If $f:S\to \Bbb R$ be defined by $f(x,y,z)=x+2y+3z$ where $S=\{(x,y,z):x^2+y^2+z^2 \le 1\}$.
Show that $f$ can't attain it's minimum in the interior of $A$.
Also show that $\min f=-\max f$.
I though that this maybe a special case of maximum modulus theorem But in that case we require f to be analytic .
I don't know how to show that f is analytic in 3 dimensions.
On
(1). If $f$ had a local max or min at $p=(x',y',z')\in $int($S$) then the partial derivatives of $f$ with respect to the co-ordinates $x,y,z$ would all vanish at $p,$ but the partials are $1,2,3.$
From a more basic view, if $p=(x',y',z')\in $ int ($S)$ then for some $r>0$ we have $(x'-r,x'+r)\times(y'-r,y'+r)\times(z'-r,z'+r)\subset $ int ($S$). Then $\forall s\in (0,r)\;(f(x',y',z'-s)<f(p)<f(x,y',z'+s)$, so $p$ cannot be a local extremum for $f.$
(2). $\{-p: p\in S\}=S.$ And $f(-p)=-f(p)$ for all $p\in S.$ Let $A= \{f(p):p\in S\}.$ Then $A=\{f(-p):p\in S\}=\{-f(p):p\in S\}.$ For any $A\subset \Bbb R,$ if $A=\{-a:a\in A\}$ and if $\min A$ exists, then $\max A=-\min A.$
On
Here is simple answer that requires no theorem at all. If $f$ attains minimum at a point $(a,b,c)$ with $a^{2}+b^{2}+c^{2}<1$ then we can replace $a$ by a slightly smaller number to get a value of $f$ lower than the value at $(a,b,c)$, which is a contradiction. Next, note that $(x,y,z) \in S$ implies $(-x,-y,-z) \in S$ and $f(-x,-y,-z)=-f(x,y,z)$. Hence the range of $f$ is a set $A$ of real numbers which is symmetric in the sense $-t \in A$ whenever $t \in A$. For any such set $\min A =- \max A$.
You actually want to use that $f$ is harmonic, which means $\Delta f =0$. Harmonic functions share many properties with analytic functions, including the maximum modulus principle. So, I would advise you to study up on that.
The other part of the problem follows from symmetry.