Exercise A12.5 of Hubbards' Vector Calculus, Linear Algebra, and Differential Forms

Question

I have a question about the solution to part (a) of exercise A12.5 of Hubbards' Vector Calculus, Linear Algebra, and Differential Forms. Here is the exercise:

Let $f $ be the function $f \begin{pmatrix} x \\ y \\ \end{pmatrix}=sgn(y) \sqrt{ \frac {-x+\sqrt{x^2+y^2}}{2}} $ where $sgn(y)$ is the sign of y (i.e., $+1$ when $y>0$ , $0$ when $y=0$ ,$-1$ when $y<0$ )

a. show that $f$ is continuously differentiable on the complement of the half line $y=0, x\le 0$.

I've included part of the solution on the student solution manual to this exercise below:

To show that $f$ is continuously differentiable on the locus where $y=0,x>0$:

In a neighborhood of a point $\begin{pmatrix} x_0 \\ y_0 \\ \end{pmatrix}$satisfying $x_0>0 $ , $y_0=0$ , we can write
$-x+\sqrt{x^2+y^2}=-x+x\sqrt{1+\frac{y^2}{x^2}}= -x+x(1+\frac{1}{2}\frac{y^2}{x^2}+o(\frac{y^2}{x^2}))$
$ =\frac{y^2}{2x}+o(\frac{y^2}{x})$,
and since $sgn(y)y^2$ is of class $C^1$, the function is of class $C^1$ on the half axis $y=0, x> 0$.

I don't understand the bold part of the solution. Could you explain why the fact that $sgn(y)y^2$ is of class $C^1$ means that the function $f$ is continuously differentiable on the half-axis $y=0, x>0$ ?

(The book uses the following definition for little $o$ notation: a function $f$ is in $o(h)$ if $\lim \limits_{x \to 0}\frac{f(x)}{h(x)}=0.$ )

Answer 1

In order to better understand what's going on here introduce polar coordinates for the moment. Then $$x=r\cos\phi,\qquad\sqrt{x^2+y^2}=r$$ and therefore $${1\over2}\bigl(\sqrt{x^2+y^2}-x\bigr)=r{1-\cos\phi\over2}=r\sin^2{\phi\over2}\ .$$ It follows that $$\sqrt{{1\over2}\bigl(\sqrt{x^2+y^2}-x\bigr)}=\sqrt{r}\left|\sin{\phi\over2}\right|\ .$$ As $\ {\rm sgn}\bigl(\sin{\phi\over2}\bigr)={\rm sgn}(y)$ we therefore obtain $$f(x,y)=\sqrt{r}\sin{\phi\over2}={\root 4\of {x^2+y^2}}\ \sin{{\rm Arg}(x,y)\over2}\ .$$ As $$\nabla{\rm Arg}(x,y)=\left({-y\over x^2+y^2},{x\over x^2+y^2}\right)$$the function $f$ is even real analytic in the indicated domain.

Answer 2

Definitely strange. In fact, $f_y(x_0,y_0)$ does not exists for $x_0 > 0$, $y_0 = 0$:

First, ignore the irrelevant 2 and consider $$f(x,y) = \hbox{sgn}(y)\sqrt{\sqrt{x^2+y^2} - x}.$$ $$ \lim_{y\to 0}\frac{\sqrt{\sqrt{x_0^2+y^2} - x_0} - 0}{y - 0} = \lim_{y\to 0}\sqrt{\frac1{2x_0} + \frac{o(y^2/x_0)}{y^2}} = \frac1{\sqrt{2x_0}} ,$$ so the lateral limits of $$ \lim_{y\to 0}\frac{\hbox{sgn}(y)\sqrt{\sqrt{x_0^2+y^2} - x_0} - 0}{y - 0} $$ are different. Using $$ \sqrt{x_0^2+y^2} - x_0 = \frac{(\sqrt{x_0^2+y^2} - x_0)(\sqrt{x_0^2+y^2} + x_0)}{\sqrt{x_0^2+y^2} + x_0} = \frac{y^2}{\sqrt{x_0^2+y^2} + x_0} $$ gives the same result.