1549301742-The_Theory_of_Difference_Schemes__Samarskii

48 Basic Concepts of the Theory of Difference Schernes

for all x. As a matter of fact,

(Ax, x) = (A 0 x, x),

where A 0 = (A+ A* )/2 is a self-adjoint operator. Hence, (Ax, x) > bl I x 112,

where c5 is the smallest eigenvalue of A 0. The number c5 cannot vanish due

to the positiveness of the operator A.

For the sake of simplicity we will assume in the body of this book that

H is a finite-dimensional space.

Recall that the norm of an operator A is defined by

llAll= sup llAxll·

11x11=1

With this, for any self-adjoint operator A the following relations occur:

l(Ax,x)I

11x11

( 11) llAll= sup l(Ax,x)l=llAll= sup

II x 11=1 II x ll;to

Lemma 3 If S = S* is a linear bounded operator and n is a positive

integer, tlren

Proof \!Ve proceed to prove this assertion by induction on n. Let n = 2.

Then

1152 11= sup l(S^2 x,x)I= sup llSxii2=11Sll^2 ,

11x11=1 11x11=1

meaning 11 52 11 = 11 S 112. Assuming that relation (12) is valid with n = k - l

and n = k, we are going to show that it continues to hold for n = k + 1,

k > 1. Indeed,

llS^2 kll= sup l(S^2 kx,:i:)I= sup l(Sk+^1 x,sk-^1 x)I

11 x lt=1 11x11=1

< sup II 5k+^1 x II · II sk-lx II< II s"+^1 II · II sk-l II

11x11=1

or, what amounts to the same,

11 sk+l II · II sk-^1 II> II s^2 k II= II sk 112 =II s ll^2 k.

This provides support for the view that 115k+l II> II S llk+l, since

II 5k-^1 II= II s llk-l.

On the other hand,

so that
II sk+l II= II s llk+l.
As far as (12) holds for n = 1 and n = 2, it will be true for any n.