Borell concentration for the norm of a separable Banach Gaussian

By [F1], ${\sigma }^2\le E\mathrm{\parallel }W{\mathrm{\parallel }}^2<\mathrm{\infty }$. Choose a countable norming family $({\mathrm{\ell }}_j{)}_{j\ge 1}$ from [F2], and let $F_n={\max }_{1\le j\le n}\mathrm{\mid }{\mathrm{\ell }}_j(W)\mathrm{\mid }$. Then $F_n\uparrow \mathrm{\parallel }W\mathrm{\parallel }$ and $E{F}_n\to E\mathrm{\parallel }W\mathrm{\parallel }$ by [F5].

If $\sigma >0$ and $r=0$, the probability is at most $1=e^0$, which is the required bound.

Suppose $\sigma >0$ and $r>0$. The vector $({\mathrm{\ell }}_1(W),\dots ,{\mathrm{\ell }}_n(W))$ is centered Gaussian. By [F3] it has the law of $AZ$ for a matrix $A$ and a standard Gaussian vector $Z$. The function $z\mapsto {\max }_{j\le n}\mathrm{\mid }(Az{)}_j\mathrm{\mid }$ is ${\max }_j\mathrm{\parallel }{\mathrm{row}}_j(A){\mathrm{\parallel }}_2$-Lipschitz, and $\mathrm{\parallel }{\mathrm{row}}_j(A){\mathrm{\parallel }}_2^2=E[{\mathrm{\ell }}_j(W{)}^2]\le {\sigma }^2$. Applying [F4] gives $P(F_n>E{F}_n+r)\le e^{-r^2\mathrm{/}(2{\sigma }^2)}$. Since $E{F}_n\le E\mathrm{\parallel }W\mathrm{\parallel }$, the event $\{F_n>E\mathrm{\parallel }W\mathrm{\parallel }+r\}$ is contained in that tail event.

If $\sigma >0$ and $r>0$, the events $\{F_n>E\mathrm{\parallel }W\mathrm{\parallel }+r\}$ increase to $\{\mathrm{\parallel }W\mathrm{\parallel }>E\mathrm{\parallel }W\mathrm{\parallel }+r\}$. Continuity from below [F6] and step 2.1 give the claimed bound.

If $\sigma =0$, each ${\mathrm{\ell }}_j(W)$ has zero variance, so countably many probability-one equalities give ${\mathrm{\ell }}_j(W)=0$ for every $j$ almost surely; [F2] then gives $W=0$ almost surely. Hence the displayed tail probability is zero for every $r\ge 0$.