# Thread: Showing a piecewise function is smooth

1. ## Showing a piecewise function is smooth

I have a function,

$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 $g$ and $h$ are smooth. Does this imply that $f$ is smooth? If so, why?

2. Originally Posted by Swlabr
I have a function,

$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 $g$ and $h$ are smooth. Does this imply that $f$ 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

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

would that help?...

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

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

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

CB

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

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

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

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

CB

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

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

CB
Hmm...I had forgotten about the conditions for l'Hopital's rule!

I understand now. That is quite annoying!

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