# Thread: Showing a piecewise function is smooth

1. ## Showing a piecewise function is smooth

I have a function,

$\displaystyle f:= \left\{ \begin{array}{lr} (g(x, y, z), h(x, y, z)) & x \neq 0 \text{ or } y \neq 0 \\ (0, 0) & x=y=0 \end{array} \right.$

and I want to show that it is smooth. I know $\displaystyle g$ and $\displaystyle h$ are smooth. Does this imply that $\displaystyle f$ is smooth? If so, why?

2. Originally Posted by Swlabr
I have a function,

$\displaystyle f:= \left\{ \begin{array}{lr} (g(x, y, z), h(x, y, z)) & x \neq 0 \text{ or } y \neq 0 \\ (0, 0) & x=y=0 \end{array} \right.$

and I want to show that it is smooth. I know $\displaystyle g$ and $\displaystyle h$ are smooth. Does this imply that $\displaystyle f$ is smooth? If so, why?
No, not without further conditions, for instance g(x,y,z)=1, h(x,y,z)=1 are smooth, but f(x,y,z) is not even continuous.

CB

3. Originally Posted by CaptainBlack
No, not without further conditions, for instance g(x,y,z)=1, h(x,y,z)=1 are smooth, but f(x,y,z) is not even continuous.

CB
Good point. If I told you the function was

$\displaystyle \left( \frac{2xz-y+yx^2+y^3+yz^2}{2(x^2+y^2)}, \frac{-2yz-x+x^3+xy^2+xz^2}{2(x^2+y^2)} \right)$

would that help?...

4. Originally Posted by Swlabr
Good point. If I told you the function was

$\displaystyle \left( \frac{2xz-y+yx^2+y^3+yz^2}{2(x^2+y^2)}, \frac{-2yz-x+x^3+xy^2+xz^2}{2(x^2+y^2)} \right)$

would that help?...
That does not look continuous to me, put $\displaystyle z=7$, and $\displaystyle x=y$ and consider the limit as $\displaystyle x \to 0$.

CB

5. Originally Posted by CaptainBlack
That does not look continuous to me, put $\displaystyle z=7$, and $\displaystyle x=y$ and consider the limit as $\displaystyle x \to 0$.

CB
Forgetting the $\displaystyle z=7$, we have $\displaystyle \left(\frac{z_1 + 2x^2}{4x}, \frac{z_2+2x^2}{4x}\right)$. If we apply l'Hopital's rule we surely get that this tends to $\displaystyle (0,0)$ as $\displaystyle x \rightarrow 0$. So that isn't a counter-example...

6. Originally Posted by Swlabr
Forgetting the $\displaystyle z=7$, we have $\displaystyle \left(\frac{z_1 + 2x^2}{4x}, \frac{z_2+2x^2}{4x}\right)$. If we apply l'Hopital's rule we surely get that this tends to $\displaystyle (0,0)$ as $\displaystyle x \rightarrow 0$. So that isn't a counter-example...
Coming in along a ray where $\displaystyle x=y$ the limit is $\displaystyle (\infty, \infty)$ for any value of $\displaystyle z \ne 0$. (and how come you now have two $\displaystyle z$'s?)

L'Hopital's rule is not applicable if $\displaystyle z \ne 0$ as the numerators are not zero.

CB

7. Originally Posted by CaptainBlack
Coming in along a ray where $\displaystyle x=y$ the limit is $\displaystyle (\infty, \infty)$ for any value of $\displaystyle z \ne 0$. (and how come you now have two $\displaystyle z$'s?)

L'Hopital's rule is not applicable if $\displaystyle z \ne 0$ as the numerators are not zero.

CB
(my $\displaystyle z_1$ and $\displaystyle z_2$ are functions of $\displaystyle z$ which I couldn't be bothered writing out).