Fourier spectrum of $\sqrt{1 + f(x)} - \sqrt{1- f(x)}$ where $f(x): \{0, 1\}^n \to [-1,1]$ Consider a function $f(x): \{0, 1\}^{n} \to [-1,1]$ defined over the Boolean cube $\{0,1\}^n$ mapping to the interval $[-1,1]$. The Fourier transform, or more precisely Walsh-Hadamard transform, of $f(x)$ is given by
$$\hat{f}(s) = \frac{1}{2^n} \sum_{x\in \{0,1\}^n} (-1)^{s\cdot x} f(x)$$
Now, consider the function $g(x)=\sqrt{1 + f(x)} - \sqrt{1- f(x)}$ where $g(x):\{0,1\}^n \to [-\sqrt{2}, \sqrt{2}]$ and its Fourier transform
$$\hat{g}(s) = \frac{1}{2^n} \sum_{x\in \{0,1\}^n} (-1)^{s\cdot x} \left(\sqrt{1 + f(x)} - \sqrt{1- f(x)}\right)$$
I want to understand if (or under which conditions) the spectra $\hat{f}(s) $ and $\hat{g}(s) $ are related.
In particular, I would like to know if

*

*the ordering in terms of absolute value is the same for $f$ and $g$ so that
$$|\hat{f}(s)| \geq |\hat{f}(s')| \Rightarrow |\hat{g}(s)| \geq |\hat{g}(s')|$$ for all pairs $s,s' \in \{0,1\}^n$

*or if at least the largest Fourier coefficient is the same: $$\mathrm{argmax}_s |\hat{f}(s)| = \mathrm{argmax}_s |\hat{g}(s)| $$
As a first observation, I looked at the series expansion of $g(x)$ in terms of $f(x)$, and find
$$g(x) = f(x) + \frac{1}{3}[f(x)]^3+ \frac{7}{128}[f(x)]^5 + \dots$$
and so hand-wavyly
$$ \hat{g}(s) = \hat{f}(s) + \frac{1}{2^n} \sum_{x\in \{0,1\}^n} (-1)^{s\cdot x} \left( \frac{1}{3}[f(x)]^3 + \frac{7}{128}[f(x)]^5 + \dots \right)$$
One also has, in particular,
$$f(x) > 0 \Rightarrow g(x)> 0$$.
Not sure how to proceed though to answer the above questions 1) and/or 2) based on this. Any ideas would be greatly appreciated.
 A: Here is a partial result for the problem, showing that 1) and 2) hold for $n = 1, 2$.
Let $h(t) \triangleq \sqrt{1 + t} - \sqrt{1 - t}$ for $t \in [-1, 1]$. Then we have $g(x) = h(f(x))$. For simplicity, let $u_x \triangleq f(x)$.
If $n = 1$, we have $$\hat{f}(0) = \frac{u_0 + u_1}{2}, \quad \hat{f}(1) = \frac{u_0 - u_1}{2}.$$
Similarly,
$$\hat{g}(0) = \frac{h(u_0) + h(u_1)}{2}, \quad \hat{g}(1) = \frac{h(u_0) - h(u_1)}{2}.$$
Therefore, we obtain
$$|\hat{f}(0)|^2 - |\hat{f}(1)|^2 = u_0 u_1, \quad |\hat{g}(0)|^2 - |\hat{g}(1)|^2 = h(u_0) \cdot h(u_1).$$
Note that $h(x)$ is odd, i.e., $h(-x) = -h(x)$, and $h(x) \geq 0$ when $x \geq 0$. Hence, we have $x \cdot h(x) \geq 0$. This implies that
$$h(u_0)\cdot h(u_1) \geq 0 \iff u_0u_1 \geq 0,$$
which gives 1) [and thus 2)].
We can prove the same result for $n = 2$. Take $s = (0, 0), s' = (1, 0)$ as an example, from
$$
\hat{f}(0, 0) = \frac14 (u_{00} + u_{01} + u_{10} + u_{11})\\
\hat{f}(1, 0) = \frac14 (u_{00} + u_{01} - u_{10} - u_{11})
$$
we obtain
$$
|\hat{f}(0, 0)|^2 - |\hat{f}(1, 0)|^2 = \frac14 \cdot (u_{00} + u_{01})(u_{10} + u_{11})
$$
and
$$
|\hat{g}(0, 0)|^2 - |\hat{g}(1, 0)|^2 = \frac14 \cdot (h(u_{00}) + h(u_{01}))\cdot (h(u_{10}) + h(u_{11})).
$$
It can be shown that
$$h(t_1) + h(t_2) \geq 0 \iff t_1 + t_2 \geq 0.$$
Hence,
$$|\hat{f}(0, 0)| \geq |\hat{f}(1, 0)|
\iff |\hat{g}(0, 0)| \geq |\hat{g}(1, 0)|.
$$
