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.




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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 $e^{Cs^2\mathsf{E}\|X\|^2}$.



Gaussian concentration inequality
This inequality is false. E.g., consider the random vector $X\_n:=(Z\_1,\dots,Z\_n)/\sqrt n$ in $\mathbb R^n$ with the Euclidean norm $\|\cdot\|$, where $Z\_1,Z\_2,\dots$ are independent standard normal random variables. Then $E\|X\_n\|^2=1$ and, by the law of large numbers,
$$\|X\_n\|^2=\frac1n\,\sum\_1^n Z\_i^2\to1$$
in probability (as $n\to\infty$), so that $P(\|X\_n\|\ge u)\to1$ for any $u\in(0,1)$. So, for any real constant $M>0$, any $u\in(0,1)$, and all large enough $n$
$$P(\|X\_n\|\ge u)\not\le \exp\Big(-\frac{u^2}{M}\Big)=\exp\Big(-\frac{u^2}{M\,E\|X\_n\|^2}\Big).$$




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On the other hand, according to [formula (3.5) in the Ledoux--Talagrand book](https://michel.talagrand.net/book.ps.gz), one has e.g. the inequality 
$$P(\|X\|\ge u)\le 4\exp\Big(-\frac{u^2}{8E\|X\|^2}\Big)$$
for $u\ge0$. 


The constants here can be a bit improved by using [Corollary 3](https://epubs.siam.org/doi/abs/10.1137/1130013?mobileUi=0), which states that 
$$P(\|X\|-E\|X\|\ge x)\le \exp\Big(-\frac{x^2}{2E\|X\|^2}\Big)\tag{1}$$
for $x\ge0$, which implies e.g. that 
$$P(\|X\|\ge u)\le\sqrt e\,\exp\Big(-\frac{u^2}{8E\|X\|^2}\Big)\tag{2}$$
for $u\ge0$. 


**Details on how to get (2) from (1):** If $u^2<4E\|X\|^2$, then the upper bound on $P(\|X\|\ge u)$ in (2) is $>1$ and thus trivial. So, without loss of generality, $u^2\ge4E\|X\|^2\ge4(E\|X\|)^2$, so that $E\|X\|\le u/2$. Letting now $x:=u-E\|X\|\ge u/2$, we see that the left-hand of (1) becomes that of (2), and the right-hand of (1) is less than that of (2). 


