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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?

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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. –  Charles Matthews Jan 18 '11 at 11:52
    
In the definition, do you want $t\in\mathbb{R}$ or $t\in\mathbb{R}_+$? –  Willie Wong Jan 18 '11 at 15:20
    
@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 $\lambda\in\mathbb{R}_{+}$. –  Patrick Jan 19 '11 at 15:23

2 Answers 2

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.

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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.

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