[proofplan] The proof is a dyadic peeling argument. The shells $S_{-1},S_0,S_1,\dots$ partition $\mathcal F$ according to the size of $R(f)$, and each shell is contained in the localized class at its upper radius $r_k$. We choose a deterministic deviation level $u_k$ large enough that the assumed localized tail bound gives probability at most $e^{-a_k}$ on the $k$-th shell, then sum these bounds over all shell indices by [countable subadditivity](/theorems/1108). [/proofplan]

[step:Choose the universal shell deviation constant]Define \begin{align*} c_B:=\max\{1,B\}+1. \end{align*} Then $c_B>0$ depends only on $B$, and \begin{align*} c_B^2\ge B \end{align*} and \begin{align*} c_B^2\ge c_B. \end{align*} Fix $s>0$ and $k\ge -1$. Define the shell deviation level $u_k\in(0,\infty)$ by \begin{align*} u_k:=c_B\left(r_k\sqrt{\frac{a_k}{n}}+\frac{M a_k}{n}\right). \end{align*} Since $r_k\ge r_0>0$ and $a_k>0$, this number is strictly positive. Set \begin{align*} x_k:=r_k\sqrt{\frac{a_k}{n}} \end{align*} and \begin{align*} y_k:=\frac{M a_k}{n}. \end{align*} Then $x_k>0$, $y_k\ge0$, and $u_k=c_B(x_k+y_k)$. Also, \begin{align*} r_k^2=\frac{n x_k^2}{a_k} \end{align*} and, when written in terms of $y_k$, the identity $M=n y_k/a_k$ holds if $a_k>0$. Therefore \begin{align*} B r_k^2+M u_k =\frac{n}{a_k}\left(Bx_k^2+c_B y_k(x_k+y_k)\right). \end{align*} Using $c_B^2\ge B$ and $c_B^2\ge c_B$, we obtain \begin{align*} B x_k^2+c_B y_k(x_k+y_k) \le c_B^2 x_k^2+c_B^2 y_k(x_k+y_k) \le c_B^2(x_k+y_k)^2. \end{align*} Consequently, \begin{align*} \frac{n u_k^2}{B r_k^2+M u_k} = \frac{a_k c_B^2(x_k+y_k)^2}{B x_k^2+c_B y_k(x_k+y_k)} \ge a_k. \end{align*}[/step]

[guided]The only algebraic point in the proof is to choose the constant multiplying the deviation level so that it dominates both terms in the denominator of the assumed Bernstein-type bound. We make this explicit by defining \begin{align*} c_B:=\max\{1,B\}+1. \end{align*} This gives two inequalities that will be used below: \begin{align*} c_B^2\ge B \end{align*} and \begin{align*} c_B^2\ge c_B. \end{align*} For a fixed shell index $k\ge -1$, define \begin{align*} u_k:=c_B\left(r_k\sqrt{\frac{a_k}{n}}+\frac{M a_k}{n}\right). \end{align*} This is the deviation level we will feed into the localized tail estimate. It is positive because $r_k\ge r_0>0$, $a_k>0$, and $n\in\mathbb N$. To verify that this choice produces the exponent $a_k$, introduce the auxiliary nonnegative quantities \begin{align*} x_k:=r_k\sqrt{\frac{a_k}{n}} \end{align*} and \begin{align*} y_k:=\frac{M a_k}{n}. \end{align*} Then $u_k=c_B(x_k+y_k)$. The point of this substitution is that it expresses both pieces of the Bernstein denominator in the same scale. Since $a_k>0$, \begin{align*} r_k^2=\frac{n x_k^2}{a_k} \end{align*} and \begin{align*} M=\frac{n y_k}{a_k}. \end{align*} Hence \begin{align*} B r_k^2+M u_k =\frac{n}{a_k}\left(Bx_k^2+c_B y_k(x_k+y_k)\right). \end{align*} Now use the two inequalities built into the definition of $c_B$. Since $x_k\ge0$ and $y_k\ge0$, \begin{align*} B x_k^2\le c_B^2 x_k^2 \end{align*} and \begin{align*} c_B y_k(x_k+y_k)\le c_B^2 y_k(x_k+y_k). \end{align*} Adding these inequalities gives \begin{align*} B x_k^2+c_B y_k(x_k+y_k) \le c_B^2 x_k^2+c_B^2 y_k(x_k+y_k). \end{align*} Finally, \begin{align*} x_k^2+y_k(x_k+y_k)\le (x_k+y_k)^2, \end{align*} so \begin{align*} B x_k^2+c_B y_k(x_k+y_k) \le c_B^2(x_k+y_k)^2. \end{align*} Substituting this estimate into the exponent yields \begin{align*} \frac{n u_k^2}{B r_k^2+M u_k} = \frac{a_k c_B^2(x_k+y_k)^2}{B x_k^2+c_B y_k(x_k+y_k)} \ge a_k. \end{align*} This is the uniform inequality needed for every shell index, and it depends only on $B$ through the single constant $c_B$.[/guided]

