# smooth function

#### rayman

show that

$$\displaystyle f(x,y,z)=\frac{1}{z^2+1}(x^2-y^2,2xy,2z)$$ defines a smooth function $$\displaystyle f: S^{2}\to S^{2}$$ where $$\displaystyle S^{2}=\{p\in \mathbb{R}^3, |p|=1\}$$ so I am guessing it is a unit sphere.

How do we proceed with such problem?

f is a smooth function if all partial derivatives of all possible orders are defined in all points of the domain of f right?

I started with computing these partial derivatives

$$\displaystyle \frac{\partial f}{\partial x}=\frac{1}{z^2+1}(2x,2y,0)$$

$$\displaystyle \frac{\partial f}{\partial y}=\frac{1}{z^2+1}(-2y,2x,0)$$

$$\displaystyle \frac{\partial f}{\partial z}=\frac{-2(z^2-1)}{(z^2+1)^2}$$

and these partial derivatives are all defined at any point from the domain of f

so I compute second order partial derivatives

$$\displaystyle \frac{\partial^2 f}{\partial x^2}=\frac{2}{z^2+1}$$

$$\displaystyle \frac{\partial^2 f}{\partial y^2}=\frac{-2}{z^2+1}$$

$$\displaystyle \frac{\partial^2 f}{\partial z^2}=\frac{4z(z^2-3)}{(z^2+1)^3}$$

and these 3 are also defined for all the points of the domain.

if I continue differentiating

$$\displaystyle \frac{\partial^3 f}{\partial x^3}=\frac{\partial^3 f}{\partial y^3}=0$$ and

$$\displaystyle \frac{\partial^3 f}{\partial z^3}=-\frac{12(z^4-6z^2+1)}{(z^2+1)^4}$$ which is defined for all points

What should I do next? can I conclude based on the calculations that f is certainly smooth?

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#### xxp9

f is obviously smooth on R^3. So what you need to show is, when f is restricted to S^2, its image is in S^2.
That is, to show that [(x^2-y^2)^2 + (2xy)^2 + (2z)^2]/(1+z^2)^2 = 1, given that x^2+y^2+z^2=1.
Which is easy to show.

#### rayman

so showing all the computations above were not necessary if I am supposed to restrict myself to $$\displaystyle S^{2}$$ right?
But I still don't get how to show $$\displaystyle \Biggl[(x^2-y^2)^2+(2xy)^2+(2z)^2\Biggr]\frac{1}{(z^2+1)^2}=1$$ how does this imply that f will be smooth on the sphere???

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#### xxp9

$$\displaystyle (x^2-y^2)^2 + (2xy)^2 + (2z)^2=(x^2+y^2)^2 + (2z)^2$$
=$$\displaystyle (1-z^2)^2+(2z)^2=(1+z^2)^2$$

This shows that $$\displaystyle f|_{S^2}$$ indeed maps to $$\displaystyle S^2$$.

Then $$\displaystyle f$$ is smooth because "the restriction of a smooth function on a smooth sub-manifold is smooth".
To prove the above statement, let $$\displaystyle i: S^2 \rightarrow R^3$$ is the smooth embedding of $$\displaystyle S^2$$ in $$\displaystyle R^3$$,
$$\displaystyle f : R^3 \rightarrow S^2$$ is a smooth map on $$\displaystyle R^3$$, then $$\displaystyle f|_{S^2} = f \circ i$$ is smooth on $$\displaystyle S^2$$.