Differentibility of a function

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Can someone help me with this?
Let f: R into R be differentiable
1) If there is an M strictly less than 1 for each x in R, f'(x) strictly less than M,
prove that there exists a unique point x such that f(x)=x. ( Note: x is a fixed point for f)
2) Give counter example to show 1) fails if M=1.
 

Answers and Replies

1,005
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What have you tried? (Hint: Consider the function g(x) = f(x) - x)
 
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What have you tried? (Hint: Consider the function g(x) = f(x) - x)
Thanks for your help. I just have no clue how to approach this problem.

By assumption, we have for each x in R, there exists M < 1 such that f '(x) < M . I just don't know what to do with this given info. Can you elaborate a bit?
Consider g(x) = f(x) -x , then g(x) is also differentiable. So g(x) has the given property as f(x) . For each x in R, there exists M < 1 such that g '(x) = f '(x) - 1 < M. Am I on the right track? I don't know how to get to f(x) = x from here.
 
mathwonk
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well for starters, what if there were two such points? what would the mvt say?
 
1,005
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Try looking at the derivative of g. f'(x) < M < 1 implies that f'(x) - 1 < ?
 
mathwonk
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for existence just take any value of x at all, say a, and look at the point (a, f(a)).

now suppose the derivative of f is always less than say 1/2.

pass a line with slope 1/2 through your point (a,f(a)). how is the graph of your function situated wrt this line?

wrt the line y=x?
 
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well for starters, what if there were two such points? what would the mvt say?
If there were two such points x, y. Wlog, let x < y, then by MVT, there exists z in (x,y) such that
f(y) -f(x) = f '(z) * (y-x) . Then for each z , there exists M < 1 such that f '(z) < M . So, f(y) - f(x) < M * (y-x) , i.e.( ( f(y) - f(x) ) / (y-x) ) < M ?
 
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Try looking at the derivative of g. f'(x) < M < 1 implies that f'(x) - 1 < ?
Derivative of g implies that f '(x) - 1 < M , which I already stated above. But the inequality f '(x) < M < 1 implies that f '(x) - 1 < M -1 < 0 right?
 
1,005
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Derivative of g implies that f '(x) - 1 < M , which I already stated above. But the inequality f '(x) < M < 1 implies that f '(x) - 1 < M -1 < 0 right?
Right. Now what does that tell you about the global behavior of the graph of g?
 
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Right. Now what does that tell you about the global behavior of the graph of g?
I don't quite get the term "global behavior" . Can I say that the graph of g assumes max and min?
 
1,005
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I don't quite get the term "global behavior" . Can I say that the graph of g assumes max and min?
Can g have a max and min with a derivative function that is always negative and never approaches 0?
 
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Can g have a max and min with a derivative function that is always negative and never approaches 0?
No, if g' is always negative, then g is increasing. What does this say about f(x)?
 
mathwonk
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if a point on the graph of f is above the line y=x, and the graph of f is always below the line y = cx where c < 1, doesn't that say the graph of f must cross the line y=x?

(since the graph of y = cx does so.)
 
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if a point on the graph of f is above the line y=x, and the graph of f is always below the line y = cx where c < 1, doesn't that say the graph of f must cross the line y=x?

(since the graph of y = cx does so.)
Yes, it must cross the line y=x. So from here I get the existence of x where f(x)= x ?
 
1,005
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No, if g' is always negative, then g is increasing. What does this say about f(x)?
No, it says that g is always decreasing. Since g is defined on all of R, can a such a function be defined that never passes through 0?
 
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No, it says that g is always decreasing. Since g is defined on all of R, can a such a function be defined that never passes through 0?
Sorry, I meant decreasing. A typo. So we have [ g(x) - g(0) ] / x is approaching 0 this means there exists some y such that g'(y) = 0. A contradiction since above we assume that g'(x) < 0 for all x.

Can you help me with the counterexample when M=1?
 
1,005
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Sorry, I meant decreasing. A typo. So we have [ g(x) - g(0) ] / x is approaching 0 this means there exists some y such that g'(y) = 0. A contradiction since above we assume that g'(x) < 0 for all x.
Can you explain this argument more distinctly? I'm not sure why we are looking at the derivative of g at x=0.

Can you help me with the counterexample when M=1?
Consider why you are given the information that f'(x)<M<1 and not simply f'(x)<1. Why is M necessary for the proof to work? Can you think of a type of function that is defined on all of R and always decreases but never crosses the x-axis?
 
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Can you explain this argument more distinctly? I'm not sure why we are looking at the derivative of g at x=0.


Consider why you are given the information that f'(x)<M<1 and not simply f'(x)<1. Why is M necessary for the proof to work? Can you think of a type of function that is defined on all of R and always decreases but never crosses the x-axis?
You told me to consider the function g(x) = f(x) -x. So, if there exists x such that f(x) =x , then g(x) must be equal to 0 for some x. By contradiction, I assume that g(x) doesn't equal 0 for all x. Pick x= 0 , g(0) does not equal 0 by the assumption I just made. There are two cases: g(0) < 0 and g(0) > 0. By MVT, I reach something that contradicts the fact g'(x) < 0 for all x.
I still need to show uniqueness of x. Do you think I can use the typical approach by assuming that there exist distinct x, y such that f(x)=x and f(y)=y ?

P.S: Showing uniqueness is simpler than I thought, and I got it too.
 
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