Jaynes (2003) writes on p. 220:
Any smooth function with a single rounded maximum, if raised to higher and higher powers, goes into a Gaussian function.
Here I presume that he means that the function is also non-negative, as this is stated in the context of probability theory. (Otherwise the power operation might cause sign-flipping and the location of the maximum could depend on the evenness of the integer power.)
Assume $f(x)$ has a single rounded maximum at $x=0$ and $f(x) \geq 0$. We expand $g(x) = \log f(x)^\alpha$ around the maximum:
$$ \log f(x)^\alpha = \alpha \log f(0) + \sum_{n=2}^\infty g^{(n)}(0) \frac{x^n}{n!} \label{a}$$
Now if the terms for which $n \geq 3$ in this expansion would become smaller for increasing $\alpha$, we have local Gaussian behavior around the maximum and the quote is proved. (At least for my purposes.)
However, I do not find this behavior. The expansion coefficients up to fourth order are: \begin{align} g^{(1)}(0) &= 0 \\ g^{(2)}(0) &= \alpha \frac{f^{(2)}(0)}{f(0)} < 0\\ g^{(3)}(0) &= \alpha \frac{f^{(3)}(0)}{f(0)} \\ g^{(4)}(0) &= \alpha \frac{f^{(4)}(0) + [f^{(2)}(0)]^2}{f(0)^2} \end{align} I am unsure whether the qualifier "rounded" (from "rounded maximum") might imply something here.
My question: How can one (dis)prove the quote? Can one say something about $f(x)^\alpha$ far away from the maximum?
Edit: Intuitively, I feel this statements should hold in many useful cases. When raising $f(x)$ to higher powers, the peak not only gets higher but also thinner. Therefore, in the Taylor expansion around the maximum the deviation from zero, i.e. $x$, lives on a smaller and smaller scale for each increment in the power $\alpha$. This will suppress the higher order terms in the Taylor expansion, not by virtue of the coefficients $g^{(n)}(0)$ but instead by multiplication with $x^n$. This is more or less what happens in the "Concepts" subsection of Wiki's General theory of Laplace's method.