# Thread: Proving properties of bumb function

1. ## Proving properties of bump function

Let be $f(x):=g(1-g(x)); x \in \mathbb{R}$, where $g: \mathbb{R} \rightarrow \mathbb{R}\text{; } x \mapsto g(x):= \begin{cases}e^{1-\frac{1}{x^2} & \text{if }x>0\\0 & \text{if }x\leq 0\end{cases}$
Let be $a, b \in \mathbb{R}$ and $h(x)=f(a-x)f(x-b)$.
Prove that: $h \in C^\infty;\: h\geq0; h(x)=1 \text{ if } x\in [a,b] \text{ and } h(x)=0 \text{ if } x \not\in ]a-1,b+1[$.

I know that $g \in C^\infty$ but not analyitic in 0. I also know that $f$ is a monotonically decreasing function $f(x)=1 \text{ if } x \leq 0 \text{ and } f(x)=0 \text{ if } x>0$.
Thank you very much in advance!

2. ## Re: Proving properties of bumb function

T
Let be $f(x):=g(1-f(x)); x \in \mathbb{R}$, where $g: \mathbb{R} \rightarrow \mathbb{R}\text{; } x \mapsto g(x):= \begin{cases}e^{1-\frac{1}{x^2} & \text{if }x>0\\0 & \text{if }x\leq 0\end{cases}$
Hey, this doesn't make sense. Are you sure you wrote right?

3. ## Re: Proving properties of bump function

Originally Posted by Drexel28
T

Hey, this doesn't make sense. Are you sure you wrote right?
Thank you so much that you warned me. You are right, f(x)=g(1-g(x)), I have corrected it. Thanks!

I have made another mistake that I have just noticed. In the title it is not bumb, but of course bump function. Sorry!