Definition 1.2.1.label Let $E$ be a separable Banach space over $\real$ and $\wien$ be a Borel measure on $E$, then $\wien$ is a Gaussian measure if $x \mapsto \angles{x, x^*}_{E}$ is a centred Gaussian random variable for each $x^{*} \in E^{*}$. If each $x \mapsto \angles{x, x^*}_{E}$ is non-degenerate, then $\wien$ is non-degenerate.
Lemma 1.2.2.label Let $(\Omega, \cf, \bp)$ be a probability space, $E$ be a separable Banach space over $\real$, and $X, X': \Omega \to E$ be i.i.d Gaussian random variables, then $((X + X')/\sqrt{2}, (X - X')/\sqrt{2})$ have the same distribution as $(X, X')$.
Proof. Let $x^{*}, y^{*} \in E^{*}$, $Y(\omega) = \angles{X(\omega), x^*}_{E}$ and $Y'(\omega) = \anglesn{X'(\omega), x^*}_{E}$, then $Y$ and $Y'$ are independent Gaussian random variables, so
\begin{align*}\ev^{\bp}\braks{e^{i\angles{X, x^*}_E + i\anglesn{X', y^*}_E}}&= \ev^{\bp}\braks{e^{i\angles{X, x^*}_E}}\ev^{\bp}\braks{e^{i\anglesn{X', y^*}_E}}\\&= \exp\braks{-\frac{1}{2}\paren{\ev^\bp\braks{\angles{X, x^*}_E^2} + \ev^\bp\braks{\angles{X', y^*}_E^2}}}\end{align*}
On the other hand,
\begin{align*}&\ev^{\bp}\braks{e^{i\anglesn{(X + X')/\sqrt{2}, x^*}_E + i\anglesn{(X - X')/\sqrt{2}, y^*}_E}}\\&= \ev^{\bp}\braks{e^{i\anglesn{X/\sqrt{2}, x^* + y^*}_E}}\ev^{\bp}\braks{e^{i\anglesn{X'/\sqrt{2}, x^* - y^*}_E}}\\&= \exp\braks{-\frac{1}{4}\paren{\ev^\bp\braks{\angles{X, x^* + y^*}_E^2} + \ev^\bp\braks{\angles{X', x^* - y^*}_E^2}}}\end{align*}
where
\begin{align*}\angles{X, x^* + y^*}_{E}^{2}&= \angles{X, x^*}_{E}^{2} + \angles{X, y^*}_{E}^{2} + 2\angles{X, x^*}_{E}\angles{X, y^*}_{E} \\ \angles{X', x^* - y^*}_{E}^{2}&= \angles{X', x^*}_{E}^{2} + \angles{X', y^*}_{E}^{2} - 2\angles{X', x^*}_{E}\angles{X', y^*}_{E}\end{align*}
Since $X$ and $X'$ are identically distributed,
\begin{align*}&\ev^{\bp}\braks{\angles{X, x^* + y^*}_E^2}+ \ev^{\bp}\braks{\angles{X', x^* - y^*}_E^2}\\&= \ev^{\bp}\braks{\angles{X, x^*}_E^2 + \angles{X', x^*}_E^2}+ \ev^{\bp}\braks{\angles{X, y^*}_E^2 + \angles{X', y^*}_E^2}\\&+2\ev^{\bp}\braks{\angles{X, x^*}_E\angles{X, y^*}_E}- \ev^{\bp}\braks{2\angles{X', x^*}_E\angles{X', y^*}_E}\\&= 2\ev^{\bp}\braks{\angles{X, x^*}_E^2}+ 2\ev^{\bp}\braks{\angles{X', y^*}_E^2}\end{align*}
Thus the characteristic functions of $(X, X')$ and $((X + X')/\sqrt{2}, (X - X')/\sqrt{2})$ are equal. By Definition 1.1.5, their distributions are equal.$\square$
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
\[\ev^{\bp}\braks{e^{\norm{X}_E^2/K}}\le e^{t_0^2/K}+ \sum_{n = 0}^{\infty} e^{2^{n+1}}\cdot 3^{-2^{n+1}}\le e^{1/2}+ \sum_{n = 0}^{\infty} \paren{\frac{e}{3}}^{2^n}\]
$\square$