One way to characterize the Gaussian $ae^{b x^2}$ is that its a $C^1$ function $h$ that is radial $h(x,y) = h(\sqrt{x^2+y^2})$ and also separable, that is expressible as a product of one-dimensional functions $h(x,y)=f(x)f(y)$.
How can I show this?
I can write $h(x,y) = f(x)f(y) = f(r)f(0)$. If $h \in C^2$ then I can use the Laplacian to $\Delta h = \partial^2_r h + \frac 1 r \partial_r h$ to get $f''(x)f(y) + f(x)f''(y) = f(0) f''(r) + \frac 1 r f(0)f'(r)$. At the point $y=0$ this becomes $f(0)f'(x) = x f(x) f''(0)$, an ODE whose solution is exactly what I want.
But is there a simpler argument which uses only the first derivative, or perhaps an integral? My textbook suggests the $C^2$ assumption isn't needed here
Let's just take $\mathbf{x} \in \mathbb{R}^n$ with $h: \mathbb{R}^n \mapsto \mathbb{R}$. Since $h$ is separable, $$ h(\mathbf{x}) = \prod_jf(x_j) \tag{1} $$ and since $h$ is radially symmetric, $$ h(\mathbf{x}) = g(||\mathbf{x}||^2) \tag{2} $$ for some $f: \mathbb{R} \mapsto \mathbb{R}$, and $g: \mathbb{R} \mapsto \mathbb{R}$.
Observe the partial derivative using $(1)$: $$ h_{x_j} = \frac{f'(x_j)}{f(x_j)}h(\mathbf{x}) $$ and using $(2)$: $$ h_{x_j} = g'(||\mathbf{x}||^2)2x_j $$
Now equate the two: $$ \frac{f'(x_j)}{f(x_j)}h(\mathbf{x}) = g'(||\mathbf{x}||^2)2x_j $$ or... $$ \frac{f'(x_j)}{2x_jf(x_j)} = \frac{g'(||\mathbf{x}||^2)}{g(||\mathbf{x}||^2)} $$ Since this must hold for all $j$, the above must equal to a constant. This yields the result.