# Thread: Proving properties of bumb function

1. ## Proving properties of bump function

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

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

2. ## Re: Proving properties of bumb function

T
Originally Posted by zadir
Let be $\displaystyle f(x):=g(1-f(x)); x \in \mathbb{R}$, where $\displaystyle g: \mathbb{R} \rightarrow \mathbb{R}\text{; } x \mapsto g(x):= \begin{cases}e^{1-\frac{1}{x^2} & \text{if }x>0\\0 & \text{if }x\leq 0\end{cases}$
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!

4. ## Re: Proving properties of bump function

Originally Posted by zadir
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!
The problem is right and clear after correction, isn"t it?
Thank you very much in advance!