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Smooth function's set of critical values

  1. Jun 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Assume that [tex] f:[a,b] \rightarrow \mathbb{R} [/tex] is continuously differentiable. A critical point of f is an x such that [tex] f'(x) = 0 [/tex]. A critical value is a number y such that for at least one critical point x, y = f(x). Prove that the set of critical values is a zero set.

    2. Relevant equations
    A set Z is said to be a zero set if for each [itex] \epsilon [/itex] there is a countable covering of Z by open intervals [tex] (a_i , b_i) [/tex] such that [itex] \sum_{n=1}^\infty} b_i - a_i \leq \epsilon [/itex]

    3. The attempt at a solution
    I'm not even sure really where to start with this one. Normally when I start a problem, I always check to make sure that it intuitively makes sense. If we let C be the set of critical points, then the goal is to show f(C) has measure zero, but I really don't see how to approach this problem. I don't really see how to use the fact that the derivative is continuous. I'd love a hint in the right direction to start or something, as this has me pretty baffled. Thanks!
  2. jcsd
  3. Jun 3, 2008 #2


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    Try showing that C can be covered by a set of intervals on which |f'(x)|<e for any e>0 for x in any of those intervals. Now if [c,d] is one of those intervals what can you say about |f(x)-f(y)| for x and y in [c,d]? Think mean value theorem. So what can you say about the measure of f([c,d])? Now what happens as you take e->0?
  4. Jun 4, 2008 #3
    Ah! That makes perfect sense! For each e>0 we know the set of points such that |F '(x)| < e , contains our set C. Let [c,d] be a subset of the aforementioned set. Now, if we let x, y, be points in [c,d], we know by the MVT [tex] | f(x) - f(y)| = |f'(a) (x - y)| < \epsilon |(x-y)| [/tex] (for some a in (x,y) Since [itex] \epsilon [/itex] was arbitrary, we see this goes to zero. This shows that the distance |f(x) - f(y)| is arbitrarily small. So, now for each interval [f(x), f(y)] has a length given by the above inequality, which should go to zero as epsilon can be made arbitrarily small. So, we can make each critical value be contained in an arbitrarily small interval, which suggests that the set of critical values has measure zero.

    Is that the right reasoning? Thank you so much for your help! This problem was driving me crazy, and I was approaching it previously in a far more complicated manner.
  5. Jun 4, 2008 #4


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    Almost. |f(x)-f(y)|<e*|x-y| probably isn't quite enough. But |f(x)-f(y)|<e*|d-c| or even |f(x)-f(y)|<e*|b-a| would do it. You want to bound the distance by a constant that doesn't depend on x and y. Do you see why? But those are trivial to prove. So yes, I think you've got it.
  6. Jun 4, 2008 #5
    Since we know |f(x) - f(y)| < e|x-y| and we know x,y are in [c,d] we can then state that |x-y| < |c-d| giving us |f(x) - f(y)| < e|c-d| which doesn't depend on the variables x or y, which makes sense we'd want to bound it with a constant, not by some variable distance. This problem makes a whole lot of sense now and wasn't anywhere near as difficult as I was making it! Thanks for the help.
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