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I've attempted a problem in Spivak's Calculus on Manifolds and I can't find a solution online. This is question 2-27, which is as follows

Define $g,h:\{x\in\mathbb{R}^2:|x|\le1\}\rightarrow\mathbb{R}^3$ by $$\begin{split} g(x,y)&=(x,y,\sqrt{1-x^2-y^2})\\ h(x,y)&=(x,y,-\sqrt{1-x^2-y^2}) \end{split}$$ Show that the maximum of $f$ on $\{x\in\mathbb{R}^3:|x|=1\}$ is either the maximum of $f\circ g$ or the maximum of $f\circ h$ on $\{x\in\mathbb{R}^2:|x|\le1\}$.

My solution is as follws

Assume that the maximum of $f$ occurs at some point $g(a)\in\mathbb{R}^3$ where $a\in\{x\in\mathbb{R}^2:|x|<1\}$, and that $D_i\left(f\circ g\right)(a)$ exists. Then by the chain rule $$D_i(f\circ g)(a)=D_if(g(a))\cdot D_ig(a)$$ since the maximum of $f$ appears at $g(a)$, by theorem 2-6 $D_if(g(a))=0$, and so $D_i(f\circ g)(a)=0$. This shows that the maximum of $f$ occurs at the maximum of $f\circ g$.

I have some issues with my solution, mainly that since the theorem only works in the interior so for points $|x|=1$ where $x\in\mathbb{R}^2$ the conclusion I've drawn doesn't hold on the boundary. However, since it would seem that $f=g$ on the boundary it might not count?

Also I'm not sure whether I've essentially done "$D_if(b)=0$ so the maximum is at $b$", but I've tried to word it as similarly as the theorem to check that I haven't.

Redcrazyguy
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1 Answers1

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There is no need to take derivatives. Actually, in the statement, there are no assumptions about the regularity of $f$.

All you need to show is that $$g(D)\cup h(D)=S$$ where $D=\{x\in\mathbb{R}^2:|x|\le1\}$ and $S=\{x\in\mathbb{R}^3:|x|=1\}$.

Then the claim follows because $$\{f(x): x\in S\}=\{f(g(x)): x\in D\}\cup\{f(h(x)): x\in D\},$$ and, in general, $\sup(X\cup Y)$ is the largest one between $\sup(X)$ and $\sup(Y)$ (see for instance Let $A,B \subset \mathbb{R}$. Show that $\sup(A \cup B) = \max\{\sup A, \sup B\}$) .

Robert Z
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  • So are you saying that essentially the proof is just that $g$ and $h$ map $D$ onto the domain of $f$, so the maximum of $f$ must be a maximum of the mapping from $R^2$ to $R^3$ followed by $f$? I'm surprised that you don't need to take the derivative because Spivak says this is supposed to show how to find the maximum on the boundary of a domain, since the theorem he gives in the text applies only to the interior. – Redcrazyguy Dec 08 '22 at 09:46
  • The maps $g$ and $h$ transfer the problem from $S={x\in\mathbb{R}^3:|x|=1}$, the boundary of the domain $B={x\in\mathbb{R}^3:|x|\leq 1}$, into the unit disc $D={x\in\mathbb{R}^2:|x|\le1}$ (which has a non empty interior). – Robert Z Dec 08 '22 at 09:53
  • It seems to me that when $|x|=1$ that is a boundary on both the sphere and the disc, how would that work? 2. So the method is basically to map it onto the disc then use conventional methods to find the max/min?
    • – Redcrazyguy Dec 08 '22 at 09:56
  • For the "equator" you can do a similar thing by considering $g_1(x,y)=(x,\sqrt{1-x^2})$ and $h_1(x,y)=(x,-\sqrt{1-x^2})$ for $x=[-1,1]$. Yes for 2).
    • – Robert Z Dec 08 '22 at 09:58