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Let $$ f(t,y)=te^{-y^2}, \quad\Big\{ (t,y): \, |t| \leq 1, \, y\in \mathbb{R} \Big\} $$ In order for $f$ to satisfy a Lipschitz condition in respect to $y$, we must have $$ \Big|f(t,y_1)-f(t,y_2)\Big|\leq k\,\Big|y_1-y_2\big| $$ $$ \bigg|t\big(e^{-{y_1}^2}-e^{-{y_2}^2}\big)\Big|\leq k\, \Big|y_1-y_2\Big| $$ Now, if we exclude the trivial case of $y_1=y_2$, which gives $k=0$, we could write $$ |t|\Bigg|\frac{e^{-{y_1}^2}-e^{-{y_2}^2}}{y_1-y_2}\Bigg|\leq k \iff |t|\cdot T(y_1,y_2)\leq k $$ It's obvious that $T(y_1,y_2)$ is a function that gives the tangent of the angle between the line which connects the points $(y_1,f(y_1))$ $(y_2,f(y_2))$ and the $y$-axis. $T$ tends to $0$ for $y$-values that tend to $+\infty$ but also tends to $+\infty$ for $y$-values that tend to $-\infty$.

Given that, how can I find a $k$ that satisfies the above inequality?

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  • $\begingroup$ Could you please check the sign in the exponent? What about using the mean value theorem? $\endgroup$ – LutzL Oct 29 '18 at 14:50
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Let $e^{-y^2}=:g(y)$. Then one has $$|f(t,y_1)-f(t,y_2)|\leq |t|\>\bigl|g(y_1)-g(y_2)\bigr|\leq \sup_{y\in{\mathbb R}} \bigl|g'(y)\bigr|\cdot|y_1-y_2|\ .$$ Now compute this $\sup_{y\in{\mathbb R}} \bigl|g'(y)\bigr|$ solving a simple maximum problem.

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