Is the function $e^{-Bx^2}$ a contraction on $\mathbb{R}$? ($B>0$)

Question

The tittle says it all. I think it's true, and I tried to prove it by showing that the derivative of this function: $-2Bxe^{-Bx^2}$ is bounded from above with a bound less than 1, in order to do that, I tried to use Taylor series of $e^{-Bx^2}$, but it seems that leads nowhere. Any suggestion?

Here $B>0$ is a real number and we consider the euclidean norm.

Answer 1

You want to check whether the maximum of $|f'|$ is less than $1$. So take one more derivative:

$$f''(x)=-2Be^{-Bx^2}+4B^2x^2e^{-Bx^2}.$$

This is zero if and only if $4B^2x^2-2B=0$, i.e. $x^2=1/(2B)$. At these points you have $|f'(x)|=2^{1/2} B^{1/2} e^{-1/2}$. You can check that these must be the points where $|f'|$ is largest since $f'(0)=0$ and $f'(x) \to 0$ as $x \to \pm \infty$. It is clear that this grows without bound as a function of $B$, so $e^{-Bx^2}$ cannot be a contraction for all $B$.

Answer 2

Suppose $f$ is a contraction with rank $\lambda <1$. In particular, we must have $|f'(x)| \le \lambda$ for all $x$, or equivalently ${2B|x| \over e^{B x^2}} \le \lambda$ for all $x$.

If we pick $B> ({e \over 2})^2$, then $f'(-{1 \over \sqrt{B}}) = { 2 \sqrt{B} \over e} > 1$.

More specifically, let $B=e^2$, then $f'(-{1 \over e}) = 2$.