Theorem 1.2.3 (Fernique, [Theorem 8.2.1, Str24]).label Let $(\Omega, \cf, \bp)$ be a probability space, $E$ be a separable Banach space over $\real$, and $X: \Omega \to E$ be a centred Gaussian random variable. Let $R > 0$ such that $\bp\bracs{\norm{X}_E \le R}\ge 9/10$, then
\[\ev^{\bp}\braks{e^{\frac{\norm{X}_{E}^{2}}{18R^{2}}}}\le e^{1/2}+ \sum_{n = 0}^{\infty} \paren{\frac{e}{3}}^{2^n}\]
Proof. By enlarging the sample space, let $X': \Omega \to E$ be independent from $X$ and with the same distribution as $X$. For any $0 \le s \le t < \infty$,
\begin{align*}&\bp\bracs{\norm{X}_E \le s}\bp\bracs{\norm{X}_E \ge t}\\&= \bp\bracs{\norm{X}_E \le s}\bp\bracsn{\normn{X'}_E \ge t}= \bp\bracs{\norm{X}_E \le s, \normn{X'}_E \ge t}\\&= \bp\bracs{\norm{X - X'}_E \le \sqrt{2}s, \normn{X + X'}_E \ge \sqrt{2}t}\\&\le \bp\bracs{\abs{\norm{X}_E - \normn{X'}_E} \le \sqrt{2}s, \norm{X}_E + \normn{X'}_E \ge \sqrt{2}t}\end{align*}
Here, if $\abs{\norm{X}_E - \normn{X'}_E}\le \sqrt{2}s$, then
\[\sqrt{2}t \le \norm{X}_{E} + \norm{X'}_{E} \le 2(\norm{X}_{E} \wedge \norm{X'}_{E}) + \sqrt{2}s\]
Thus $\norm{X}_{E} + \norm{X'}_{E} \ge \sqrt{2}(t - s)$, so
\begin{align*}\bp\bracs{\norm{X}_E \le s}\bp\bracs{\norm{X}_E \ge t}&\le \bp\bracs{\norm{X}_E \wedge \norm{X'}_E \ge (t - s)/\sqrt{2}}\\&= \bp\bracs{\norm{X}_E \ge (t - s)/\sqrt{2}}^{2}\end{align*}
Let $t_{0} = R$, and $t_{n} = R + \sqrt{2}t_{n - 1}$ for all $n \in \nat$. From here, for each $n \in \nat$,
\begin{align*}\bp\bracs{\norm{X}_E \le R}\bp\bracs{\norm{X}_E \ge t_n}&\le \bp\bracs{\norm{X}_E \ge (t_n - R)/\sqrt{2}}^{2} \\&= \bp\bracs{\norm{X}_E \ge t_{n-1}}^{2} \\ \frac{\bp\bracs{\norm{X}_E \ge t_n}}{\bp\bracs{\norm{X}_E \le R}}&\le \braks{\frac{\bp\bracs{\norm{X}_E \ge t_{n - 1}}}{\bp\bracs{\norm{X}_E \le R}}}^{2}\end{align*}
Thus
\begin{align*}\frac{\bp\bracs{\norm{X}_E \ge t_n}}{\bp\bracs{\norm{X}_E \le R}}&\le \braks{\frac{\bp\bracs{\norm{X}_E \ge R}}{\bp\bracs{\norm{X}_E \le R}}}^{2^n}\le 9^{-2^n}\\ \bp\bracs{\norm{X}_E \ge t_n}&\le 9^{-2^{n}}\le 3^{-2^{n+1}}\end{align*}
so for any $K > 0$,
\begin{align*}\ev^{\bp}\braks{e^{\norm{X}_E^2/K}}&\le e^{t_0^2/K}+ \sum_{n = 0}^{\infty} e^{t_{n+1}^2/K}\bp\bracs{\norm{X}_E \ge t_n}\\&\le e^{t_0^2/K}+ \sum_{n = 0}^{\infty} e^{t_{n+1}^2/K}\cdot 9^{-2^n}\le e^{t_0^2/K}+ \sum_{n = 0}^{\infty} e^{t_{n+1}^2/K}\cdot 3^{-2^{n+1}}\end{align*}
Now, since
\begin{align*}t_{N}&= R + \sqrt{2}t_{N - 1}= R + \sqrt{2}(R + \sqrt{2}t_{N - 2}) \\&= R + \sqrt{2}R + \sqrt{2}^{2}(R + \sqrt{2}t_{N - 3}) = R\sum_{n = 0}^{N}\sqrt{2}^{n} = \frac{R(1 - \sqrt{2}^{N+1})}{1 - \sqrt{2}}\\&\le 3R\sqrt{2}^{N+1}= 3R \cdot 2^{(N+1)/2}\end{align*}
so if $K = 18R^{2}$,
\[\frac{t^{2}_{n+1}}{K}\le \frac{9R^{2} \cdot 2^{(N+2)}}{18R^{2}}= 2^{N+1}\]
Thus
\[\ev^{\bp}\braks{e^{\norm{X}_E^2/K}}\le e^{t_0^2/K}+ \sum_{n = 0}^{\infty} e^{2^{n+1}}\cdot 3^{-2^{n+1}}\le e^{1/2}+ \sum_{n = 0}^{\infty} \paren{\frac{e}{3}}^{2^n}\]
$\square$