1.3. Heat equation 27
In fact, the φj can be chosen to be orthonormal,
φj,φk :=
x∈A
φj(x) φk(x) = δ(k j).

Here we have introduced the delta function notation, δ(z) = 1 if z = 0
and δ(z) = 0 if z = 0.
Since pn(x) 0 as n ∞, we know that the eigenvalues have
absolute value strictly less than one. We can order the eigenvalues
1 λ1 λ2 · · · λN −1.
We will write p(x, y; A) to be the solution of the heat equation with
initial condition equal to one at x and 0 otherwise. In other words,
pn(x, y; A) = P{Sn = y, TA n | S0 = x}, x, y A.
Then if #(A) = N,
pn(x, y; A) =
N
j=1
cj(x) λj
n
φj(y),
where cj(x) have been chosen so that
N
j=1
cj(x)φj(y) = δ(y x).
In fact, this tells us that cj(x) = φj(x). Hence,
pn(x, y; A) =
N
j=1
λj
n
φj(x) φj(y).
Note that the quantity on the right is symmetric in x, y. One can
check that the symmetry also follows from the definition of pn(x, y; A).
The largest eigenvalue λ1 is often denoted λA. We can give a
“variational” definition of λA as follows. This is really just a theorem
about the largest eigenvalue of symmetric matrices.
Theorem 1.8. If A is a finite subset of Zd, then λA is given by
λA = sup
f
Qf, f
f, f
,
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