how do I find a Lipschitz constant for $e^{-x^2}$? This is homework so I'd just like some tips on how to proceed. I've got $|e^{-x^2}-e^{-y^2}| = |(e^{-x^2+y^2} -1)||e^{-y^2}| < a$ where $a = -x^2 +y^2$.
I'm not sure how to proceed so any help will be appreciated! Thanks!
Thanks that makes sense!
I would solve it like this: you have that $f(x) = e^{-x^2}$. This is the graph of your function:
As you can see there are two points around $-0.8$ and $0.8$ where the slope is the highest one. The definition of Lipschitz continuous says:
A function $f: \mathbb{R} \to \mathbb{R}$ is Lipschitz continuous if there exists some constant L such that:
$$\vert f(x) - f(y)\vert \le L |x-y|$$
Since your $f$ is differentiable, you can use the mean value theorem, $$ \frac{f(x) - f(y)}{x-y} \leq f'(z) \quad\text{for all $x < z < y$,} $$ and deduce that it is sufficient to find a bound $L$ with $|f'(z)| \leq L$ for all $z \in \mathbb{R}$.
Now from your graph the important thing that you have to understand is if the slope in the points around $-0.8$ and $0.8$ is lower or bigger than $1$ (because there the slope assumes obviously the biggest values). If it's lower than $1$ you can choose $L=1$ and you have $|f(x) - f(y) \le |x-y|$ because the slope of the function is lower than $1$ and this means that the distance between the images of $f(x)$ is lower than the distance between the arguments in every point $x,y \in \mathbb{R}$.
So let's look after the maximal value of the slope of your function $f(x)$ to do that we calculate the derivative of $f(x)$ given by:
$$f'(x) = -2x e^{-x^2}$$
This is the graf of the function $f'(x)$:
You see that as we said she assumes the maximal values around the points $-0.8$ and $0.8$ to get the exact value we have to derivate once again $f'(x)$ and look for which point the functions is equal to $0$:
$$f''(x) = -2e^{-x^2} + 4x^2e^{-x^2} = e^{-x^2} * (-2 + 4x^2)$$
Now
$$f''(x) = 0 \iff (4x^2 -2) =0$$
since $e^{-x^2}$ will never be $0$
$$\iff x = +\frac{1}{\sqrt(2)} \lor x = -\frac{1}{\sqrt(2)}$$
Now you notice that $f'(\frac{1}{\sqrt(2)}) = -\sqrt{\frac{2}{e}} \gt -1$ and $f'(-\frac{1}{\sqrt(2)}) = \sqrt{\frac{2}{e}} \lt 1$
Now you are allowed to take $L = 1$ and you have your Lipschitz continuity :-)