[step:Embed each shell into its localized class] For $k=-1$, the definitions give \begin{align*} S_{-1}=\{f\in\mathcal F:R(f)\le r_0\}=\mathcal F(r_{-1}). \end{align*} For $k\ge0$, if $f\in S_k$, then \begin{align*} R(f)\le 2^{k+1}r_0=r_k. \end{align*} Thus \begin{align*} S_k\subseteq\mathcal F(r_k) \end{align*} for every $k\ge -1$. [/step]

[step:Apply the localized tail bound to each shell] For each $k\ge -1$, define the event \begin{align*} E_k:=\left\{\sup_{f\in S_k}|(P_n-P)f|>A(r_k)+u_k\right\}. \end{align*} Since $S_k\subseteq\mathcal F(r_k)$, monotonicity of the supremum over nested index sets gives \begin{align*} \sup_{f\in S_k}|(P_n-P)f| \le \sup_{f\in\mathcal F(r_k)}|(P_n-P)f|. \end{align*} Therefore \begin{align*} E_k\subseteq \left\{\sup_{f\in\mathcal F(r_k)}|(P_n-P)f|>A(r_k)+u_k\right\}. \end{align*} Because $r_k\ge r_0$ and $u_k>0$, the assumed localized deviation inequality applies with $r=r_k$ and $u=u_k$. Hence \begin{align*} \mathbb P(E_k) \le \exp\left(-\frac{n u_k^2}{B r_k^2+M u_k}\right). \end{align*} By the exponent estimate from the previous step, \begin{align*} \mathbb P(E_k)\le e^{-a_k}. \end{align*} [/step]

[step:Sum the shell probabilities by countable subadditivity] Define the exceptional event \begin{align*} E:=\bigcup_{k=-1}^{\infty}E_k. \end{align*} By countable subadditivity of probability measures, \begin{align*} \mathbb P(E)\le \sum_{k=-1}^{\infty}\mathbb P(E_k). \end{align*} Using the shellwise estimates, \begin{align*} \mathbb P(E)\le \sum_{k=-1}^{\infty}e^{-a_k}. \end{align*} By the definitions of $a_{-1}$ and $a_k$ for $k\ge0$, \begin{align*} \sum_{k=-1}^{\infty}e^{-a_k} = e^{-(s+1)}+\sum_{k=0}^{\infty}e^{-(s+k+2)}. \end{align*} The second term is a geometric series: \begin{align*} \sum_{k=0}^{\infty}e^{-(s+k+2)} = e^{-s}\sum_{k=0}^{\infty}e^{-(k+2)} = e^{-s}\frac{e^{-2}}{1-e^{-1}}. \end{align*} Thus \begin{align*} \sum_{k=-1}^{\infty}e^{-a_k} = e^{-s}\left(e^{-1}+\frac{e^{-2}}{1-e^{-1}}\right). \end{align*} Since \begin{align*} e^{-1}+\frac{e^{-2}}{1-e^{-1}} = \frac{1}{e-1}, \end{align*} we obtain \begin{align*} \mathbb P(E)\le \frac{e^{-s}}{e-1}. \end{align*} This is precisely \begin{align*} \mathbb P\left( \exists k\ge -1:\sup_{f\in S_k}|(P_n-P)f| >A(r_k)+c_B\left(r_k\sqrt{\frac{a_k}{n}}+\frac{M a_k}{n}\right) \right) \le \frac{e^{-s}}{e-1}. \end{align*} [/step]

[step:Pass from the exceptional event to the simultaneous functionwise bound] It remains to justify the equivalent formulation. Since $P f^2<\infty$ for every $f\in\mathcal F$, the number $R(f)$ is finite for every $f\in\mathcal F$. Because $r_0>0$, every finite nonnegative number belongs to exactly one of the intervals \begin{align*} [0,r_0] \end{align*} or \begin{align*} (2^k r_0,2^{k+1}r_0] \end{align*} with $k\ge0$. Therefore every $f\in\mathcal F$ belongs to exactly one shell $S_k$ with $k\ge -1$. On the complement $E^c$, no event $E_k$ occurs. Hence, for every $k\ge -1$, \begin{align*} \sup_{f\in S_k}|(P_n-P)f| \le A(r_k)+u_k. \end{align*} Substituting the definition of $u_k$ gives \begin{align*} \sup_{f\in S_k}|(P_n-P)f| \le A(r_k)+c_B\left(r_k\sqrt{\frac{a_k}{n}}+\frac{M a_k}{n}\right). \end{align*} If $f\in\mathcal F$ and $k$ is its unique shell index, then $f\in S_k$, and so \begin{align*} |(P_n-P)f| \le A(r_k)+c_B\left(r_k\sqrt{\frac{a_k}{n}}+\frac{M a_k}{n}\right). \end{align*} Since $\mathbb P(E^c)\ge 1-e^{-s}/(e-1)$, the simultaneous shellwise bound holds with the asserted probability. [/step]