The question: Determine if $f(t,y) = \frac{t^2y}{1+y^2}$ satisfies a local or a uniform Lipschitz condition.
My thoughts: Well from my work, I have $$ |f(t,x) - f(t,y)| \leq K |x-y|. $$
I can determine what $K$ is by the mean value theorem. For any $z \in [x,y]$,
$$ \Big|\frac{f(t,x)-f(t,y)}{x -y}\Big| = \Big|\frac{\partial f}{\partial y} \Big |. $$
Therefore $$\frac{\partial f}{\partial y} = \frac{t^2 (y^2-1)}{(y^2 + 1)^2}$$. My question is how can we show this function is either locally Lipschitz and not globally Lipschitz. I am guessing this function is not uniformly Lipschitz because of the $t^2$ term.
Thank you very much!
$Df(t,y)=\begin{pmatrix} \frac{2ty}{1+y^{2}} &\frac{1-y^2}{(y^2+1)^2} \end{pmatrix}$ so $\|Df(x,y)\|\le \sqrt{4t^2+1}.$
Let $(t,y),(t_1,y_1),(t_2,y_2)\in \overline J\times \overline U$, where $J,U$ is some bounded open sets in $\mathbb R$
Then, the MVT (in two variables) and compactness show that
$|f(t_1,y_1)-f(t_2,y_2)| \le \sup_{(t,y)\in\overline J\times \overline U}\{\sqrt{4t^2+1}\}\cdot\|(t_1,y_1)-(t_2,y_2)\|\le$ $ K\|(t_1,y_1)-(t_2,y_2)\|$, where $K$ depends on $J$ and $U$. Thus, $f$ is locally Lipschitz.
Now, $f$ is differentiable eveywhere, so
$f(t+h,y+k)-f(t,y)=Df(t,y)(h,k)+r(h,k)$ where $\frac{r(h,k)}{|(h,k)|}\to 0$ as $(h,k)\to (0,0)$.
Set $y=1$ and consider the points $(t,1).$ We have then,
$\frac{|Df(t,1)(h,k)|}{|(h,k)|}=\frac{|th|}{|h|+|k|}$ so $\frac{|f(t+h,1+k)-f(t,1)|}{|(h,k)|}=\frac{|th|}{|h|+|k|}+\frac{r(h,k)}{|(h,k)|}\to |t|$ as $(h,k)\to (0,0)$
which implies that $f$ is not globally Lipschitz.