Infinitely differentiable functions

[AxiomOfChoice]

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

 

[lugita15]

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]? Because if it were continuous on the whole close interval [-b,b], then it would be bounded, so the conclusion would follow pretty trivially.

 

[jgens]

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 think that OP is saying that $f$ has support $[-b,b]$. In any case, assuming that my interpretation is correct, I think that the claim is true. If $f$ is supported in $[-b,b]$, then certainly $f^{(k)}(-b) = 0$ for all $k \in \mathbb{N}$. Since the $k$-th derivative of $(x+b)^k$ is $k!$, it should be possible to show that $|f(x)| \leq |x+b|^k$ in some neighborhood of $-b$. Then by choosing the constant $C$ 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.

 

[Skrew]

Ooops, I ignored that it was nonzero only on [-b, b]. So the function is zero everywhere else.

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

 

[AxiomOfChoice]

If $f$ is supported in $[-b,b]$, then certainly $f^{(k)}(-b) = 0$ for all $k \in \mathbb{N}$.

I feel a little bad about this, but I don't see why this is. Could you explain, please? Is it just because, since $f(x) = 0$ for $x \leq -b$, we must have $f'(x) = 0$ for $x < -b$, so by continuity of the derivative, $f'(-b)$ can't possibly be anything other than zero? (And then the same argument is repeated for the higher order derivatives.)

 

[deluks917]

This can basically be done for any k >= 1. Let y = x + b. Then we get F(y) is zero outside of [0,2b].

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'hospital's 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.