the answer below? –  Did Apr 7 '11 at 7:58

A short introduction was given last year at PIMS Summer School in Probability by Zhenqing Chen talking about Dirichlet Form Theory and Invariance Principle, see the slides of the first lecture.

A translation from the continuous setting you know to the discrete setting is as follows. Consider a graph $G=(V(G),E(G))$ with vertex set $V(G)$ and edge set $E(G)$ and a Markov chain on $G$ of kernel $K$. This means that $K$ is defined on $V(G)\times V(G)$ and that $K(x,y)=0$ if $(x,y)\notin E(G)$. The gradient of a function $f$ defined on $V(G)$ is the function $\nabla f$ defined on $E(G)$ by $$\nabla f(x,y)=f(y)-f(x).$$ Then $\mathcal{E}(f)$ is simply the $\ell^2$ norm of $\nabla f$ with respect to the measure $\mu$ defined on $E(G)$ by $\mu(x,y)=\pi(x)K(x,y)$.

In the continuous case $\mathcal{E}(f)$ for a function $f$ defined on $\mathbb{R}^n$ (which is the analogue of $V(G)$ in the discrete setting) is the $L^2$ norm of the usual gradient $\nabla f$, defined on the tangent space $T_x\mathbb{R}^n$ (which is the analogue of $E(G)$), but the difference is not visible because at every $x$ in $\mathbb{R}^n$ the tangent space $T_x\mathbb{R}^n$ is $\mathbb{R}^n$ itself.

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Also another thing I don't understand is why $$\frac{E_{\mu}(\left\|\triangledown e^{2\lambda F}\right|_2^2)}{E_{\mu}(e^{2\lambda F})} \leq \sup \left\|\triangledown F\right\|_2^2$$ –  Clark Kent Mar 23 '11 at 20:40
This one is easy: write $\nabla\mathrm{e}^{2\lambda F}$ as $G\nabla F$ for a well chosen scalar function $G$. –  Did Mar 23 '11 at 20:57
That's where I'm stuck. I'm trying to figure out what $G$ should be. (Someone who is very unfamiliar with gradients :/...) –  Clark Kent Mar 23 '11 at 21:40
Come on! You write a whole post referring to gradients (in Euclidean spaces) as the easy case, and you do not know the definition of the gradient of a function... You must be joking. –  Did Mar 23 '11 at 21:47
OK. So what is $G$? –  Did Mar 23 '11 at 21:52