I am currently studying for an exam about differential equations and in one exercise I have to prove that the function $f(t,x)=te^{-x^{2}}$, $|t|\leq 1$, $x\in \mathbb{R}$ is Lipschitz-continous with respect to the second variable x on the domain $\mathbb{R}$.
My professor has shown that $|e^{-x^{2}}-e^{-y^{2}}|\leq K|x-y|$, but I don't see this result. Can anyone enlighten me, and explain me why is that?
My professors reasoning, $$|te^{-x^{2}}-te^{-y^{2}}|= |t||e^{-x^{2}}-e^{-y^{2}}| \leq |e^{-x^{2}}-e^{-y^{2}}| \leq K|x-y|\hspace{3mm}\forall |t|\leq 1, x\in \mathbb{R}$$ Hence $f(t,x)$ is globally Lipschitz-continous with respect to the second variable x on $\mathbb{R}$.
I understand each and every equality/inequality except for the last one.
I have to point out that if I use the mean value theorem, then $K$ depends on $x$, so $K_x$ $$|e^{−x^2}−e^{−y^2}|=∣\frac{−2r}{e^{r^2}}∣|x−y|\leq2r|x−y|$$ with $r\in[x,y]$ so that would mean that my function $f(t,x)$ is locally Lipschitz-continous with respect to $x$. But my professor is saying a stronger thing, he is saying that $f(t,x)$ is globally Lipschitz-continous.
From L'Hopital's rule: $$\lim_{r \to \infty}\frac{r}{\exp(r^2)}=\lim_{r \to - \infty}\frac{r}{\exp(r^2)}=0$$ Thus, we may pick $N$ large enough such that $|r|>N$ implies $$\big| \frac{r}{\exp(r^2)}\big| \leq 1$$ By the Extreme value theorem, since $r \mapsto r/ \exp(r^2)$ is continuous, and $[-N,N]$ is a closed and bounded interval, we know this function attains a maximum and minimum there. In other words, we can pick $M$ large enough such that $|r| \leq N$ implies $$\big| \frac{r}{\exp(r^2)}\big| \leq M$$ Putting the pieces together, we have that, for all $r \in \mathbb{R}$:$$\big| \frac{r}{\exp(r^2)}\big| \leq \max(1,M)$$ So we can use $K= \max(1,M)$