I have to find out whether the equation $$y' = \sqrt{y} $$ by the condition $$y(1) = 1 $$ satisfies the assumptions of the Existential theorem.
Now I claim that the function is continuous around x = 1, which means then we have one exact solution for $y' = \sqrt{y}; y(1) = 1$. Is this enough?
Because if I go with the Lipschitz under assumption that $x = 1, y_1 = y, y_2 = 1$ I get:
$|f(x,y_1) - f(x,y_2) |\leq k(x)|y_1 - y_2|$; $$|\sqrt{y} - 1 |\leq k(x)|y - 1|$$ From where I am not able to make any conclusion.
Hint: $$ \frac{\partial f}{\partial y} = \frac{1}{2\sqrt{y}}, $$ and by the MVT it follows that this derivative is bounded in a neighborhood of $(1,1)$. Hence it is locally Lipschitz. Of course, since $y \mapsto \sqrt{y}$ is not globally Lipschitz, you can apply your theorem only to get a unique local solution.
Moreover, remark that the mere continuity of $f$ is not enough to ensure existence and uniqueness. However it does ensure local existence (Peano's theorem).