Uniqueness of solution of functional equation

I have a function $f(x_1,x_2) \colon \mathbb{R}^2_{+} \to \mathbb{R}_{+}$ positive homogenous: $$f(\lambda x_1, \lambda x_2) = \lambda f(x_1,x_2), \; \forall \lambda > 0$$ and such that $f(x_1,x_2)$ permits decomposition $$f(x_1,x_2) = h(g(x_1)+g(x_2))$$ where $h,g$ are some continuous functions. One of appropriate functions is $$f_{0}(x_1,x_2) = C(x_1^{\gamma}+x_{2}^{\gamma})^{\frac{1}{\gamma}}, \; \gamma > 0, \; C \geq 0$$ Are there some other functions that satisfy specified conditions or $f_{0}(x_1,x_2)$ is unique?

Update If $h = g^{-1}$ then $f_{0}$ is a unique family of solutions (Hardy, G. H.; Littlewood, J. E.; Pólya, G. (1952) Inequalities. 2nd ed, page 68).

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Well, $f_0$ isn't unique, because you can choose $C$ and $\gamma$. – Stefan Nov 21 '12 at 10:03

The condition $f( \lambda x_1, \lambda x_2) = \lambda f(x_1, x_2)$ for $\lambda > 0$ for $x_1, x_2 > 0$ means $f$ is uniquely determined by its restriction to $S$, the part of the unit circle in the first quadrant, i.e. $\{(\cos \theta, \sin \theta): \theta \in (0, \frac{\pi}{2}) \}$. Working in polar coordinates, it is easy to see that $f$ is smooth iff it's restriction to $S$ is.
With that, on $S$, the second condition is $f = h(g(\cos \theta) + g(\sin \theta))$, for any choice of continuous $h,g$.
It suffices to find such an $f$ which is not smooth. Indeed, one should be able to do it by taking $g = id$, note the map $(0, \frac{\pi}{2}) \rightarrow \mathbb{R}$ given by $\theta \mapsto \sin \theta + \cos \theta$ is a local diffeomorphism for $\theta \neq \frac{\pi}{4}$, so it suffices to take $h$ to be any continuous function $\mathbb{R} \rightarrow \mathbb{R}$ which isn't differentiable at say the image of $\frac{\pi}{8}$.
Note that $g$ cannot be $x\mapsto ax$ and $x \mapsto bx$ at the same time, unless $a = b$, which makes your $f$ equal to $Ca\bigl(x_1^\gamma + x_2^\gamma)^{1/\gamma}$. So, nothing new. – martini Nov 21 '12 at 9:39