1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another contraction question

  1. Apr 22, 2007 #1
    Here is the question
    --------
    Give an example of a function [itex]f:[a,b] \to \mathbb{R}[/itex] which is continuously differentiable and which is a strict contraction, but such that [itex]|f'(x)| = 1[/itex] for at least one value of [itex]x \in [a,b][/itex].
    --------

    Continuously differentiable means that the first derivative is continuous.

    Strict contraction means that [itex]d(f(x),f(y)) < d(x,y)[/itex] for all [itex]x,y \in [a,b][/itex] (in our case the metric would just be the standard metric on R (d(x,y) = |x-y|)).


    ------

    When I was looking at the problem I figured we need something that has slope less than 1, then hits 1 for exactly one point, and then the slope goes back down below 1 again.

    My first guess was sin (and cos).

    For example. [itex]f:[0,\pi]\to \mathbb{R}[/itex] defined by [itex]f(x) = -cos(x)[/itex]. This gives us [itex]f'(x) = sin(x)[/itex]. Which will give us that [itex]f'(\pi/2) = 1[/itex].

    I think this example should work, it is obviously continuously differentiable, but my problem is proving that f is a STRICT contraction. We know that f is a contraction by a previous exercise (if f is diff and [itex]|f'(x)| \leq 1[/itex] then f is a contraction). But getting the strict part I am having trouble with.

    First, is it even true that it is a strict contraction? If so, any hints on how to prove it. Also, are there any interesting, or simpler, examples for this problem? Thanks!
     
  2. jcsd
  3. Apr 22, 2007 #2

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    I'm confused with the question.

    If [tex]f:[a,b] \rightarrow \mathbb{R}[/tex] is continuously differentiable, then using the mean value theorem,

    [tex]\frac{f(x)-f(y)}{x-y} = f'(c) , c \in [x,y][/tex]

    So, won't f be a strict contraction only if [tex]f'(x)<1 \quad \forall x \in [a,b][/tex]?
     
    Last edited: Apr 22, 2007
  4. Apr 22, 2007 #3
    I am not 100% sure, but the problem is as I stated it, I just double checked a minute ago.

    I had thought of the same thing, but I think there is something with taking the limit that allows for equals. The argument you gave is similar to the one I gave for the previous problem (which was that if f is differentiable and [itex]|f'(x)| \leq 1[/itex] then f is a contraction).
     
  5. Apr 22, 2007 #4

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    You're arguing backwards. What you say would only be true if for all c in [a,b], there exists x,y in [a,b] with [f(x)-f(y)]/[x-y] = f'(c).
     
  6. Apr 22, 2007 #5

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    -cos(x) should work. If it didn't, there would be a point (x,f(x)) on the graph such that either the line with slope 1 or the line with slope -1 through (x,f(x)) intersects the graph of -cos(x) at another point. See graphically why this cannot be, and use this understanding to motivate a proof using calculus as to why this cannot be.
     
  7. Apr 22, 2007 #6

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    Yeah, I understood my error, Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Another contraction question
  1. Another question (Replies: 3)

  2. Another question (Replies: 3)

Loading...