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Define $g,h:\{x\in \mathbb{R}^2: ||x|| \leq 1\} \rightarrow \ \mathbb{R}^3$ by: $$g(x,y)=(x,y,\sqrt{1-x^2-y^2}) $$ $$h(x,y)=(x,y,-\sqrt{1-x^2-y^2}) $$ 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 $\{x\in \mathbb{R}^2: ||x||\leq1\}$

My attempt:

I am not really sure how to start here except I know that:

If $A \subset\mathbb{R}^n$ then the max (or min) of $f: A \rightarrow \mathbb{R}^n$ occurs at a point $a$ in the interior of $A$ and $D_if(a)=0$ (if it exists).

If we let $A=\{x\in \mathbb{R}^2: ||x|| \leq 1\}$ then $int(A)=||x|| < 1$

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    You're making this too hard. Assuming $f$ is continuous, by the maximum value theorem it must have a maximum on the unit sphere $S=\{x:\|x\|=1\}$. $g$ and $h$ give parametrizations of the two closed hemispheres, whose union is $S$. Now proceed. – 2017-02-07
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    You mean "$f g$ or $f h$ on $\{x\in R^2:\|x\|=1\}$" not "on $\{x\in R^3:\|x\|\leq1\}$". – 2017-02-07
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    @TedShifrin so the maximum must lie in either one of those 2 hemispheres which is f on g or f on h? – 2017-02-07
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    @user25466524: You misread the question. – 2017-02-07

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