I have $f(t, y) = e^{t - y}$. I want to see if $f$ satisfies a Lipschitz condition on $D = \{(t, y): 0 \le t \le 1, \; - \infty < y < \infty \}$.

I am using this definition:

A function $f(t, y)$ satisfies a LC in the variable $y$ on the set $D \subset \mathbb R^2$ if there is a constant $L > 0$ such that $|f(t, y_1) - f(t, y_2)| \le L |y_1 - y_2|$ with $(t, y_1)$ and $(t, y_2)$ in $D$.

I tried:

$$ |e^{t - y_1} - e^{t - y_2} |$$ $$= |e^t(e^{-y_1} - e^{-y_2})|$$ $$= |e^t||e^{-y_1} - e^{-y_2}|$$ $$\le e^1 |y_1 - y_2| $$

At this step, can I say that this does not satisfy a LC, since if $y_1 = -1$ and $y_2 = -2$, the inequality would not be true?


Suppose $y <0$, then $|f(0,y)-f(0,0)| = |e^{-y}-1| = \sum_{k=1}^\infty \frac{1}{n!} |y|^n\geq \frac{1}{2} y^2$, so $f$ cannot be uniformly Lipschitz on the domain specified (for any $L>0$, you can choose $|y|>L$ to violate the Lispschitz bound).

If $f$ satisfies a Lipschitz bound in the second variable, then you would have $|f(0,y)-f(0,0)| \leq L |y-0| = L|y|$, for some $L>0$. However, we have shown that $|f(0,y)-f(0,0)|\geq \frac{1}{2} y^2$, so if both bounds hold, then you would have $\frac{1}{2} y^2 \leq L|y|$ for $y \in \mathbb{R}$. Choosing $y\neq 0$ and dividing across by $|y|$ gives $\frac{1}{2}|y| \leq L$, which clearly cannot hold for all $y \in \mathbb{R}$.

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  • $\begingroup$ Why did you choose $\frac{1}{2} y^2$? $\endgroup$ – badjr Jan 30 '13 at 4:02
  • $\begingroup$ Show I can show that any Lipschitz bound will be violated, any of the exponential terms will do, the $y^2$ is the first nonlinear term... $\endgroup$ – copper.hat Jan 30 '13 at 4:08
  • $\begingroup$ Can you go into more detail for this answer? I understand everything up to $|f(0,y)-f(0,0)| = |e^{-y}-1| = \sum_{k=1}^\infty \frac{1}{n!} |y|^n$ but I'm not seeing how $\geq \frac{1}{2} y^2$ relates to the Lipschitz bound. $\endgroup$ – badjr Jan 30 '13 at 4:40
  • $\begingroup$ I have added more detail to the answer... $\endgroup$ – copper.hat Jan 30 '13 at 4:49

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