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Suppose a positive function $M$ satisfies the displayed fourth-power identity and $\log M(h)=h^2/2+o(h^2)$ at zero. Iteration gives $\log M(t)=4^m\log M(t/2^m)$, which tends to $t^2/2$. A related first-power symmetry step is: if $\psi (t)=\psi (t/2{)}^2$ and $\psi (h)=1+o(h^2)$, then $\log \psi (t)=2^m\log \psi (t/2^m)\to 0$. These two different dyadic exponents must not be interchanged in the Gaussian characterization by independent sum and difference.

Let $S=X+Y$ and $D=X-Y$. Since they are independent, applying the product rule for moment-generating functions to $S+D=2X$ givesAll factors are finite and strictly positive for every real $t$ by the stated hypothesis. Applying the same identity to $-t$ and dividing yields $\psi (2t)=\psi (t{)}^2$, orThe centered unit-variance Taylor expansion from part (ii) gives $M(h)=1+h^2/2+o(h^2)$ and the same expansion for $M(-h)$. Consequently $\psi (h)=1+o(h^2)$.

For the first step of dyadic rigidity of a moment-generating function, take logarithms, which are allowed because $\psi >0$. For each fixed $t$ and every integer $m$,Since $\log \psi (h)=o(h^2)$, the right side tends to zero: it is $2^{m}o(t^2/4^m)$. Thus $\psi (t)=1$ for every $t$, and the moment-generating function is even.

The original identity now becomes $M(2t)=M(t{)}^4$. Iterating this time with the correct fourth-power scaling givesNear zero, $\log M(h)=h^2/2+o(h^2)$, so the right side tends to $t^2/2$. ThereforeThis is the moment-generating function of a standard normal variable. By the uniqueness theorem for moment-generating functions, each of $X$ and $Y$ has the standard normal distribution. The proof establishes the Gaussian characterization by independent sum and difference directly from the two functional identities and the first two moments.