Results 1 to 10 of 10

Math Help - second-differentiable function proving help

  1. #1
    Junior Member
    Joined
    Mar 2013
    From
    israel
    Posts
    70

    second-differentiable function proving help

    f is a second-differentiable function at  (0,\infty) so f''(x)>0 to every  x\in(0,\infty)
    i need to prove that if:  lim{}_{x\rightarrow\infty}f(x)=\ell (\ell is finite), so -
    (1)  f'(x)<0 to every  x\in(0,\infty)
    (2)  sup\: f'((0,\infty))=0
    (3)  lim{}_{x\rightarrow\infty}f'(x)=0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2010
    Posts
    980
    Thanks
    236

    Re: second-differentiable function proving help

    Here's some ideas on how to prove them:

    1. If f'(x_0) = r > 0 for some x_0\in(0,\infty), then f'(x) would have to be greater than r for all x > x_0, which would mean it is unbounded as x \to \infty.

    2. If the sup is not zero, then it must be less than zero. So the derivative is less than some fixed negative number, which again means f(x) is unbounded.

    3. This follows immediately from #1 and #2.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2013
    From
    israel
    Posts
    70

    Re: second-differentiable function proving help

    it's greating getting ideas and clues.. that's what's the best. thx!
    but i didn't quite understand you well.. can you repeat your ideas in other way?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2010
    Posts
    980
    Thanks
    236

    Re: second-differentiable function proving help

    Let's start with #1. I should have said f'(x_0)=r\ge{0}, by the way, and "it" at the end is f(x):
    1. If f'(x_0) = r \ge 0 for some x_0\in(0,\infty), then f'(x) would have to be greater than r for all x > x_0, which would mean f(x) is unbounded as x \to \infty.

    The proof is by contradiction, so you assume that f'(x_0)\ge{0} for some real number x_0.

    If you look at the graph of f'(x) vs. x, you know that it always goes up as it goes to the right (that's what f''(x)>0 means). So if (x_0,r) is a point on the graph, every point to the right of x=x_0 is above y=r. To prove this rigorously, use the Mean Value Theorem (for f'(x) between x_0 and x, where x is some real number greater than x_0). So in this first part, you should prove that if f'(x_0)\ge{0} for some real number x_0, then f'(x)>0 for all x>x_0.

    The second half is essentially the same argument using f(x) instead of f'(x). You choose x_1>x_0, so f'(x_1)>0 and prove that f(x)>f(x_1)+f'(x_1)(x-x_1) for all x>x_1. Since f'(x_1) is positive, the right-hand side is unbounded, so the left-hand side is unbounded. This contradicts the hypothesis that \lim_{x\to\infty}f(x)=\ell. Therefore the assumption that f'(x_0)\ge{0} for some real number x_0 is false.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2013
    From
    israel
    Posts
    70

    Re: second-differentiable function proving help

    great! i got all, thank you.. except -
    Quote Originally Posted by hollywood View Post
    You choose x_1>x_0, so f'(x_1)>0 and prove that f(x)>f(x_1)+f'(x_1)(x-x_1) for all x>x_1. Since f'(x_1) is positive, the right-hand side is unbounded, so the left-hand side is unbounded.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Mar 2010
    Posts
    980
    Thanks
    236

    Re: second-differentiable function proving help

    You need to move from x_0 to x_1 because f'(x_0) could be zero, and we need the derivative to be greater than zero.

    You can use the Mean Value Theorem from x_0 to x_1 to prove that f'(x_1)>0. Then use the Mean Value Theorem (again!) to prove my inequality. Remember f''(x) is always positive, so if c>x_1, f'(c)>f'(x_1).

    As x goes to infinity, f'(x_1) and f(x_1) are fixed, and f'(x_1) is positive. So for any M, you can determine an x for which f(x)>M. This is inconsistent with \lim_{x\to\infty}f(x)=\ell.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2013
    From
    israel
    Posts
    70

    Re: second-differentiable function proving help

    ok.. got it!
    and what about number 2 - sup f'((0,infty))=0
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Mar 2010
    Posts
    980
    Thanks
    236

    Re: second-differentiable function proving help

    By (1) you have \sup{f'(x)}\le{0}. So you need to show that f'(x)>-\epsilon for all \epsilon>0. Suppose this were not true - that there is some \epsilon>0 for which f'(x)\le-\epsilon for all x \in (0,\infty). You can use the same argument as #1 to show that f(x) goes to -\infty, which contradicts \lim_{x\to\infty}f(x)=\ell.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Mar 2013
    From
    israel
    Posts
    70

    Re: second-differentiable function proving help

    great! and now, how #3 follows immediately from #1 and #2?
    sorry i can't figure it out by my own...
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Mar 2010
    Posts
    980
    Thanks
    236

    Re: second-differentiable function proving help

    Well, maybe not so immediately. Let \epsilon>0. From #2, you can conclude that there is an N such that f'(N)>-\epsilon. If x>N, then you can use the mean value argument to prove that f'(x)>-\epsilon (since f''(x)>0). Of course, you have (from #1) that f'(x)<0. So you have what you need to prove the limit - for all \epsilon, there is an N such that if x>N, |f(x)|<\epsilon.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. proving that arctanh(x) is differentiable
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 20th 2011, 04:24 AM
  2. proving the exponential function is differentiable
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 21st 2010, 08:00 AM
  3. Replies: 0
    Last Post: October 3rd 2010, 07:03 AM
  4. Proving an equation is non-differentiable?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 23rd 2009, 09:51 AM
  5. Proving x|x| is differentiable
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 16th 2009, 05:39 PM

Search Tags


/mathhelpforum @mathhelpforum