A Gaussian tail bound implies a sub-Gaussian MGF bound

The tail bound and [F1] imply $\mathbb{E}{e}^{a\mathrm{\mid }X\mathrm{\mid }}=1+a{\int }_0^{\mathrm{\infty }}{e}^{at}\mathbb{P}(\mathrm{\mid }X\mathrm{\mid }>t)dt\le 1+2a{\int }_0^{\mathrm{\infty }}{e}^{at-t^2\mathrm{/}(2\tau )}dt<\mathrm{\infty }$ for every $a>0$, since completing the square gives $at-t^2\mathrm{/}(2\tau )=-(t-a\tau {)}^2\mathrm{/}(2\tau )+a^2\tau \mathrm{/}2$. Thus all MGFs below are finite.

Let $X_1,X_2$ be the coordinate variables on the product of two copies of the law of $X$; [F2] gives this probability space, and its rectangle formula makes them independent with law equal to that of $X$. Set $D=X_1-X_2$. Independence [F3] gives $\mathbb{E}{e}^{\lambda D}=(\mathbb{E}{e}^{\lambda X})(\mathbb{E}{e}^{-\lambda X})$. By [F4], the second factor is at least $1$, so $\mathbb{E}{e}^{\lambda X}\le \mathbb{E}{e}^{\lambda D}$. The product is unchanged when $\lambda$ is replaced by $-\lambda$, hence $\mathbb{E}{e}^{\lambda D}=\mathbb{E}\cosh (\lambda D)$.

A union bound gives $\mathbb{P}(\mathrm{\mid }D\mathrm{\mid }>t)\le 4e^{-t^2\mathrm{/}(8\tau )}$ because $\mathrm{\mid }D\mathrm{\mid }>t$ implies $\mathrm{\mid }{X}_1\mathrm{\mid }>t\mathrm{/}2$ or $\mathrm{\mid }{X}_2\mathrm{\mid }>t\mathrm{/}2$. For $m\ge 1$, [F1] and the substitution $u=t^2\mathrm{/}(8\tau )$ yield $\mathbb{E}\mathrm{\mid }D{\mathrm{\mid }}^{2m}=2m{\int }_0^{\mathrm{\infty }}{t}^{2m-1}\mathbb{P}(\mathrm{\mid }D\mathrm{\mid }>t)dt\le 4(8\tau {)}^{m}m!\mathrm{.}$ The integral is evaluated using ${\int }_0^{\mathrm{\infty }}{t}^{2m-1}{e}^{-t^2\mathrm{/}(8\tau )}dt=\frac{1}{2}(8\tau {)}^m(m-1)!$.

By [F5], the nonnegative power series for $\cosh$ may be integrated term by term. With $x=8\tau {\lambda }^2$, step 2.1 and $(2m)!\ge (m!{)}^2$ give $\mathbb{E}\cosh (\lambda D)\le 1+4{\sum }_{m\ge 1}\frac{x^m}{m!}=4e^x-3\le e^{4x}\mathrm{.}$ For the last inequality, $f(x)=e^{4x}-4e^x+3$ satisfies $f(0)=0$ and $f^{\mathrm{\prime }}(x)=4(e^{4x}-e^x)\ge 0$ for $x\ge 0$. Therefore $\mathbb{E}{e}^{\lambda X}\le e^{32\tau {\lambda }^2}$ for every $\lambda$, which is the proxy-$64\tau$ assertion by [F6].