# Differentibility of a function

## Main Question or Discussion Point

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.

What have you tried? (Hint: Consider the function g(x) = f(x) - x)

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
Homework Helper
well for starters, what if there were two such points? what would the mvt say?

Try looking at the derivative of g. f'(x) < M < 1 implies that f'(x) - 1 < ?

mathwonk
Homework Helper
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?

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 ?

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?

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?

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?

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?

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
Homework Helper
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.)

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 ?

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?

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?

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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