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Mean Value Theorem Applications

  1. Nov 1, 2007 #1
    Q: Prove that if f: R^n -> R is defined by f(x)=arctan(||x||), then |f(x)-f(y)| <= ||x-y|| FOR ALL x,y E R^n.
    [<= means less than or equal to]

    Theorem: (a corollary to the mean value theorem)
    Suppose f is differentiable on an open, convex set S and ||gradient [f(x)]|| <= M for all x E S. Then |f(b) - f(a)| <= M ||b-a|| for all a,b E S.

    Now the trouble is that f(x)=arctan(||x||) is not differentiable at x=0 E R^n since the partial derivatives doesn't exist at x=0. Even worse, notice that S = R^n \ {0} is not convex.

    Then how I can show that "f is differentiable on an open, convex set S"? (what is S in this case?) I strongly believe that this is an important step because if the conditions in the theorem are not fully satisifed, then there is no guarantee that the conclusion will hold. But this seems to be the only theorem that will help. What should I do?

    Next, how can I prove that the conclusion is true FOR ALL x,y E R^n ?


    Thanks for explaining!:smile:
     
    Last edited: Nov 1, 2007
  2. jcsd
  3. Nov 1, 2007 #2

    Dick

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    If you want to use that theorem, just break the problem into cases. Suppose x and y are points such that 0 is not on the line between x and y. Then you shouldn't have any trouble finding an open convex set containing x and y that doesn't contain 0. Now what happens if 0 is on the line? You could break things up more, but I think it's easier to think about a sequence x_n->x where 0 is not on the line connecting x_n and y.
     
    Last edited: Nov 1, 2007
  4. Nov 2, 2007 #3
    How can I break the problem into cases such that in each case, f is differentiable on an open, convex set S?
     
  5. Nov 2, 2007 #4

    Dick

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    Do one case at a time. If 0 is not on the line between x and y, then what? If it is, then you are going to have a lot of trouble finding an open convex set on which f is differentiable containing x and y. Obviously. So give up that hope. Please reread my post.
     
  6. Nov 2, 2007 #5

    Dick

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    If 0 is on the line, you could try breaking the problem up by arguing about the difference between x and 0 and y and 0, but that doesn't involve a single convex domain of differentiability. And it's certainly more complicated.
     
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