Define $f:\mathbb R^3\to \mathbb R$ as $f(x,y,z)=x+e^yz$. Is $f$ Lipschitz?
I'm having a hard time with this question. Simply chugging $(x,y,z),(x',y',z') \in \mathbb R^3$ and calculating $$|f(x,y,z)-f(x',y',z')|$$ seems like a dead end. However, I don't see any combination of variables in $\mathbb R^3$ that would disprove that $f$ is Lipschitz. I tried $(x,x,x),(y,y,y)$ but that doesn't work. I know $e^x$ is Lipschitz, but I'm not sure if this means that $f$ is Lipschitz. Any hints?
The function $$f(x,y,z)=x+e^yz$$ is not Lipschitz.
Consider $$f(1,y_1,1)-f(1,y_2,1)= e^{y_1 } -e^{y_2}$$
There is no constant $k$ which makes $$ | e^{y_1 } -e^{y_2}|<k|y_1-y_2|$$ for all $y_1$ and $y_2$
Thus the function is not Lipschitz.
If $f(x,y,z)$ were Lipschitz then $g(x) =f(0,y,1)=e^y$ should be Lipschitz but it is not true since the inequality $$|e^y -1|\leq M|y|$$ is not true for any $M$ for all $y\in\mathbb{R}$
Hint: You can write $$x_2+e^{y_{2}} z_2-x_1-e^{y_1}z_1=x_2-x_1+e^{y_2} z_2-e^{y_1}z_2+e^{y_1} z_2-z_1e^{y_2}=x_2-x_1+(e^{y_2}-e^{y_1})z_2+e^{y_1}(z_2-z_1)$$
