Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f_i : R \rightarrow R$ and $g_j: R \rightarrow R$ be unknown functions, for $i = 1, \cdots, N$ and $j = 1, \cdots, K$. Let $A$ be a $K \times N$ matrix whose columns are unit-length vectors ${\mathbf{a}}_i$. Consider the functional equation $\displaystyle\sum_i f_i ({\mathbf{a}}_i^\top {\mathbf{x}} ) = \displaystyle\sum_j g_j(x_j)$, where ${\mathbf{x}} = [x_1, \cdots, x_K]^\top$.

For some $r > 0$, when the domain of ${\mathbf{x}}$ contains the set $B$, where $B = ( \mathbf{x} : |x_i| < r, i = 1, \cdots, K )$, it is known that when ${\mathbf{a}}_i$ are linearly independent, with at least two non-zero components in each ${\mathbf{a}}_i$, then the functions $f_i$ and $g_j$ must all be quadratic polynomials. This is proved in, for example, this paper: http://www.jstor.org/stable/25049527 .

My question is: are similar results known for functional equations of the above type when the ${\mathbf{x}}$ are restricted to be on the unit sphere, that is, $||{\mathbf{x}}|| = 1$?

share|improve this question
add comment

1 Answer

How does the statement of the problem rule out the following: Let N=K, let $a_i$ be the ith unit vector, so $a_i^Tx=x_i$, and then choose $f_i=g_i$, with the functions otherwise arbitrary?

share|improve this answer
    
Hi Michael, thanks for your input. The condition in the linked paper actually mentions that each a_i must have at least two non-zero components. I have added that to the question now. But in general, please feel free to assume any reasonable conditions that produce "non-trivial" conditions on f and g. –  user11443 Feb 9 '11 at 2:07
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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