Fred wrote: << I wrote: << In differential topology it's crucial that there's a C^oo function f: R -> R that's 0 for x <= 0, 1 for x >= 0, and strictly increasing in between. That implies each derivative is bounded.

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I think this should have read " 0 for x <= 0, 1 for x >= 1 "; and given its stated purpose, and the subsequent construction, presumably my constraints on the derivatives vanishing at the end-points are also involved.

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Since the function f is C^oo, and f(x) = 0 for x <= 0 and f(x) = 1 for x >= 1, it immediately follows that all derivatives f^(k)(0) = f^(k)(1) = 0 for all k >= 1. So, no additional conditions are required. I continued: << (This often allows for glueing together local things on a manifold to create a global thing. There's can't be such a function that's real analytic (C^w) everywhere, which is why it's often much harder to show that global C^w things exist on manifolds, and often they don't.) The standard example f is given as follows: { 0, x <= 0 Let g(x) = { { exp(-1/x), x > 0 Then define f(x) = g(x) / (g(x) + g(1-x)).

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This looks essentially similar to the naive construction I mentioned earlier, and suffers from the same (practical) defect: for example, its first derivative has x^2 (1-x)^2 in the denominator.

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I'm not sure what this means, but the proof that this function has the stated properties can be found as example 4.1b on this web page: < http://www.math.lsu.edu/~lawson/Chapter4.pdf > (though Lawson's definitions of f and g are the reverse of mine.) --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx