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Recently I found a concentration inequality for infinite dimensional Gaussian r.v.s in this paper. 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 some $s>0$. Then the desired inequality holds if $\mathsf{E}e^{s\|X\|}$ is bounded by $e^{Cs^2\mathsf{E}\|X\|^2}$.

Recently I found a concentration inequality for infinite dimensional Gaussian r.v.s in this paper. 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 $e^{Cs^2\mathsf{E}\|X\|^2}$.

Recently I found a concentration inequality for infinite dimensional Gaussian r.v.s in this paper. 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.

Recently I found a concentration inequality for infinite dimensional Gaussian r.v.s in this paper. 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 some $s>0$. Then the desired inequality holds if $\mathsf{E}e^{s\|X\|}$ is bounded by $e^{Cs^2\mathsf{E}\|X\|^2}$.

Recently I found a concentration inequality for infinite dimensional Gaussian r.v.s in this paper. 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 does one prove this result.

Recently I found a concentration inequality for infinite dimensional Gaussian r.v.s in this paper. 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.