[proofplan]
Fix an outcome $\omega \in E$ and prove a deterministic inequality for the realized numbers $\hat f_h(x)(\omega)$. The oracle bandwidth $h^*$ is first shown to be acceptable for the Lepski rule by comparing it with all finer bandwidths $h' \le h^*$ and using the assumed slack in the thresholds. Since the Lepski rule chooses the largest acceptable bandwidth, this forces $\hat h(x)(\omega) \ge h^*$, so acceptability of $\hat h(x)(\omega)$ gives a direct comparison between the selected estimator and the oracle estimator. The threshold upper bound, monotonicity of $s_h(x)$, and the oracle property of $h^*$ then convert this comparison into the desired oracle inequality.
[/proofplan]
[step:Freeze an outcome in the high-probability event]
Fix $\omega \in E$. For each $h \in \mathcal H_n$, define the realized estimator value
\begin{align*}
Y_h:\mathcal H_n \to \mathbb R,\qquad
h \mapsto \hat f_h(x)(\omega).
\end{align*}
By the defining property of $E$, for every $h \in \mathcal H_n$,
\begin{align*}
|Y_h-\mathbb E[\hat f_h(x)]| \le s_h(x).
\end{align*}
All estimates below are deterministic estimates for this fixed $\omega$.
[/step]
[step:Show that the oracle bandwidth is acceptable]
Let $h' \le h^*$. By adding and subtracting the two expectations and applying the triangle inequality in $\mathbb R$,
\begin{align*}
|Y_{h^*}-Y_{h'}| \le |Y_{h^*}-\mathbb E[\hat f_{h^*}(x)]|+|\mathbb E[\hat f_{h^*}(x)]-\mathbb E[\hat f_{h'}(x)]|+|\mathbb E[\hat f_{h'}(x)]-Y_{h'}|.
\end{align*}
The concentration bounds on $E$ and the oracle-scale pairwise bias bound give
\begin{align*}
|Y_{h^*}-Y_{h'}| \le s_{h^*}(x)+B b_{h^*}(x)+s_{h'}(x).
\end{align*}
The final slack hypothesis gives
\begin{align*}
s_{h^*}(x)+B b_{h^*}(x)+s_{h'}(x)
\le
\lambda(h^*,h').
\end{align*}
Therefore
\begin{align*}
|Y_{h^*}-Y_{h'}| \le \lambda(h^*,h')
\end{align*}
for every $h' \le h^*$, so $h^*$ belongs to the acceptable set in the definition of $L_x((Y_h)_{h \in \mathcal H_n},\lambda)$.
[guided]
We must verify that the Lepski rule can choose at least the oracle bandwidth. The rule accepts a bandwidth $h$ exactly when its realized estimator value is close, within threshold $\lambda(h,h')$, to every realized estimator value at a bandwidth $h' \le h$. Thus, for $h^*$, we fix an arbitrary $h' \le h^*$ and estimate the realized difference $|Y_{h^*}-Y_{h'}|$.
The useful decomposition is obtained by inserting the two deterministic means:
\begin{align*}
Y_{h^*}-Y_{h'}
=
\bigl(Y_{h^*}-\mathbb E[\hat f_{h^*}(x)]\bigr)
+
\bigl(\mathbb E[\hat f_{h^*}(x)]-\mathbb E[\hat f_{h'}(x)]\bigr)
+
\bigl(\mathbb E[\hat f_{h'}(x)]-Y_{h'}\bigr).
\end{align*}
Applying the triangle inequality in $\mathbb R$ gives
\begin{align*}
|Y_{h^*}-Y_{h'}| \le |Y_{h^*}-\mathbb E[\hat f_{h^*}(x)]|+|\mathbb E[\hat f_{h^*}(x)]-\mathbb E[\hat f_{h'}(x)]|+|\mathbb E[\hat f_{h'}(x)]-Y_{h'}|.
\end{align*}
Because $\omega \in E$, both stochastic deviations are controlled:
\begin{align*}
|Y_{h^*}-\mathbb E[\hat f_{h^*}(x)]| \le s_{h^*}(x),
\qquad
|\mathbb E[\hat f_{h'}(x)]-Y_{h'}| \le s_{h'}(x).
\end{align*}
The oracle-scale pairwise bias hypothesis controls the deterministic middle term:
\begin{align*}
|\mathbb E[\hat f_{h^*}(x)]-\mathbb E[\hat f_{h'}(x)]|
\le
B b_{h^*}(x).
