A one-dimensional Brownian motion $B_t$ has exponential moments of all orders, i.e.
$$Ee^{\zeta B_t}=e^{t\zeta ^2/2}\; \text{for all} \; \zeta \in \mathbb{C}. (2.6)$$
This is given as a corollary to the following proposition.
Proposition. Let $(B_t)_{t\ge 0}$ be a one-dimensional Brownian motion. Then $B_t, t\ge 0$, are Gaussian random variables with mean $0$ and variance $t$:
$$Ee^{i\xi B_t}=e^{-t\xi ^2/2} \; \text{for all} \; t\ge 0, \xi \in \mathbb{R}. (2.5)$$
The text says that since $e^{cy} e^{-y^2 /2}\le e^{c^2}e^{-y^2/4}$ for all $c,y \in \mathbb{R}$, it follows that considering real and imaginary parts separately, the integrals in (2.5) converge for all $\xi \in \mathbb{C}$ and define an analytic function.
However, I don't see how the above shows that for all complex $\xi$ the integrals in $(2.5)$ converge, and why $(2.6)$ is a corollary of this result. I would greatly appreciate it if anyone could explain the above to me.
You have to prove the convergence of the integral $$\int_{-\infty}^{+\infty}\left|e^{i\xi \sqrt tu}\right|e^{-u^2/2}\mathrm du$$ for each complex number $\xi$. Since we can write $\xi=a+ib$ where $a$ and $b$ are real numbers, it suffices to prove the convergence of the integral $$\int_{-\infty}^{+\infty}e^{b\sqrt tu}e^{-u^2/2}\mathrm du$$ for each real number $b$. The suggested inequality applied with $c=b\sqrt t$ and $y=u$ gives $$e^{b\sqrt tu}e^{-u^2/2}\leqslant e^{b^2t}e^{-u^2/2}.$$ The fact that (2.6) is a corollary of (2.5) follows from the fact that if two analytic functions agree on the real line then they agree on the set of complex numbers. Here, the two analytic functions are $z\mapsto \mathbb E\left[\exp\left(izB_t\right)\right]$ and $z\mapsto \exp\left(tz^2/2\right)$.