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Real Analysis / Advanced Calc Puzzler

  1. Aug 4, 2011 #1
    Let [itex]f:[a,b] \rightarrow R[/itex] be a continuous function such that [itex]f(a)=f(b)=0[/itex] and [itex]f'[/itex] exists on [itex](a,b)[/itex]. Prove that for every real [itex]\lambda[/itex] there is a [itex] c \in (a,b)[/itex] such that [itex]f'(c) = \lambda f(c)[/itex].
     
    Last edited: Aug 4, 2011
  2. jcsd
  3. Aug 4, 2011 #2

    micromass

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    So, what did you try already??
     
  4. Aug 4, 2011 #3
    I was looking at this question as well just now.. considering f ' ( x ) / f ( x ) around a and b seems to be a way that might work.. but I am interested in seeing other methods, or a fuller solution
     
  5. Aug 4, 2011 #4

    micromass

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    Hmm, I actually tried that, but I found nothing satisfying when considering [itex]\frac{f^\prime (x)}{x}[/itex]. The problem is that you don't even know that f' is continuous. Furthermore, you may divide by 0 in the numerator. So I would actually be very curious how you did this (but only post after the OP found his answer).

    I solved it by looking at [itex]f^\prime-\lambda f[/itex] and by applying Darboux's theorem (i.e. http://en.wikipedia.org/wiki/Darboux's_theorem_(analysis) ). Rolle's theorem is also used...
     
  6. Aug 4, 2011 #5
    Ahh, I was thinking about the darboux's theorem too! I never really got a full answer with that other approach, which is why I only said "seems to work".

    Dividing by 0 I don't think would be that much of an issue, if the function hits 0 before a or b, then you can consider f ' ( x ) / f ( x ) around that point instead? What I'm not sure about is the behaviour of f ' ( x ) around f when it hits zero. f can grow arbitrarily small, but can f ' be small enough so that f ' / f doesn't necessarily diverge to infinity?
     
  7. Aug 4, 2011 #6

    micromass

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    Hmm, even with Darboux's theorem this problem is fun (=nontrivial). I'm considering asking this to my students as a bonus homework problem. Anybody who can solve this certainly knows his analysis...
     
  8. Aug 5, 2011 #7
    Ok, made some progress. We will assume that [itex]f(x)>0[/itex] for [itex]x\in (a,b)[/itex]. (We could prove using a bit of topology that provided f is not identically 0 then there is a subinterval where this is true for either f or -f.)

    We want to use Darboux's theorem which requires that [itex]f'_+(a)[/itex] exists, so first assume that it fails to exist. This implies that the limit isn't 0 so there exists [itex]\epsilon>0[/itex] such that for all [itex]\delta>0[/itex] there is [itex]x\in (a, a+\delta)[/itex] such that [itex]f(x)/x>\epsilon[/itex]. Let [itex]M[/itex] be a positive real number, and choose [itex]\delta = \frac{\epsilon}{M}[/itex]. (Adjust [itex]\delta[/itex] to be smaller if necessary to ensure that [itex]x\in (a, a+\delta)\Rightarrow f(x)<1[/itex].) By assumption there is [itex]x_0\in (a, a + \delta)[/itex] such that [itex]f(x)>\epsilon[/itex]. Apply the mean value theorem on the interval [itex](a, x_0)[/itex] to find [itex]c \in (a, x_0)[/itex] such that [itex]f'(c) = \frac{f(x_0)-f(a)}{x_0-a} > \frac{\epsilon}{\epsilon / M} = M[/itex]. Thus [itex]\frac{f'(c)}{f(c)} > M[/itex] so [itex]\frac{f'}{f}[/itex] can be made arbitrarily large close to the left endpoint.

    It can be similarly shown that if [itex]f'_-(b)[/itex] fails to exist then [itex]\frac{f'}{f}[/itex] gets arbitrarily small near the right endpoint.

    Now suppose that [itex]f'_+(a)[/itex] does exist. It cannot be negative. Assume that it is positive, and let [itex]M[/itex] be a positive real number. Let [itex]\delta>0[/itex] be such that [itex]x\in (a, a+\delta) \Rightarrow f(x)<\frac{f'_+(a)}{2M}[/itex]. It follows from the intermediate value property that there is [itex]c\in (a, a+\delta)[/itex] such that [itex]f'(c)>f'_+(a)/2[/itex]. Thus [itex]\frac{f'(c)}{f(c)} >M[/itex] so [itex]\frac{f'}{f}[/itex] can be made arbitrarily large close to the left endpoint.

    It can be similarly shown that if [itex]f'_-(b)[/itex] is negative then [itex]\frac{f'}{f}[/itex] gets arbitrarily small near the right endpoint.

    I guess there are a couple cases left but I'm going to bed now. Thanks for your help everyone.
     
  9. Aug 5, 2011 #8
    micromass, I was able to get the f'(x)/f(x) approach to work. The trick is
    f'(x)/f(x) = (log f(x))'
     
  10. Aug 5, 2011 #9
    Citan, that's a clever observation but how do you use it to solve the problem?
     
  11. Aug 5, 2011 #10

    disregardthat

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    Problem is that f is not non-zero, so log(f(x)) doesn't necessarily make sense.
     
  12. Aug 5, 2011 #11
    Ha, I get it now. Thats why it works. As I pointed out earlier we can assume that f is positive on (a, b). Since f(a) = f(b) = 0 and [itex]ln(0) = -\infty[/itex] we have that [itex]ln(f(x)) \rightarrow -\infty[/itex] as [itex]x \rightarrow a[/itex] (or b). Now apply the mean value theorem twice, once on the interval [itex](a+ \epsilon, (a+b)/2)[/itex] to show that f'(x)/f(x) gets arbitrarily small and again on the right side to show that it gets large. Then we can apply the intermediate value property.
     
  13. Aug 5, 2011 #12

    disregardthat

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    Well, f(x) can be zero anywhere on (a,b) as well. e.g. if f(x) = x^2sin(2pi/x) on [0,1] (with f(0) = 0), it doesn't matter how close you are to a = 0, you just cannot use the argument that lim_(x->a) log(f(x)) --> -infty in that situation. Of course the function could behave like this at b as well. For this argument to work, I would suggest to you to start out with an open interval where f(x) is non-zero (you must prove that such an interval exists), and narrow the domain to the closure of that open interval, and then extend the boundaries of this closed interval so that f is zero on the boundaries.
     
    Last edited: Aug 5, 2011
  14. Aug 5, 2011 #13

    micromass

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    The thing is, of course, that we can always pick a segment where only the boundary points are zero. So we can restrict to that case.
     
  15. Aug 5, 2011 #14

    disregardthat

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    Which is exactly what I explained in my post above.
     
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