# Finding a Lipschitz constant $k$ for this function

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?

• Could you please check the sign in the exponent? What about using the mean value theorem? – LutzL Oct 29 '18 at 14:50

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.