Facebook Page
Twitter
RSS
Thanks Thanks:  0
+ Reply to Thread
Results 1 to 3 of 3
  1. MHB Craftsman

    Status
    Offline
    Join Date
    Dec 2012
    Posts
    290
    Thanks
    29 times
    Thanked
    72 times
    #1
    If:

    f is a real value function continuous over [a,b] ,a<b and f(a)<f(b)

    ,then prove that for all η,
    f(a)<η<f(b) ,there exists an ξ ,such that : f(ξ)=η

    Proof.

    1)Let S be a set of all xε[a,b] such that :f(x)<η.

    Since a belongs to S and b is an upper bound then S has a Supremum

    2)Let ξ= SupS


    3) Will show that : a) $f(\xi)\geq\eta $ ,b) $ f(\xi)\leq\eta $ and hence : $f(\xi)=\eta$


    3a) By the definition of supremum, for every positive ε, there is a number x' of S with
    $\xi-\epsilon\leq x'\leq\xi$ .
    For this x', f(x')<η.
    Since f is contiguous at ξ, $f(\xi)\leq\eta$ which is : 3a

    3b) Any x greater than ξ is not in S and so $f(x)\geq \eta$.By continuity ,f(ξ) is the limit of f(x) as x tends to ξ through values greater than ξ ,and so :$f(\xi)\geq\eta$

    I have no idea how he gets $f(\xi)\geq \eta $ in part 3b and $f(\xi)\leq\eta$ in part 3a

  2. # ADS
    Circuit advertisement
    Join Date
    Always
    Posts
    Many
     

  3. MHB Craftsman

    Status
    Offline
    Join Date
    Mar 2012
    Location
    Jordan
    Posts
    275
    Thanks
    488 times
    Thanked
    213 times
    #2
    Quote Originally Posted by solakis View Post
    If:

    f is a real value function continuous over [a,b] ,a<b and f(a)<f(b)

    ,then prove that for all η,
    f(a)<η<f(b) ,there exists an ξ ,such that : f(ξ)=η

    Proof.

    1)Let S be a set of all xε[a,b] such that :f(x)<η.

    Since a belongs to S and b is an upper bound then S has a Supremum

    2)Let ξ= SupS


    3) Will show that : a) $f(\xi)\geq\eta $ ,b) $ f(\xi)\leq\eta $ and hence : $f(\xi)=\eta$


    3a) By the definition of supremum, for every positive ε, there is a number x' of S with
    $\xi-\epsilon\leq x'\leq\xi$ .
    For this x', f(x')<η.
    Since f is contiguous at ξ, $f(\xi)\leq\eta$ which is : 3a

    3b) Any x greater than ξ is not in S and so $f(x)\geq \eta$.By continuity ,f(ξ) is the limit of f(x) as x tends to ξ through values greater than ξ ,and so :$f(\xi)\geq\eta$

    I have no idea how he gets $f(\xi)\geq \eta $ in part 3b and $f(\xi)\leq\eta$ in part 3a
    3a) $ \displaystyle f(\xi) \leq \eta $
    Pick $ \displaystyle \epsilon = \eta - f(\xi) $ as you should know
    f is continuous at ξ that means for any $ \displaystyle \epsilon >0 $ there exist $ \displaystyle \delta >0 $
    such that whenever $ \displaystyle \mid x - \xi \mid < \delta $ we have $ \displaystyle \mid f(x) - f(\xi) \mid < \epsilon $
    remember that epsilon>0 so if you choose a wrong epsilon the criteria for the continuity will be affected, back to our proof after we choose epsilon sub it
    $ \displaystyle \mid f(x) - f(\xi) \mid < \eta - f(\xi) $

    $ \displaystyle -\eta + f(\xi) < f(x) - f(\xi) < \eta - f(\xi) $

    $ \displaystyle f(x) - f(\xi) < \eta - f(\xi) \Rightarrow f(x) < \eta $ for

    $ \displaystyle \mid x - \xi \mid < \delta \Rightarrow \xi - \delta < x < \xi + \delta $

    but $ \displaystyle \xi $ is an upper bound for S, $ \displaystyle \xi < \xi +\frac{\delta}{2} $
    and
    $ \displaystyle \xi + \frac{\delta}{2} \in S $ since
    $ \displaystyle f\left(\xi + \frac{\delta}{2}\right) < \eta $ but $ \displaystyle \xi $ is an upper bound for S contradiction!! so $ \displaystyle \eta - f(\xi) $ is not positive so still two choices less than or equal zero

