Show that $f(x,y)=y^{1/2}$
(a) does not satisfy a Lipschitz condition on the rectangle $|x| \le 1$ and $0 \le y \le 1$.
Answer:
My book says that $\frac{f(x,y_1)-f(x,y_2)}{y_1-y_2}$ must be bounded on R to satisfy the Lipschitz condition in the variable y.
Thus, I tried $\frac{f(0,y)-f(0,0)}{y-0}=\frac{1}{\sqrt{y}}$ which is unbounded for $o \le y \le 1$ therefore $f$ does not satisfy Lipschitz continuous on R.
(b) does satisfy a Lipschitz condition on the rectangle $|x| \le 1$ and $c \le y \le d$, where $0 \lt c \lt d$.
Again I used the definition given and got $\frac{f(x,y_1)-f(x,y_2)}{y_1-y_2}=\frac{\sqrt{y_1}-\sqrt{y_2}}{(\sqrt{y_1}-\sqrt{y_2})(\sqrt{y_1}+\sqrt{y_2})}=\frac{1}{\sqrt{y_1}+\sqrt{y_2}}$ which is bounded for $|x| \le 1$ and $c \le y \le d$, where $0 \lt c \lt d$ with maximum of $\frac{1}{2\sqrt{c}}$ and minimum $\frac{1}{2\sqrt{d}}$.
Are these right?