Bound on the sup norm for derivatives of a particular $C^\infty$ function I'm reading textbook "A Primer of Real Analytic Functions" and on page 86 the following "obvious" claim is made:


  Let $|| \cdot ||$ be the sup norm on $[0, 2 \pi]$ and define function  $f$ to be $e^{\frac{1}{x(x-2\pi)}}$ on $(0, 2 \pi)$ and $0$ at points $0$ and $2 \pi$. Note that $f \in C^\infty([0,2\pi])$. Show that there exists a constant $C&gt;0$ s.t. for all $n \in \mathbb{N}$ we have $$ || f^{(n)}|| \leq C \cdot 2^{2^n}$$


I suspect that this bound can be obtained by noticing that $f$ is in some class that can be represented by some integral kernal - however I know of no such thing.  The book mentions a couple facts from Fourier analysis so perhaps that is relevant - I've included it as a tag.
This is only a hint.

Try the following problem first: Define $g(t) = e^{1/t}$ on $(0,1]$ and $g(0) = 0.$ Can you show that $g(t) \in C^\infty([0,1])?$ What estimates can you make on the sup norm of the derivatives of $g.$ 

Now the problem at hand. To show $f$ is $C^\infty$ on an interval like $[0, \epsilon]$ with $\epsilon &lt; 2\pi,$ write $f(t) = g(t)^{1/(t-2\pi)}.$ You can use your solution from the above part, to obtain the conclusion. For $f$ near $2\pi,$ repeat the same argument with the two singularities switched around.
This is only a hint.

Try the following problem first: Define $g(t) = e^{1/t}$ on $(0,1]$ and $g(0) = 0.$ Can you show that $g(t) \in C^\infty([0,1])?$ What estimates can you make on the sup norm of the derivatives of $g.$ 

Now the problem at hand. To show $f$ is $C^\infty$ on an interval like $[0, \epsilon]$ with $\epsilon &lt; 2\pi,$ write $f(t) = g(t)^{1/(t-2\pi)}.$ You can use your solution from the above part, to obtain the conclusion. For $f$ near $2\pi,$ repeat the same argument with the two singularities switched around.