6

1. Jacobi operators

Both are readily verified. Nevertheless, let me remark that 9, 9* are no derivations

since they do not satisfy Leibnitz rule. This very often makes the discrete case

different (and sometimes also harder) from the continuous one. In particular, many

calculations become much messier and formulas longer.

There is much more to say about relations for the difference expressions (1.15)

analogous to the ones for differentiation. We refer the reader to, for instance, [4],

[87], or [147] and return to (1.13).

Associated with r is the eigenvalue problem ru = zu. The appropriate setting

for this eigenvalue problem is the Hilbert space

£2(Z).

However, before we can

pursue the investigation of the eigenvalue problem in ^2(Z), we need to consider

the Jacobi difference equation

(1.19) TU = zu, u ef(Z), z e C.

Using a(n) ^ 0 we see that a solution u is uniquely determined by the values u(no)

and u(no + 1) at two consecutive points no, n0 + 1 (you have to work much harder

to obtain the corresponding result for differential equations). It follows, that there

are exactly two linearly independent solutions.

Combining (1.16) and the summation by parts formula yields Green's formula

n

(1-20) ]T (/(r5) - (rf)g)(j) = Wn(f,g) - Wm-iU,9)

j-m

for f,g e f'{1), where we have introduced the (modified) Wronskian

(1.21)

Wn (/, g) = a(n) (f(n)g(n + 1) - g(n)f(n + 1)).

Green's formula will be the key to self-adjointness of the operator associated with

r in the Hilbert space

£2(Z)

(cf. Theorem 1.5) and the Wronskian is much more

than a suitable abbreviation as we will show next.

Evaluating (1.20) in the special case where / and g both solve (1.19) (with the

same parameter z) shows that the Wronskian is constant (i.e., does not depend on

n) in this case. (The index n will be omitted in this case.) Moreover, it is nonzero

if and only if / and g are linearly independent.

Since the (linear) space of solutions is two dimensional (as observed above) we

can pick two linearly independent solutions c, s of (1.19) and write any solution u

of (1.19) as a linear combination of these two solutions

(1-22) u(n) = — -c(n) - — -s{n).

W(c,s) W{c,s)

For this purpose it is convenient to introduce the following fundamental solutions

c,se£(Z)

(1.23) Tc(z,.,n0) = : c ( : , , n

0

) ,

fulfilling the initial conditions

(1.24) c(*,no,no) = l,

TS(Z,

., no) = z s(;

c(z,n{) + l,n

0

) = 0,

s(z,n0 + l,n

0

) = 1.

Most of the time the base point no will be unessential and we will choose no

for simplicity. In particular, we agree to omit no whenever it is 0, that is,

(1.25) c(z,n) = c(z,n,0), s(z,n) = s(z, n,0).