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- Thread starter AxiomOfChoice
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jgens

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I think that OP is saying that [itex]f[/itex] has support [itex][-b,b][/itex]. In any case, assuming that my interpretation is correct, I think that the claim is true. If [itex]f[/itex] is supported in [itex][-b,b][/itex], then certainly [itex]f^{(k)}(-b) = 0[/itex] for all [itex]k \in \mathbb{N}[/itex]. Since the [itex]k[/itex]-th derivative of [itex](x+b)^k[/itex] is [itex]k![/itex], it should be possible to show that [itex]|f(x)| \leq |x+b|^k[/itex] in some neighborhood of [itex]-b[/itex]. Then by choosing the constant [itex]C[/itex] appropriately, this would prove the claim. I have not checked to see if this works, but it sounds like a decent way to attack the problem.Are you saying that F is infinitely differentiable on the open interval (-b,b) but not necessarily continuous at the two endpoints of the closed interval [-b,b]?

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To the OP, you likely have to do something with looking at the maximum value of the derivatve on [-b, b].

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I feel a little bad about this, but I don't see why this is. Could you explain, please? Is it just because, since [itex]f(x) = 0[/itex] for [itex]x \leq -b[/itex], we must have [itex]f'(x) = 0[/itex] for [itex]x < -b[/itex], so by continuity of the derivative, [itex]f'(-b)[/itex] can't possibly be anything other than zero? (And then the same argument is repeated for the higher order derivatives.)If [itex]f[/itex] is supported in [itex][-b,b][/itex], then certainly [itex]f^{(k)}(-b) = 0[/itex] for all [itex]k \in \mathbb{N}[/itex].

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We need to show F(y) <= Cy^k

Let g(y) = 0 if y=0

= f(y)/(x^k) if y =/=0

Note g is obviously continuous except possibly at y=0. However F is differentible at y=0 and zero to the left of 0. So all the derivatives of F at 0 are 0. Hence By L'hopitals rule lim y ->0 g(y) = 0. So g is continuous. Hence |g(y)| <= C for all y in [0,2b]. So |f(y)|/(x^k) <= C so

|f(y)| <= Cx^k for all y in [0,2b]. Assuming k is even we get that |f(y)| <= Cx^k for all x. If k is odd we can only get |f(y)| <= Cx^k for all positive x.

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