3
$\begingroup$

Let $\mathbb{R}^n$ be the $n$-dimensional real vector space with Cartesian coordinates $x=(x^1,\ldots, x^n)\in \mathbb{R}^n$. I'm searching for a non-trivial example of a function $A:\mathbb{R}^n \rightarrow \mathbb{R}$, which is continuously differentiable on $\mathbb{R}^n\backslash \lbrace 0\rbrace$ (i.e. $A\in C^\infty(\mathbb{R}^n\backslash \lbrace 0\rbrace)$), such that it satisfies \begin{align} A(tx)&=tA(x),\newline \sum^{n}_{i,j=1}\xi^i A(x)\frac{\partial^2 A(x)}{\partial x^i \partial x^j} \xi^j &\le 0, \end{align} for all $t\in \mathbb{R}$, $x, \xi \in \mathbb{R}^n$.

Obviously, if $A$ is chosen to be a linear function, it is a solution to the problem above. But are there any non-linear solutions for dimension $n\ge 3$? Is there some topological argument against the existence of such solutions?

$\endgroup$
3
  • $\begingroup$ Can you phrase what you want in terms of real projective space of dimension one less? The Hessian would need to be "reduced" but there is one degree of freedom that can be removed, e.g. by means of Euler's theorem on homogeneous functions. The topology of the projective space is of course well known, and given that it has only torsion in homology in the middle dimensions, it isn't immediately apparent that there is an obstruction. $\endgroup$ Commented Jan 18, 2011 at 11:52
  • $\begingroup$ In the definition, do you want $t\in\mathbb{R}$ or $t\in\mathbb{R}_+$? $\endgroup$ Commented Jan 18, 2011 at 15:20
  • $\begingroup$ @Charles: I have already tried to rewrite it as a problem on the unit-sphere (for $m=3$) to get rid of the homogenity condition, so I think it would also work for the real projective space. In the unit sphere case, I used a polarcoordinate representation. But the resulting polar coordinate version of the differential inequality condition does not have a nice form. @Willie: I do actually want $t\in \mathbb{R}$. So $A$ is an function for which holds \begin{align*} A(-x)&=-A(x)\\ A(\lambda x)&= \lambda A(x) \end{align*} for all $x\in\mathbb{R}^n$ and $\lambda\in\mathbb{R}_{+}$. $\endgroup$
    – Patrick
    Commented Jan 19, 2011 at 15:23

2 Answers 2

3
$\begingroup$

You can define $A(x_1,x_2,\dots,x_n)=B(x_1,x_2)$ where $B$ is a solution for $n=2$. As for $B$, one can e.g. define it on the unit circle by $B(\sin t,\cos t)=\cos 3t$ and extend by homogeneity.

$\endgroup$
1
$\begingroup$

Take $A(x)=\vert x\vert$ the Euclidean norm in $\mathbb R^n$, which is obviously homogeneous of degree 1. We have $$ A'(x)=\frac{x}{A(x)},\quad A''(x)=\frac{Id}{A(x)}-\frac{x\otimes x}{A(x)^3}=A(x)^{-1} \underbrace{\bigl(Id-\frac{x\otimes x}{\vert x\vert^2}\bigr)}_{L(x)}.$$ The matrix $L(x)$ is the matrix of the projection onto the orthogonal of $x$ and is thus a non-negative matrix.

That example does not answer your question, but nevertheless gives an example of an homogeneous $A$ with degree 1 such that $$ A(x) A''(x)\ge 0, $$ which has also the linear forms as solutions.

Bazin.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.