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?
Best Answer
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 $y \in \mathbb{R}$.