\end{align*}
Combining these three bounds yields
\begin{align*}
|Y_{h^*}-Y_{h'}|
\le
s_{h^*}(x)+B b_{h^*}(x)+s_{h'}(x).
\end{align*}
The slack assumption says precisely that this upper bound is no larger than the Lepski threshold:
\begin{align*}
B b_{h^*}(x)+s_{h^*}(x)+s_{h'}(x)
\le
\lambda(h^*,h').
\end{align*}
Hence $|Y_{h^*}-Y_{h'}| \le \lambda(h^*,h')$ for the arbitrary $h' \le h^*$. Since every required comparison has been verified, $h^*$ is acceptable.
[/guided]
[/step]
[step:Use maximality of the Lepski choice to compare it with the oracle]
Since $h^*$ is acceptable, the acceptable set is nonempty. Because $\mathcal H_n$ is finite and totally ordered, the maximum defining
\begin{align*}
\hat h(x)(\omega)=L_x((Y_h)_{h \in \mathcal H_n},\lambda)
\end{align*}
exists. By maximality of this choice and acceptability of $h^*$,
\begin{align*}
h^* \le \hat h(x)(\omega).
\end{align*}
Since $\hat h(x)(\omega)$ is acceptable and $h^* \le \hat h(x)(\omega)$, the defining comparison for the acceptable bandwidth $\hat h(x)(\omega)$ with $h'=h^*$ gives
\begin{align*}
|Y_{\hat h(x)(\omega)}-Y_{h^*}|
\le
\lambda(\hat h(x)(\omega),h^*).
\end{align*}
[/step]
[step:Bound the selected-oracle comparison by the oracle stochastic scale]
The upper threshold assumption applied to the pair $h^* \le \hat h(x)(\omega)$ gives
\begin{align*}
\lambda(\hat h(x)(\omega),h^*)
\le
A\bigl(s_{\hat h(x)(\omega)}(x)+s_{h^*}(x)\bigr).
\end{align*}
Since $s_h(x)$ is nonincreasing along the order and $h^* \le \hat h(x)(\omega)$,
\begin{align*}
s_{\hat h(x)(\omega)}(x) \le s_{h^*}(x).
\end{align*}
Therefore
\begin{align*}
|Y_{\hat h(x)(\omega)}-Y_{h^*}|
\le
2A\,s_{h^*}(x).
\end{align*}
[/step]
[step:Convert the comparison estimate into the oracle inequality]
Using the triangle inequality in $\mathbb R$, the comparison estimate, the stochastic deviation bound on $E$, and the bias bound at $h^*$,
\begin{align*}
|\hat f_{\hat h(x)(\omega)}(x)(\omega)-f(x)| = |Y_{\hat h(x)(\omega)}-f(x)|.
\end{align*}
The triangle inequality gives
\begin{align*}
|Y_{\hat h(x)(\omega)}-f(x)| \le |Y_{\hat h(x)(\omega)}-Y_{h^*}|+|Y_{h^*}-\mathbb E[\hat f_{h^*}(x)]|+|\mathbb E[\hat f_{h^*}(x)]-f(x)|.
\end{align*}
Substituting the three available bounds yields
\begin{align*}
|\hat f_{\hat h(x)(\omega)}(x)(\omega)-f(x)| \le 2A\,s_{h^*}(x)+s_{h^*}(x)+b_{h^*}(x).
\end{align*}
Equivalently,
\begin{align*}
|\hat f_{\hat h(x)(\omega)}(x)(\omega)-f(x)| \le b_{h^*}(x)+(2A+1)s_{h^*}(x).
\end{align*}
Since $b_{h^*}(x) \ge 0$ and $s_{h^*}(x) \ge 0$,
\begin{align*}
|\hat f_{\hat h(x)(\omega)}(x)(\omega)-f(x)| \le (2A+1)\bigl(b_{h^*}(x)+s_{h^*}(x)\bigr).
\end{align*}
By the oracle assumption on $h^*$,
\begin{align*}
b_{h^*}(x)+s_{h^*}(x)
\le
2\inf_{h\in\mathcal H_n}\{b_h(x)+s_h(x)\}.
\end{align*}
Combining the last two displays yields
\begin{align*}
|\hat f_{\hat h(x)(\omega)}(x)(\omega)-f(x)|
\le
(4A+2)\inf_{h\in\mathcal H_n}\{b_h(x)+s_h(x)\}.
\end{align*}
Thus the asserted inequality holds on $E$ with $C=4A+2$, which depends only on $A$ and hence only on the allowed constants $A$ and $B$.
[/step]
