Results 1 to 6 of 6

Thread: continuous functions on intervals 2

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    62

    continuous functions on intervals 2

    Let I:=[a,b] and let f: be a continuous function on I such that for each x in I there exists y in I such that . Prove that there exists a point c in I such that f(c)=0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2008
    Posts
    147
    I think you can say something along the lines of take some f(x) in R, and w/ $\displaystyle d(0, f(x)) \leq \epsilon$. We can find a $\displaystyle y_1 $ such that $\displaystyle d(0,f(y_1))\leq \epsilon /2$. Suppose there was minimal value c, such that $\displaystyle f(c) \geq 0.$ But we know we can find another value that is even smaller than c, and by the Archimedean Property, it must be 0. This is not a very rigorous proof though, since I didn't really use continuity.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    62
    Quote Originally Posted by terr13 View Post
    I think you can say something along the lines of take some f(x) in R, and w/ $\displaystyle d(0, f(x)) \leq \epsilon$. We can find a $\displaystyle y_1 $ such that $\displaystyle d(0,f(y_1))\leq \epsilon /2$. Suppose there was minimal value c, such that $\displaystyle f(c) \geq 0.$ But we know we can find another value that is even smaller than c, and by the Archimedean Property, it must be 0. This is not a very rigorous proof though, since I didn't really use continuity.
    yeh i think it has to be more rigorous and using continuity. Any idea how to do it that way??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,742
    Thanks
    2814
    Awards
    1
    There is a theorem say that “A continuous function on a closed interval has a minimum and a maximum”.
    So $\displaystyle \left( {\exists m \in I} \right)\left( {\forall z \in I} \right)\left[ {f(m) \leqslant f(z)} \right]$.
    So suppose that $\displaystyle \left( {\forall z \in I} \right)\left[ {0 \ne f(z)} \right]$ then from the given $\displaystyle f(m) > 0$.
    A contradiction follows at once from the minimum.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2008
    Posts
    62
    Quote Originally Posted by Plato View Post
    There is a theorem say that “A continuous function on a closed interval has a minimum and a maximum”.
    So $\displaystyle \left( {\exists m \in I} \right)\left( {\forall z \in I} \right)\left[ {f(m) \leqslant f(z)} \right]$.
    So suppose that $\displaystyle \left( {\forall z \in I} \right)\left[ {0 \ne f(z)} \right]$ then from the given $\displaystyle f(m) > 0$.
    A contradiction follows at once from the minimum.
    I have been trying to do this problem as well along with the other problem, yet i still cant do it. So frustrated . Being trying to read the book but I just dont understand this material.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,742
    Thanks
    2814
    Awards
    1
    You are simply over-thinking all of this.
    If the absolute minimum is $\displaystyle f(m)$ there cannot be any $\displaystyle y$ such that $\displaystyle \left| {f(y)} \right| \leqslant \frac{1}{2}\left| {f(m)} \right|$
    Otherwise, $\displaystyle f(m)$ would not be the absolute minimum!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Jul 7th 2011, 04:14 AM
  2. Replies: 1
    Last Post: Oct 11th 2010, 07:18 PM
  3. Intervals and Continuous Functions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 4th 2010, 02:28 PM
  4. Replies: 8
    Last Post: Jun 27th 2010, 11:42 AM
  5. Replies: 1
    Last Post: Jan 29th 2010, 04:44 PM

Search Tags


/mathhelpforum @mathhelpforum