# Restriction of a linear functional equation to surface of a sphere

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, $||{\mathbf{x}}|| = 1$?

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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?

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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