Is my determination of this maximum correct?

Question

Consider $\Omega:=B_1(0)\subset\mathbb{R}^n$ (it is the open unit ball), $\mathbb{R}^n$ is provided with the euclidean norm $\lVert\cdot\rVert_2$. Now I want to determine the following maximum: $$ \max\left\{\max_{x\in\partial\Omega}\left\{\lvert \sin^3(x_1)\rvert\right\},\sup_{x\in\Omega}\left\{\frac{\lvert\sum_{i=1}^{n}x_i^2\rvert}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert}\right\}\right\} $$

Here is my result:

$$ \max_{x\in\partial\Omega}\left\{\lvert \sin^3(x_1)\rvert\right\}=\sin^3(1)\approx0,596 $$ Now to the supremum $$ \sup_{x\in\Omega}\left\{\frac{\lvert\sum_{i=1}^{n}x_i^2\rvert}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert}\right\}: $$

I estimated as follows: $$ \frac{\lvert\sum_{i=1}^{n}x_i^2\rvert}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert}<\frac{\sum_{i=1}^{n}\lvert x_i\rvert^2}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert}\\<\frac{n}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert} $$

Because of $$ \sum_{i=1}^{n}\frac{1}{1+x_i^2}>\frac{n}{2} $$ on $B_1(0)$, it is $$ \left\lvert \frac{n}{2}-1-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\right\rvert >1 $$ and therefore $$ \frac{n}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert}<n, $$ so the supremum is $n$.

Because of $n\geq 1$ the searched maximum is $n$.

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Please tell me if I am right! Thank you very much!

Sincerely yours,

math12

Answer 1

Notice that: $$\sup\limits_{x\in\Omega}\;\dfrac{\lvert\sum_{i=1}^{n}x_i^2\rvert}{\lvert (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\rvert}=\sup\limits_{x\in\Omega}\;\left\{\dfrac{\sum_{i=1}^{n}x_i^2}{ (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}},-\dfrac{\sum_{i=1}^{n}x_i^2}{ (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}}\right\}$$ Moreover, fix $j$ and let $f_j(x_j)=\dfrac{\sum_{i=1}^{n}x_i^2}{ (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}}$ and so the first-order condition for an interior maximum is: $$f_j^\prime(x_j)=\dfrac{2x_j\cdot\left( (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\right)-\left(\sum_{i=1}^n x_i^2\right)\cdot\left(\dfrac{2x_j}{\left(1+x_j^2\right)^2}\right)}{\left( (\frac{n}{2}-1)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}\right)^2}=0\quad \forall j\tag{FOC}$$ The "symmetry" of the problem suggests (or you can prove if $x_i>0$ and $x_j>0$) that $x_i=x_j$ for all $i$ and $j$ at the solution. So the FOC becomes: $$2x\left(\left(\frac{n}{2}-1\right)-\frac{n}{1+x^2}\right)-\dfrac{2x^3}{(1+x^2)^2}=0\Rightarrow x^2=\dfrac{4+\sqrt{n^2+12}}{n-2}.$$ Maybe you can solve from here? Here is a list of what is left to do:

1. Show that $n\cdot \dfrac{4+\sqrt{n^2+12}}{n-2}\le 1$ so our candidate to a solution is on the unit ball.
2. Solve the case $n=2$ aside. The above works only for $n\ge 3$.
3. If a point in the boundary say $z=(z_i)$ (I ignored the boundary so far) is optimal then $$f_i^\prime(z_i)=f_j^\prime(z_j)\ge 0\text{ for all } i \text { and } j \text{ such that } z_i>0 \text { and } z_j>0 \text{ and } f_k^\prime(z_k)\le 0 \text{ if } z_k=0.$$
4. If there is a point in the boundary that is optimal then I think none of its coordinates can be zero. But in this case all the $x_i^2$ will also be equal, you just have to solve $f_i^\prime(z_i)=\lambda>0$ for all $i$. But before one must prove that no point can be optimal if the square of the coordinates is not the same.
5. Notice (for my work above) that the FOC for $f_j(x_k)$ and the FOC for $-f_k(x_k)$ are the same. So I don't have to consider two cases. Only after you compute the $x_i^2$ do you have to figure out the sign of $\left(\frac{n}{2}-1\right)-\sum_{i=1}^{n}\frac{1}{1+x_i^2}$.