Results 1 to 3 of 3

Math Help - Continuity and differentiability answer check.

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Continuity and differentiability answer check.

    Let f:[-1,1] \rightarrow \mathbb{R} be continuous with f(-1)=2 and f(1)=1.
    i). If also f is injective prove that f is decreasing.
    Assume f is injective but not decreasing. This implies f is either an oscillating function, strictly increasing, increasing or constant.

    f cannot be an oscillating function because if f was an oscillating function over [-1,1] then if f(a)=b where a \in [-1,1] and b \in Im(f) then by the definition of an oscillating function \exists c \in [-1,1] such that f(a+c)=b. This would contradict the fact that f is injective.

    So we now know that f is either strictly increasing, increasing or constant.

    Once again, f cannot be a constant function. If f was a constant function (similar to before) if f(a)=b then f(a+ \delta)=b  \forall \ \delta>0.

    f is either a strictly increasing or an increasing function.

    If f is strictly increasing then if [a,b] \subset [-1,1] then f(b)>f(a).

    This gives a contradiction since f(1)<f(-1) so f cannot be strictly increasing.

    A similar argument holds for an increasing function:
    If f is increasing then if [a,b] \subset [-1,1] then f(b) \geq f(a).
    This contradicts the definition of injectivity since it is possible for f(a)=f(b).

    Hence we are forced to conclude that f is a decreasing function.

    ii) if f takes only positive values show that there exists a positive real number \delta with f(x)> \delta \ \forall x \in[-1,1], carefully stating any general resullts you use.
    If f only takes positive values then f(x) \neq 0. Since f is also continous, f must be bounded since it's domain is bounded (general result). f is bounded above by \max \{ f(x)|x \in[-1,1] \} and below by \min \{f(x)| x \in [-1,1] \}.

    Take \delta= \min \{ f(x)| x \in[-1,1] \}- \epsilon for some \epsilon>0.

    This gives f(x) \geq \min \{f(x)| x \in[-1,1] \}> \min \{f(x)| x \in[-1,1] \}- \epsilon= \delta.

    Hence f(x)> \delta as required.

    b). Suppose f: \mathbb{R} \rightarrow \mathbb{R} is increasing and satisfies the conclusion of the intermediate value theorem. Prove that f is left continuous (ie. prove that \forall \ c \in \mathbb{R} the restriction of f to (- \infty, c] is continuous at c).
    This is the part i'm really having trouble with (provided my previous two proofs are okay! ).

    I need to find |f(x)-f(c)|< \epsilon when |x-c|< \delta where x <c.

    Let x=c-k \Rightarrow |c-k-c|=|k|<\delta \ \forall k>0.

    We now need to find |f(c-k)-f(c)|< \epsilon when |k|< \delta.

    ...and this is where I get stuck

    Can someone please tell me whether my two above proofs are fine and show me how to do this last part?

    I think there's a way of doing parts i) and ii) using the intermediate value theorem, but I couldn't see that way of doing it.

    Any help is appreciated. Thankyou!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Quote Originally Posted by Showcase_22 View Post
    If f only takes positive values then f(x) \neq 0. Since f is also continous, f must be bounded since it's domain is bounded (general result). f is bounded above by \max \{ f(x)|x \in[-1,1] \} and below by \min \{f(x)| x \in [-1,1] \}.

    Take \delta= \min \{ f(x)| x \in[-1,1] \}- \epsilon for some \epsilon>0.

    This gives f(x) \geq \min \{f(x)| x \in[-1,1] \}> \min \{f(x)| x \in[-1,1] \}- \epsilon= \delta.

    Hence f(x)> \delta as required.
    Hum ...

    It is important to notice that \min \{f(x)| x \in [-1,1] \} > 0

    Of course due to the fact that f is positive \min \{f(x)| x \in [-1,1] \} \geq 0

    And \min \{f(x)| x \in [-1,1] \} \neq 0 because if it was 0 then there would exist c \in [-1,1] \mbox{ such that} f(c) = 0 (this is due to the fact that [-1,1] is a closed interval. If you take g(x) = x over )0,1], you have \forall x\in )0,1]~g(x) > 0 but g has no minimum, only a lower bound which is 0)

    Take \delta = \frac{\min \{f(x)| x \in [-1,1] \}}{2}

    \forall x\in[-1,1]~f(x) \geq \min \{f(x)| x \in [-1,1] \} > \delta > 0

    A drawing is better than a long discussion ...

    Last edited by running-gag; May 9th 2009 at 08:29 AM. Reason: Drawing added
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    I understand what you're saying, and the picture was super!

    I also think that your \delta was much better than mine!

    Do you think my proof for the injectivity was fine and do you have any ideas for the last part? I looked at it again today and it was still highly confusing!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Continuity and Differentiability
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 10th 2010, 01:32 AM
  2. Differentiability and continuity
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 30th 2010, 03:26 AM
  3. Continuity and differentiability
    Posted in the Calculus Forum
    Replies: 8
    Last Post: July 30th 2009, 10:16 PM
  4. continuity and differentiability
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 2nd 2009, 12:03 AM
  5. Continuity and Differentiability
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 29th 2008, 11:12 PM

Search Tags


/mathhelpforum @mathhelpforum