    Lets see if $ \displaystyle f(\xi) > \eta $
    f is continuous at $ \displaystyle \xi $ , now pick $ \displaystyle \epsilon = f(\xi) - \eta $
    there exist delta

    $ \displaystyle \mid f(x) - f(\xi) \mid < f(\xi) - \eta $ for $ \displaystyle \mid x - \xi \mid < \delta $

    $ \displaystyle -f(\xi) + \eta < f(x) - f(\xi) $

    $ \displaystyle \eta < f(x) $ for $ \displaystyle \xi - \delta < x < \xi + \delta $

    $ \displaystyle \xi - \delta $ is in S the definition of the upper bound which means $ \displaystyle f \left(\xi - \delta \right) $ should be less than $ \displaystyle \eta $ not larger contradiction so our epsilon is not positive

    This idea from Wiki see
    Last edited by Amer; April 10th, 2013 at 12:54.

  4. MHB Craftsman

    Status
    Offline
    Join Date
    Dec 2012
    Posts
    290
    Thanks
    29 times
    Thanked
    72 times
    #3 Thread Author
    Quote Originally Posted by Amer View Post
    3a) $ \displaystyle f(\xi) \leq \eta $
    Pick $ \displaystyle \epsilon = \eta - f(\xi) $ as you should know
    f is continuous at ξ that means for any $ \displaystyle \epsilon >0 $ there exist $ \displaystyle \delta >0 $
    such that whenever $ \displaystyle \mid x - \xi \mid < \delta $ we have $ \displaystyle \mid f(x) - f(\xi) \mid < \epsilon $
    remember that epsilon>0 so if you choose a wrong epsilon the criteria for the continuity will be affected, back to our proof after we choose epsilon sub it
    $ \displaystyle \mid f(x) - f(\xi) \mid < \eta - f(\xi) $

    $ \displaystyle -\eta + f(\xi) < f(x) - f(\xi) < \eta - f(\xi) $

    $ \displaystyle f(x) - f(\xi) < \eta - f(\xi) \Rightarrow f(x) < \eta $ for

    $ \displaystyle \mid x - \xi \mid < \delta \Rightarrow \xi - \delta < x < \xi + \delta $

    but $ \displaystyle \xi $ is an upper bound for S, $ \displaystyle \xi < \xi +\frac{\delta}{2} $
    and
    $ \displaystyle \xi + \frac{\delta}{2} \in S $ since
    $ \displaystyle f\left(\xi + \frac{\delta}{2}\right) < \eta $ but $ \displaystyle \xi $ is an upper bound for S contradiction!! so $ \displaystyle \eta - f(\xi) $ is not positive so still two choices less than or equal zero

    Lets see if $ \displaystyle f(\xi) > \eta $
    f is continuous at $ \displaystyle \xi $ , now pick $ \displaystyle \epsilon = f(\xi) - \eta $
    there exist delta

    $ \displaystyle \mid f(x) - f(\xi) \mid < f(\xi) - \eta $ for $ \displaystyle \mid x - \xi \mid < \delta $

    $ \displaystyle -f(\xi) + \eta < f(x) - f(\xi) $

    $ \displaystyle \eta < f(x) $ for $ \displaystyle \xi - \delta < x < \xi + \delta $

    $ \displaystyle \xi - \delta $ is in S the definition of the upper bound which means $ \displaystyle f \left(\xi - \delta \right) $ should be less than $ \displaystyle \eta $ not larger contradiction so our epsilon is not positive

    This idea from Wiki see
    The proof you proposed is completely different from the proof i wrote in my OP.

    Sorry but i am not asking for a proof of the IVT but an explanation of the OP's proof

Similar Threads

  1. New Theorem
    By soroban in forum Chat Room
    Replies: 1
    Last Post: March 27th, 2013, 20:23
  2. Karla's question at Yahoo! Answers (Intermediate Value Theorem).
    By Fernando Revilla in forum Questions from Other Sites
    Replies: 1
    Last Post: February 4th, 2013, 04:14
  3. lamps's question at Yahoo! Answers about the Intermediate Value Theorem
    By Fernando Revilla in forum Questions from Other Sites
    Replies: 1
    Last Post: January 30th, 2013, 05:13
  4. Theorem
    By Fernando Revilla in forum Chat Room
    Replies: 7
    Last Post: January 29th, 2013, 17:02
  5. Lagrange Theorem
    By Yankel in forum Calculus
    Replies: 3
    Last Post: December 19th, 2012, 03:07

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards