Showing a piecewise function is smooth

**Swlabr** · March 16th 2010, 02:56 AM

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?

**CaptainBlack** · March 16th 2010, 03:20 AM

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

**Swlabr** · March 16th 2010, 05:51 AM

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?...

**CaptainBlack** · March 16th 2010, 06:21 AM

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

**Swlabr** · March 16th 2010, 06:55 AM

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...

**CaptainBlack** · March 16th 2010, 08:53 AM

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

**Swlabr** · March 16th 2010, 09:12 AM

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).

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