Assume we have some function $f(t,y)$ defined on an interval $D = \{(t,y) : |t-t_0|\leq a,|y-y_0|\leq b \}$. I know from the definition of Lipschitz continuity that it requires $|f(t,y_1) - f(t,y_2)| \leq L | y_1 - y_2|$, but I don't understand why this is the same as $L \leq max_{t,y\in D} \;|\;\frac{\partial}{\partial y} f(t,y)\;|$. Is it even the same?
An explanation would be greatly appreciated, thanks!
Here I am assuming you are considering $f(t,y)$ which is differentiable in $y$. As has been pointed out, there are Lipschitz functions which are not differentiable (merely differentiable almost everywhere) but for $f(t,y)$ differentiable, the following holds:
The Lipschitz constant $L$ is usually taken to be the least positive number $C$ such that $$| f(t, y_1) - f(t, y_2) \rvert \le C |y_1 - y_2 \rvert$$ for all $y_1, y_2$ in the domain. By the mean value theorem, for any $y_1, y_2$, there is $y^*$ between $y_1$ and $y_2$ with $$|f(t, y_1) - f(t, y_2) \rvert = \left | \frac{\partial f}{\partial y}(t,y^*) \right \rvert | y_1 - y_2 \rvert \le \sup_{t,y} \left | \frac{\partial f}{\partial y}(t,y) \right \rvert | y_1 - y_2 \rvert.$$ Thus $ \sup_{t,y} \left | \frac{\partial f}{\partial y}(t,y) \right \rvert$ is such a $C$. Since $L$ is the least such $C$, we have $$L \le\sup_{t,y} \left | \frac{\partial f}{\partial y}(t,y) \right \rvert. $$