Recently I found a concentration inequality for infinite dimensional Gaussian r.v.s in [this paper](https://www.sciencedirect.com/science/article/pii/0047259X86900850). Specifically, Lemma 4 on page 307 states (without a proof) that

> There exists a universal constant $M$ such that for each Banach space valued Gaussian random variable $X$ (having zero mean):
$$
\mathsf{P}(\|X\|\ge u)\le \exp\left(-\frac{u^2}{M\mathsf{E}\|X\|^2}\right).
$$

The authors refer to an older paper which is not available online. So I'm wondering how one proves this result.

---

As a first step, applying the generic Chernoff bound, one gets
$$
\mathsf{P}(\|X\|\ge u)\le e^{-su}\mathsf{E}e^{s\|X\|}
$$
for any $s>0$. Then the desired inequality holds if $\mathsf{E}e^{s\|X\|}$ is bounded by $e^{Cs^2\mathsf{E}\|X\|^2}$.