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In the last step of the proof, I have to show that $f:(x,y)\mapsto\sqrt{1-x^2-y^2}$ is a smooth mapping, where $x,y\in\mathbb{R}$. From direct computation with the usage of induction, I know that this map is $C^{\infty}$ if we view this map with respect to $x$ and $y$, and view $y$ and $x$ as a constant, respectively. But to show that $f_{xyxyxyyxyy}$ is continuous, the only thing I can think of is clairaut theorem, which I am uncertain I can use here. Getting my hands dirty for this function seems also like an impossible mission. What should I do? Is there any general theorem regarding this?

Edit: $x,y$ is in the disc, not the entire real.

able20
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1 Answers1

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Your function is defined for $x^2+y^2\leq 1$, which represents the unit disc (with boundary) centered at the origin of $\mathbb R^2$. Now note that $f = g\circ h$, where $h(x,y) = 1-x^2-y^2$ and $g\colon \mathbb R_{\geq 0} \to \mathbb R$ is defined by $g(x) = \sqrt{x}$. Showing that $h$ is smooth is easy, because it is polynomial. The map $g$ is smooth in the open domain $(0,+\infty)$, whereas for $x \to 0^+$ its first derivative explodes to $+\infty$. So your $f$ is a composition of smooth functions in the interior of your disc, and is thus smooth (see e.g. here). There is no way to prove smoothness on a bigger subset of $\mathbb R^2$ including your disc, because such a set would contain points on the boundary of the disc where the gradient of $f$ is not defined.

Gibbs
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