Independence of Gaussian Distribution Function with Different Means

Question

Is there any way to prove that $N$ Gaussian distribution functions are linearly independent if and only if the means are different. For example, if f1(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\frac{-(x-\mu1)^2}{2\sigma^2} \vdots fN(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\frac{-(x-\muN)^2}{2\sigma^2} $f_1,...,f_N$ they are all linearly independent iff the means $(\mu_1, ... ,\mu_N)$ are not equal.

Answer 1

Assume without loss of generality that $(\mu_n)$ is decreasing. The simplest approach could be to divide any null linear combination $\sum\limits_na_nf_n=0$ by the gaussian density with parameters $(0,\sigma^2)$, yielding $$\sum_na_n\exp(x\mu_n-\mu_n^2/2)=0.$$ When $x\to\infty$, every exponential term except the first one is negligible with respect to the first one since $\exp(x\mu_1)\gg\exp(x\mu_n)$ for every $n\ne1$, hence $a_1=0$. Lather, rinse, repeat.