A characteristic function that is Gaussian near zero is globally Gaussian

By [F9], substitution $x=\sqrt{2}u$ in [F6] gives $\int g(x)dx=1$, so $g$ defines a probability law $\gamma$. These integrals can be read as Lebesgue integrals: on each $[-m,m]$, the uniform dyadic lower and upper sums of the continuous integrand converge to the same Riemann integral and bound its Lebesgue integral; increasing truncations then give the whole-line equality by [F3]. For each integer $k\ge 0$, $\mathrm{\mid }x{\mathrm{\mid }}^{k}g(x)$ is integrable: its tail is bounded by a constant times $e^{-x^2\mathrm{/}4}$, since $\mathrm{\mid }x{\mathrm{\mid }}^k{e}^{-x^2\mathrm{/}4}$ is bounded (differentiate it on $x>0$). If $G$ has law $\gamma$, integration by parts [F8] on $[-m,m]$, followed by [F3], using $g^{\mathrm{\prime }}(x)=-xg(x)$ and vanishing polynomial times Gaussian boundary terms, gives $\mathbb{E}{G}^{2n}=(2n-1)\mathbb{E}{G}^{2n-2}$ and odd moments zero. Hence $m_{2n}:=\mathbb{E}{G}^{2n}=(2n)!\mathrm{/}(2^{n}n!)$. Also ${\phi }_G^{\mathrm{\prime }}(t)=\int ix{e}^{itx}g(x)dx=-t{\phi }_G(t)$ by the same integration by parts, justified by $\int \mathrm{\mid }x\mathrm{\mid }g(x)dx<\mathrm{\infty }$. By [F11], differentiation of $e^{t^2\mathrm{/}2}{\phi }_G(t)$ gives zero and its value at zero is one: ${\phi }_G(t)=e^{-t^2\mathrm{/}2}$ for every real $t$. All integrations by parts first concern smooth functions on compact intervals, whose derivatives are continuous and therefore integrable by [F10]; the same dyadic comparison identifies their Riemann and Lebesgue integrals.

Fix an integer $n\ge 1$ and define $H_{n,t}(x)=(2\sin (tx\mathrm{/}2)\mathrm{/}t{)}^{2n}$ for $t\mathrm{\ne }0$. It is nonnegative and bounded for fixed $t$. The binomial theorem and $2i\sin (v\mathrm{/}2)=e^{iv\mathrm{/}2}-e^{-iv\mathrm{/}2}$ give $H_{n,t}(x)=(-1{)}^n{t}^{-2n}{\sum }_{k=0}^{2n}(-1{)}^k\left(\genfrac{}{}{0px}{}{2n}{k}\right)e^{i(n-k)tx}\mathrm{.}$ For $0<\mathrm{\mid }t\mathrm{\mid }<r\mathrm{/}n$, its expectation under $Y$ therefore equals its expectation under $G$, since every frequency $(n-k)t$ lies in $(-r,r)$. On the prescribed sequence $t_{\mathrm{\ell }}=r\mathrm{/}(2n\mathrm{\ell })$, $H_{n,t_{\mathrm{\ell }}}(x)\to x^{2n}$; moreover $H_{n,t}(x)\le \mathrm{\mid }x{\mathrm{\mid }}^{2n}$ because $\mathrm{\mid }\sin v\mathrm{\mid }\le \mathrm{\mid }v\mathrm{\mid }$. By [F3] applied to $G$, $\mathbb{E}{H}_{n,t_{\mathrm{\ell }}}(G)\to m_{2n}$. Fatou's lemma [F2] applied to $Y$ now gives $\mathbb{E}\mathrm{\mid }Y{\mathrm{\mid }}^{2n}\le m_{2n}<\mathrm{\infty }$. Applying [F3] again, now with the integrable majorant $\mathrm{\mid }Y{\mathrm{\mid }}^{2n}$, shows equality $\mathbb{E}{Y}^{2n}=m_{2n}$.

