Let ϕ : [a, b] → [a, b] a function C^1([a, b]) such...

Let $\phi : [a, b] \to [a, b]$ a function $C^1([a, b])$ such that $\min_{a \le x \le b} |\phi'(x)| < 1$ Then, given $x_0 \in [a, b]$, the sequence: $x_{k+1} = \phi(x_k)$, $k \in \mathbb{N}$ converges to a single point $\alpha \in [a, b]$ such that $\phi(\alpha) = \alpha$. Using the assumptions of the Fixed Point Theorem to prove the following: a) The sequence {$x_k$} converges. b) If $0 < \phi'(\alpha) < 1$ and $x_0 > \alpha$, then $x_k > \alpha$ for $k \in \mathbb{N}$. c) If $-1 < \phi'(\alpha) < 0$, then $x_k - \alpha$ and $x_{k+1} - \alpha$ have opposite signs for $k \in \mathbb{N}$.

Transcript

00:01 Hello, hello, we have a function f from real numbers to real numbers.

00:09 It has the property that if x and y are two points, then the magnitude of the difference between fx and fy is less than equal to m times the magnitude of the difference between x and y, where m is a number in the open interval 0 ,1.

00:27 Now we want to prove that f is a continuous function.

00:33 For f will be continuous at x equal to a if given epsilon greater than 0, we can find a delta greater than 0 such that magnitude of fx minus fa will be less than epsilon if magnitude of x minus a is less than delta.

00:49 Now if we take delta equal to epsilon, then the magnitude of fx minus fa is less than equal to m times the magnitude of x minus a, which is less than equal to magnitude of x minus a, since m is less than equal to 1 and magnitude of x minus a is less than equal to epsilon.

01:10 So magnitude of fx minus fa is less than equal to epsilon if magnitude of x minus a is less than equal to delta which we have taken to be epsilon.

01:26 So we can find such a delta.

01:30 So f is continuous at x equal to a and but since x is arbitrary fx is continuous on all of r.

01:47 Let us recall this notation f to the power n a1 equal to f of f of n times of a1 that is there is n compositions of f here.

01:58 Now let us choose an arbitrary point an arbitrary real number and denote it as and take it to be a1 and then let us take a2 to be f of a1, a3 to be f2 of a1, a4 to be f3 of a1 and an to be f to the power n minus 1 of a1.

02:19 Now we want to get some estimates for this sequence.

02:27 We want to prove that this sequence is cauchy and for that we want to get some estimates.

02:34 We can pause this video and see this proof but the estimates we will get is that magnitude of an plus 1 minus magnitude of an will be less than equal to m to the power n minus 1 times magnitude of a2 minus a1 and using these estimates we will get that if s and k are two numbers which are greater than equal to n positive numbers which are greater than equal to n then the magnitude of a s minus a k is less than equal to m to the power n minus 1 magnitude of a2 minus a1 over 1 minus m.

03:13 Now if given an epsilon greater than 0 we can choose a number n so that this quantity is less than epsilon then for that capital n for all sk that are greater than equal to n magnitude of a s minus a k will be less than epsilon.

03:35 So given an epsilon we can choose n such that a s minus a k magnitude is less than equal to epsilon when s and k are greater than equal to n.

03:45 So the sequence an is cauchy.

03:47 Now this is part b.

03:50 In part c let since r is a complete metric space an will converge to a point in real numbers.

03:59 So n converges to a real number a.

04:02 Now since f we have already proved that f is continuous since and if an converges to a we will have that f of an will converge to f of f of an will converge to f of a but f of an is an plus 1.

04:20 So this sequence is same as the sequence an.

04:24 So f of an which is same as the sequence an will converge to a f of an will converge to a...

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