Lecture Notes in Differential Equations - Bruce E. Shapiro

220 LESSON 26. GENERAL EXISTENCE THEORY*
Proof. We know from theorem 26.3 that a unique fixed point p exists. We
need to show that p i → p as i → ∞.
Since f maps onto a subset of itself, every point p i ∈ [a, b].
Further, since p itself is a fixed point, p = f(p) and for each i, since p i =
f(p i−1 ), we have
|p i − p| = |p i − f(p)| = |f(p i−1 ) − f(p)| (26.15)
If for any value of i we have p i = p then we have reached the fixed point
and the theorem is proved.
So we assume that p i ≠ p for all i.
Then by the mean value theorem, for each value of i there exists a number
c i between p i−1 and p such that
|f(p i−1 ) − f(p)| = |f ′ (c i )||p i−1 − p| ≤ K|p i−1 − p| (26.16)
where the last inequality follows because f ′ is bounded by K < 1 (see
equation 26.7).
Substituting equation 26.15 into equation 26.16,
|p i − p| = |f(p i−1 ) − f(p)| ≤ K|p i−1 − p| (26.17)
Restating the same result with i replaced by i − 1, i − 2, . . . ,
|p i−1 − p| = |f(p i−2 ) − f(p)| ≤ K|p i−2 − p|
|p i−2 − p| = |f(p i−3 ) − f(p)| ≤ K|p i−3 − p|
|p i−3 − p| = |f(p i−4 ) − f(p)| ≤ K|p i−4 − p|
.
|p 2 − p| = |f(p 1 ) − f(p)| ≤ K|p 1 − p|
|p 1 − p| = |f(p 0 ) − f(p)| ≤ K|p 0 − p|
⎫
⎪⎬
⎪⎭
(26.18)
Putting all these together,
|p i − p| ≤ K 2 |p i−2 − p| ≤ K 3 |p i−2 − p| ≤ · · · ≤ K i |p 0 − p| (26.19)
Since 0 < K < 1,
Thus p i → p as i → ∞.
0 ≤ lim
i→∞
|p i − p| ≤ |p 0 − p| lim
i→∞
K i = 0 (26.20)