Question

Let I be a bounded closed interval, and let f : I -> I be continuous. Show that there exists x in I such that f(x) = x. Are the hypotheses about I both necessary?

Answer 1

@Hero help?

Answer 2

@Zarkon please help?

Answer 3

I don't know where to start

Answer 4

Mind you, Real analysis isn't my best field... let's put together stuff that we know...
One, it's continuous, and on a closed and bounded interval...
that means f has an absolute max and absolute min on that interval, right?

Answer 5

thanks! and yes.

Answer 6

Actually, there may be a slight problem...

Answer 7

I did some research and I'm wondering if this would help? http://www.physicsforums.com/showthread.php?t=575366

Answer 8

hang on :D

Answer 9

May be there is a possibility to find a proof using the intermediate value theorem, without sequences.

Answer 10

My sentiments exactly :D
Just finding out how...

Answer 11

Okay, I got it now :)

Answer 12

ok i'm going to try that

Answer 13

Oh... sure... let me know if you get stuck :D

Answer 14

i actually used the fact that since f is continuous, there would exist a sequence \((x_n)\) in I with \(\lim x_n=c\) and the sequence \((f(x_n))\) converges to \(f(c)\)

Answer 15

That's messy don't you think? :)

Answer 16

that was a property given in the text though

Answer 17

It was specifically mentioned that you are to use the Sequential Criterion?

Answer 18

no.

Answer 19

i had the idea after reading that forum.

Answer 20

Then hear me out :)

Answer 21

Also, my professor hadn't gone over IVT yet

Answer 22

Oh... that is a problem...

Answer 23

But he's already gone through the sequential criterion?

Answer 24

yeah

Answer 25

Interesting... let's rethink this...

Answer 26

I have \(x_1=a_0\) and \(x_{n+1}=f(x_n)\) and then I take the limit on both sides of the second equation

Answer 27

I set \(lim(x_n)=c\) because it's continuous, then I got \(c=f(c)\) but I'm not sure if it would be so easy

Answer 28

Frustrating... IVT gets this done in nothing flat...

Answer 29

how do i do it with IVT though?

Answer 30

I'll just explain to him what I would do

Answer 31

It's actually really simple... let I = [a,b] (the closed and bounded interval)

Answer 32

Then we have f(a) and f(b)

Answer 33

Inevitably, one of these would be greater than the other, so let's assume f(b) > f(a)

Answer 34

ok, i understand that part.

Answer 35

Wait.... hold that thought... wrong assumption... rewind... start over (sorry, not myself tonight :3)

Answer 36

it's ok

Answer 37

Since I is closed and bounded, f attains an absolute minimum for some b in I and an absolute minimum for some a in I, aye? (LOL)

Answer 38

yup

Answer 39

I meant an absolute maximum for some b.

Answer 40

lol

Answer 41

Wait something's wrong again... gah... never mind, false alarm :/

Answer 42

he just sent out an email saying we're allowed to use IVT but since we haven't gone over it, I am not quite familiar with it to use it

Answer 43

Okay, first, let me introduce it to you :)

Answer 44

In its simplest form we have... suppose we have a continuous function f on a closed and bounded interval [a,b]

Answer 45

Where f(a)f(b) < 0

Answer 46

Catch me so far?

Answer 47

ok

Answer 48

yes

Answer 49

That is to say, the product of f(a) and f(b) is negative

Answer 50

then there exists a c between a and b such that f(c) = 0

Answer 51

then there would be a c in which f(c) =0?

Answer 52

oh wait you already explained it lol

Answer 53

Yup... can you visualize that? :D

Answer 54

yeah

Answer 55

The meaning of f(a)f(b) < 0
is that f(a) and f(b) must have different signs.

Answer 56

Oh okay, great :D

Answer 57

it's basically the same application of IVT in calculus, right?

Answer 58

Yup. So it turns out I was right the first time.... I just panicked. :3

Answer 59

ok, hit me! lol

Answer 60

So, we let I = [a,b] our interval.

Answer 61

And our function is f from I to I.

Answer 62

Well, in that case, first we assume f(a) is not equal to a and f(b) is not equal to b.
Of course, if this weren't true, then we'd already be done.

Answer 63

So far so good?

Answer 64

so we're going to prove by contradiction?

Answer 65

Not exactly... we assume first that f(a) \(\ne\) a and f(b) \(\ne\) b

Answer 66

ok

Answer 67

Now, since the function's codomain is also I, then a is the lowest value that f could possibly take on that interval.

Answer 68

ok i'm with you

Answer 69

So since f(a) is not a, it follows that f(a) > a

Answer 70

Similarly, b is the highest value f could possibly take on the interval I, so since f(b) is not b, it follows that g(b) < b

Answer 71

ok

Answer 72

Good... now define a new function
g(x) = f(x) - x

Answer 73

by g(b) up there, you did mean f(b) right?

Answer 74

whoops... sorry, got a little too excited :) Yeah, f(b)

Answer 75

ok

Answer 76

So, we now consider g(a) and g(b)
Since f(a) > a, then it follows that
f(a) - a = g(a) > 0

Answer 77

And since f(b) < b, then it follows that
f(b) - b = g(b) < 0

Answer 78

ow do we know g(a) would be >0?

Answer 79

Because we know that f(a) > a

Answer 80

ohh ok thanks i was stupid

Answer 81

from assuming that f(a) is not equal to a, and knowing that a is the lowest value that f could possibly take on the interval I.

Answer 82

So, we now have g(a) > 0
g(b) < 0

Answer 83

And mind you, g is continuous, being the difference of two continuous functions.

Answer 84

yes

Answer 85

That now means g(a)g(b) < 0

Answer 86

Implying there is a value x somewhere between a and b where g(x) = 0

Answer 87

Or f(x) - x = 0

Answer 88

And the rest is history :)

Answer 89

YAY you're awesome.

Answer 90

How do I know if these hypotheses were necessary, though?

Answer 91

We needed the interval I to be bounded... otherwise it could be that I is the entire set of Real numbers

Answer 92

In which case, an easy counterexample would be f(x) = x+1
Which has no value of x such that f(x) = x
And yet f is a continuous function from I to I.

Answer 93

So we needed it to be a bounded interval... we needed it to be continuous on the CLOSED interval [a,b] because we needed f(a) and f(b)

Answer 94

ok i got it! thanks so much!

Answer 95

sure :)