Every absolute moment of $Y$ is finite, since $\mathrm{\mid }y{\mathrm{\mid }}^k\le 1+\mathrm{\mid }y{\mathrm{\mid }}^{2k}$. By [F4], equality of $\phi$ and ${\phi }_G$ on a neighborhood of zero implies equality of their derivatives there, hence $\mathbb{E}{Y}^k=\mathbb{E}{G}^k$ for every $k\ge 0$. For every $a\ge 0$, the nonnegative partial sums of $\cosh (aY)$ increase to it; [F3] and step 2.1 give $\mathbb{E}\cosh (aY)={\sum }_{n=0}^{\mathrm{\infty }}\frac{a^{2n}{m}_{2n}}{(2n)!}=e^{a^2\mathrm{/}2}\mathrm{.}$ Since $e^{a\mathrm{\mid }y\mathrm{\mid }}\le 2\cosh (ay)$, $\mathbb{E}{e}^{a\mathrm{\mid }Y\mathrm{\mid }}<\mathrm{\infty }$. The identical calculation applies to $G$. For every real $t$, the partial sums of the exponential series for $e^{itY}$ have modulus at most $e^{\mathrm{\mid }t\mathrm{\mid }\mathrm{\mid }Y\mathrm{\mid }}$, so [F3] and moment equality yield $\phi (t)={\sum }_{k=0}^{\mathrm{\infty }}\frac{(it{)}^k\mathbb{E}{Y}^k}{k!}={\sum }_{k=0}^{\mathrm{\infty }}\frac{(it{)}^k\mathbb{E}{G}^k}{k!}=e^{-t^2\mathrm{/}2}\mathrm{.}$ This proves global equality without presupposing analyticity of $\phi$.

For completeness identify the law directly, without importing a general Fourier uniqueness theorem. For $\epsilon >0$ set $g_{\epsilon }(y)={\epsilon }^{-1}g(y\mathrm{/}\epsilon )$. The Gaussian transform computed in step 1.1, with the substitution $s=\epsilon t$, gives $g_{\epsilon }(y)=\frac{1}{2\pi }{\int }_{\mathbb{R}}{e}^{-ity}{e}^{-{\epsilon }^2{t}^2\mathrm{/}2}dt\mathrm{.}$ If $\mu$ is the law of $Y$, then $h_{\mu ,\epsilon }(z)=\int g_{\epsilon }(z-x)\mu (dx)$ is a probability density by Tonelli. Substituting the displayed identity and applying Fubini gives $h_{\mu ,\epsilon }(z)=\frac{1}{2\pi }\int e^{-itz}{e}^{-{\epsilon }^2{t}^2\mathrm{/}2}\phi (t)dt\mathrm{.}$ Absolute integrability for this interchange follows from $\mu (\mathbb{R})=1$ and $\int e^{-{\epsilon }^2{t}^2\mathrm{/}2}dt<\mathrm{\infty }$. By step 3.1 this density equals $h_{\gamma ,\epsilon }$ for every $z$.

For a bounded continuous real $f$, Fubini and substitution give $\int f(z)h_{\mu ,\epsilon }(z)dz=\int \text{\hspace{0.17em}⁣}\int f(x+\epsilon u)g(u)du\mu (dx)\mathrm{.}$ The product-integrand is bounded in absolute integral by $\parallel f{\parallel }_{\mathrm{\infty }}$. Along $\epsilon =1\mathrm{/}\mathrm{\ell }$, dominated convergence gives limit $\int fd\mu$; the same holds for $\gamma$. Step 4.1 therefore implies $\int fd\mu =\int fd\gamma$ for every such $f$. For $a\in \mathbb{R}$ take the continuous functions $f_{a,n}(x)$ equal to $1$ on $x\le a$, equal to $1-n(x-a)$ on $a<x<a+1\mathrm{/}n$, and zero on $x\ge a+1\mathrm{/}n$. They are bounded by one and converge pointwise to ${\mathbf{1}}_{(-\mathrm{\infty },a]}$, so [F3] gives equality of both measures on every such half-line. These half-lines are a pi-system generating the Borel sets: open intervals are countable unions and differences of half-lines with rational endpoints, and every open subset of $\mathbb{R}$ is the union of the rational-endpoint intervals contained in it. Both measures have total mass one; [F7] proves $\mu =\gamma$. This proves the assertion.