- #1
- 533
- 1
Suppose I have an infinitely differentiable function F that is nonzero exactly on a set [-b,b]. Can I say that [itex]|F(x)| \leq C(x+b)^k[/itex] for some integer [itex]k > 2[/itex]? If so, why?
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]?
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].