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This is a recent qualifying exam question from Indiana University.

Let $$X=\{(x_1,x_2,x_3,x_4)\in\mathbb{R}^4: x_1^4+x_2^4+x_3^4+x_4^4=64\ \ \text{and}\ \ x_1+x_2+x_3+x_4=8 \}.$$ For which points $p\in X$ is it possible to find a product of open intervals $V=I_1\times I_2 \times I_3 \times I_4$ containing $p$ such that $X\cap V$ is the graph of a function expressing two of the variables $x_1,x_2,x_3,x_4$ in terms of the other two? If there are any two points in $X$ where this is not possible, explain why not.

Is is clear that we can apply Implicit function theorem on all points in $\mathbb{R}^4$ except possible at $(2,2,2,2)$ where the Jacobian is singular. My question is that how do we know we really can't solve two of the variables in terms of another two at this point. The conclusion of implicit function theorem is merely sufficient and not necessary. Thanks for the help.

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Are you sure that $(2,2,2,2)$ is the only troublemaker here? I'm getting a negative vibe from $(0,0,0,0)$ as well. – user53153 Jan 2 '13 at 2:03
but (0,0,0,0) does not solved the equation. – KWO Jan 2 '13 at 2:04
All points of the form (a,a,a,a) will made the Jacobian matrix singular but only (2,2,2,2) satisfied the equations. – KWO Jan 2 '13 at 2:05
I think that $X=\{(2,2,2,2)\}$, hence the answer is at no points... – copper.hat Jan 2 '13 at 3:36
up vote 5 down vote accepted

As copper.hat wrote in comments, $X=\{(2,2,2,2)\}$. Indeed, the function $f(x) = x^4$ is strictly convex. Therefore, $$x_1^4 + \dots + x_4^4 \geq 4\left(\frac{x_1 + \dots +x_4}{4}\right)^4,$$ and the inequality is strict unless $x_1 = x_2 = x_3 = x_4$. Since for $x\in X$, $x_1^4 + \dots + x_4^4 = 64$ and $4\left(\frac{x_1 + \dots +x_4}{4}\right)^4 = 64$, we conclude that $x_1 = x_2 = x_3 = x_4 = 2$.

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+1 Nice. Much simpler than my reasoning. – copper.hat Jan 2 '13 at 4:08
Thanks for the solution. – KWO Jan 2 '13 at 9:56

To simplify life, let $y_i = \frac{1}{\sqrt{8}} x_i$, then the equations become $\sum y_i^4 = 1$, $\sum y_i = \sqrt{8}$. A standard $p$-norm inequality gives (in $\mathbb{R}^4$) $\|x\|_1 \leq \sqrt{8}\|x\|_4$, hence it is clear that $\sum y_i \leq \sqrt{8}$ whenever $\sum y_i^4 = 1$. It follows that the strictly convex $4$-ball $B_4(0,1)$ lies on one side of the hyperplane $\sum y_i = \sqrt{8}$. Since $y=\frac{1}{\sqrt{2}}(1,1,1,1)$ lies in the intersection and the $4$-ball is strictly convex, it is the only point that satisfies the equations. Since $\frac{1}{\sqrt{8}}X$ lies in the intersection of the hyperplane and the ball, we have $X=\{(2,2,2,2)\}$.

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Thanks for the solution. – KWO Jan 2 '13 at 9:55

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