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I am trying to figure out how to prove that this function is Lipschitz. I tried the following (for $h>0$):

$$ \left|e^{-(x+h)^2}-e^{-x^2}\right|=\left|\frac{e^{-(x+h)^2}-e^{-x^2}}{h}\right|\cdot|h| $$

By playing with Desmos, I noticed that $\displaystyle\left|\frac{e^{-(x+h)^2}-e^{-x^2}}{h}\right|$ increases as $h$ decreases. Thus, we get an upper bound if we let $h$ approach $0.$ So we get $\left|-2xe^{-x^2}\right|$ and, by elementary calculus, we get a uniform maximum of $\sqrt{\frac{2}{e}},$ which we may choose as our Lipschitz constant. What I wonder is if there is a better (more analytic) argument.

G.O.F.
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Following standard methods and considering the second derivative of f, one can show that $|f'(x)|\leq \sqrt{\frac{2}{e}}$. Furthermore we have that, $f(x)=f(x)-f(y)+f(y)=\int\limits_{y}^{x} f'(t)dt +f(y) \leq \int\limits_{y}^{x} \sqrt{\frac{2}{e}}dt +f(y)=\sqrt{\frac{2}{e}}(x-y)+f(y)$. Thus, $f(x)-f(y)\leq \sqrt{\frac{2}{e}}(x-y)$. Furthermore, using the exact same arguements, we also conclude that $f(y)-f(x)\leq \sqrt{\frac{2}{e}}(y-x)$ and hence $|f(x)-f(y)|\leq \sqrt{\frac{2}{e}}|x-y|$ which shows that $f$ is $\sqrt{\frac{2}{e}}$-Lip.

NOTE that $\sqrt{\frac{2}{e}}$ is sharp Lip-constant.

GaloisTH
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