Infinite Differentiability and Analyticity.

[Alone]

I want to show that the function defined as follows:

$f(x)=e^{-1/x^2}$ for $|x|>0$ and $f^{(k)}(0)=0$ for $k=0,1,2,\ldots$ is infinitely differentiable but not analytic at the point $x=0$.

For infinite-differentiability I used the fact that $\lim_{|x|\to 0^+} x^{-n} e^{-1/x^2}=0$ for every $n$ nonnegative integer, from which I can deduce that this function is infinite differnetiable at $x=0$ since all of its derivatives are continuous at $x=0$ and equal zero.

The problem I am having is with showing that it's not analytic, I need to show that for every $\delta>0, M>0$ the derivative $f^{(k)}(x)$ its absolute value is greater than $M\cdot k!/\delta^k$; not sure how to show it exactly.

 

[S.G. Janssens]

To start by picking a nit, I would like to remark that the equality $f^{(k)}(0) = 0$ for $k \ge 1$ is not part of the definition of $f$, but a consequence of the definition of $f$.

If $f$ were analytic, there would exist $r > 0$ such that $f$ can be written as a pointwise convergent Taylor series on the open interval $(-r,r)$. Now, $f(0) = 0$ and you wrote that you have proven that all derivatives of $f$ at the origin vanish. However, $f(x) \neq 0$ for any $x \neq 0$. So what could you conclude?

 

[Alone]

To start by picking a nit, I would like to remark that the equality $f^{(k)}(0) = 0$ for $k \ge 1$ is not part of the definition of $f$, but a consequence of the definition of $f$.

If $f$ were analytic, there would exist $r > 0$ such that $f$ can be written as a pointwise convergent Taylor series on the open interval $(-r,r)$. Now, $f(0) = 0$ and you wrote that you have proven that all derivatives of $f$ at the origin vanish. However, $f(x) \neq 0$ for any $x \neq 0$. So what could you conclude?

I would get a contradiction since the Taylor series is identically zero, but the function isn't zero for $x\ne 0 $.

Thanks.

 

[Alone]

@Krylov I have a follow up question:

I want to verify that for every $M,\delta>0$ in any $\delta$-neighbourhood of the origin the function $f$ that was given in my first post in this thread satisfies:

$$|f^{(k)}(x)| >Mk!/\delta^k$$

This can be concluded from the fact that $f$ isn't analytic, but can you show this explicitly?